Answer:
30.10%
Explanation:
The mass of water is:
50.00 g − 34.95 g = 15.05 g
So the percentage of water is:
15.05 g / 50.00 g × 100% = 30.10%
Final answer:
The experimental percentage of water in copper(II) sulfate pentahydrate, after heating a sample and measuring the mass loss, is calculated to be 30.10%.
Explanation:
Experimental Percentage of Water in a Hydrate
The question involves calculating the percentage of water in a hydrate, specifically copper(II) sulfate pentahydrate, which is a classic chemistry experiment. To determine the percentage of water, you have to compare the mass of water lost upon heating to the original mass of the hydrate.
First, calculate the loss of mass due to heating: 50.00 grams (initial mass) - 34.95 grams (final mass) = 15.05 grams of water lost.
Percentage of water in the hydrate is found by the formula:
Mass% of H₂O = (Mass of water lost / Original mass of hydrate) * 100%.
Substituting the known values gives us: Mass% of H₂O = (15.05 g / 50.00 g) * 100% = 30.10%.
The experimental percentage of water in the copper(II) sulfate pentahydrate is therefore 30.10%, which is determined by this experimental process of heating and weighing.
A student obtained 1.69 g of pure caffeine following recrystallization from a crude sample, which originally weighed 2.51 g. What is the percent recovery of pure caffeine from crude?
Answer:
The recovery of pure caffeine is 67.3% of the original, crude sample.
Explanation:
Step 1: Data given
Mass of pure caffeine = 1.69 grams
Original mass of sample = 2.51 grams
Step 2: What is the percent recovery of pure caffeine from crude?
Percentage = (mass of recovery / original mass) *100%
Percentage = (1.69 grams/2.51 grams) *100%
Percentage = 67.3 %
The recovery of pure caffeine is 67.3% of the original, crude sample.
Haley's parents bought her a used bicycle for her birthday. She was thrilled until she learned that her best friend received a brand-new bicycle for her birthday. Haley's declining satisfaction illustratesA)relative deprivation.B)the adaptation-level phenomenon.C)the catharsis hypothesis.D)the behavior feedback effect.Page 7
Answer:
A) relative deprivation
Explanation:
The given example is of the relative deprivation principle of sociology.
The Theory of Relative Deprivation. Throughout sociology, the principle of relative deprivation is a concept of societal change and campaigns, whereby people act for social changes in order to gain something (for example, privileges, prestige, or wealth) that others have and that they also feel they should have the same. Here Haley is feeling deprived of brand new bicycle as has a used one.
Convert grams of FeCl2 to moles. Then rearrange M = n/V to solve for V: V = n/M (in Liters) What volume of a 0.01 M solution can be made using 120 grams of FeCl2?
Answer:
The answer to your question is Volume = 214.3 ml
Explanation:
Data
mass = 120 g of FeCl₂
concentration = 0.01 M
volume = ?
Formula
Molarity = [tex]\frac{number of moles}{volume}[/tex]
Solve for volume
Volume = [tex]\frac{number of moles}{molarity}[/tex]
Process
1.- Convert grams to moles
Atomic weight = 56 g
56 g of Fe --------------- 1 mol
120 g of Fe ---------------- x
x = (120 x 1) / 56
x = 2.14 moles
2.- Calculate the volume
Volume = [tex]\frac{2.14}{0.01}[/tex]
Volume = 214.3 ml
Answer:
We need 94.67 liters if a 0.01 M solution
Explanation:
Step 1: Data given
Molarity of the solution = 0.01 M
Mass of FeCl2 = 120.0 grams
Molar mass FeCl2 = 126.75 g/mol
Step 2: Calculate moles of FeCl2
moles FeCl2 = massFeCl2 / molar mass FeCl2
Moles FeCl2 = 120.0 grams / 126.75 g/mol
Moles FeCl2 = 0.9467 moles
Step 3: Calculate volumes FeCl2
Molarity = moles / volume
Volume = moles / molarity
Volume = 0.9467 moles /0.01 M
Volume = 94.67 L
We need 94.67 liters if a 0.01 M solution
Antoine Lavoisier, the French scientist credited with first stating the law of conservation of matter, heated a mixture of tin and air in a sealed flask to produce tin oxide. Did the mass of the sealed flask and contents decrease, increase, or remain the same after the heating?
Answer:
Remain the same
Explanation:
The law of mass conservation states that matter cannot be created nor destroyed but can be converted from one form to another.
Essentially, what Lavoisier was trying to proof is that by heating the mixture, after all the change the mass still remains. That was why he used a sealed flask. If the flask was not sealed, it probably would have been that some of the mass will escape as vapor to the atmosphere which might be difficult to account for
A student conducted an experiment to determine DHrxn for the reaction between HCl(aq) and NaOH(aq). The student ran two trials using the volumes of HCl(aq) and NaOH(aq) indicated in the table above, and determined the amount of heat released. Which of the following best explains the relationship between X and Y?
Answer:
Y=X
Explanation:
The number of moles of acid and base reacting with each other is the same in both trials.
In the neutralization reaction between HCl and NaOH, a one-to-one mole ratio leads to a predictable enthalpy change (ΔHrxn). The temperature increase signifies an exothermic reaction, and ΔHrxn can be calculated using the formula q=mcΔT, considering the reaction's exothermic nature.
Explanation:The relationship between X and Y refers to the amount of heat (ΔH) released during the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). In this reaction, a one-to-one mole ratio is observed, meaning 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of sodium chloride (NaCl) and water (H₂O)
When calculating the enthalpy change, the student must use the temperature change, the mass of the solution, and the specific heat capacity (which is assumed to be the same as water's for this type of experiment). The amount of heat produced (ΔHrxn) can be calculated using the formula q = mcΔT, where 'm' is the mass of the solution, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.
In a typical calorimetry experiment, the student would need to carefully measure the temperature changes and use stoichiometry to relate the amounts of the reactants to the heat released. If the temperature increases, as in the example provided, it indicates that the reaction is exothermic, and the value for ΔHrxn would be negative, signifying that heat is released into the surroundings.
HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)ΔH°=−57.1kJ/molrxn
The chemical equation above represents the reaction between HCl(aq) and NaOH(aq). When equal volumes of 1.00MHCl(aq) and 1.00MNaOH(aq) are mixed, 57.1kJ of heat is released. If the experiment is repeated with 2.00MHCl(aq), how much heat would be released?
Answer:
57.1 kilo Joules of heat would be released.
Explanation:
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex] ΔH°=-57.1kJ/mol
Molarity of HCl = 2.00 M
Molarity of NaOH = 1.00 M
According to reaction , 1 M of HCl reacts with 1 M of NaOH. Then 2.00 M of HCl will react with:
[tex]\frac{1}{1}\times 2.00 M= 2M [/tex] of NaOH
But according to question we only have 1.00 M NaOH .So, this means that NaOH is limiting reagent and HCl is an excessive reagent.
Heat evolved will depend upon concentration of NaOH solution :
Heat evolved when 1.00 M of NaOH reacts =
[tex]1.00\times (-57.1 kJ/mol)=-57.1 kJ[/tex]
Negative sign means that heat is released during the reaction.
57.1 kilo Joules of heat would be released.
The heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.
The given reaction,
[tex]\bold {HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l) \ \ \ \ \ \ \ \ \ \Delta H = - 57.1\ kJ/mol}[/tex]
Concentration of HCl is 2 mole.
So, 2 mole of HCl react with 2 mole of NaOH but only mole of NaOH is available.
So, NaOH is a Limiting factor in the reaction and HCl is excessive factor.
Only, 1mole of HCl will react with available 1 mole of NaOH,
Therefore, the heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.
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Identify the true statement. Choose one: A. The expansion of ice sheets at the end of the Pleistocene caused glacial rebound in northern Canada. B. Regions covered by glacial ice are termed periglacial environments. C. Glaciers move by basal sliding as friction between the ice and its substrate increases. D. The Antarctic ice sheet has been calving off huge icebergs over the last few decades.
Answer:
Option (D)
Explanation:
Antarctica is located in the south pole and covers a large area of ice-covered region. It is an extremely cold environment, having a low freezing temperature. The glaciers and ice-bergs of this region has been constantly affected due to the increasing global surface temperature. This rise in the surface temperature of the earth is due to the increasing concentration of CO₂ in the atmosphere. One of the reasons for this temperature increase is also due to the introduction of the industrial revolution, which is responsible for the emission of a large number of harmful gases, including carbon into the atmosphere. Due to this global climate change, these glaciers are melting at a faster rate, resulting in the rise of the sea level.
Thus, the Antarctic ice sheets have been calving off large pieces of icebergs over the last few decades.
Hence, the correct answer is option (D), which is the true statement.
A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ), molality ( m ), and mass percent concentration of the solution
Answer:
[KOH] = 1.47 M
[KOH] = 1.22 m
KOH = 6.86 % m/m
Explanation:
Let's analyse the data
1.87 L is the volume of solution
Density is 1.29 g/mL → Solution density
155 g of KOH → Mass of solute
Moles of solute is (mass / molar mass) = 2.76 moles.
Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M
Let's determine, the mass of solvent.
Molality is mol of solute / 1kg of solvent
We can use density to find out the mass of solution
Mass of solution - Mass of solute = Mass of solvent
Density = Mass / volume
1.29 g/mL = Mass / 1870 mL
Notice, we had to convert L to mL, cause the units of density.
1.29 g/mL . 1870 mL = Mass → 2412.3 g
2412.3 g - 155 g = 2257.3 g of solvent
Let's convert the mass of solvent to kg
2257.3 g / 1000 = 2.25kg
2.76 mol / 2.25kg = 1.22 m (molality)
% percent by mass = mass of solute in 100g of solution.
(155 g / 2257.3 g) . 100g = 6.86 % m/m
The molarity, molality, and mass percent concentration of the KOH solution are approximately 1.476 M, 1.224 m, and 6.43%, respectively.
Explanation:The first step in solving this problem is to determine the number of moles of KOH in the solution. The molar mass of KOH is approximately 56.11 g/mol, so you can calculate the number of moles by dividing the mass of the KOH by its molar mass: moles of KOH = 155 g / 56.11 g/mol = 2.762 mol.
Molarity (M) is defined as the number of moles of solute per liter of solution. To find the molarity, divide the number of moles by the volume of the solution in liters: M = 2.762 mol / 1.87 L = 1.476 M.
To calculate molality (m), you need the mass of the solvent in kilograms. First, find the total mass of the solution: mass = volume x density = 1.87 L x 1.29 g/mL = 2411.3 g. Then, subtract the mass of the KOH to find the mass of the solvent: mass of solvent = 2411.3 g - 155 g = 2256.3 g = 2.2563 kg. Now, you can find the molality: m = 2.762 mol / 2.2563 kg = 1.224 m.
Finally, the mass percent concentration of the solution is the mass of the solute divided by the mass of the solution, multiplied by 100: mass percent = (155 g / 2411.3 g) x 100 = 6.43%.
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You wish to prepare 0.13 M HNO3 from a stock solution of nitric acid that is 16.6 M. How many milliliters of the stock solution do you require to make up 1.00 L of 0.13 M HNO3?
Answer:
0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]
Explanation:
According to laws of equivalence, [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are initial and final concentration respectively. [tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume respectively.
Here, [tex]C_{1}=16.6M[/tex], [tex]C_{2}=0.13M[/tex] and [tex]V_{2}=1.00L[/tex]
So, [tex]V_{1}=\frac{C_{2}V_{2}}{C_{1}}[/tex]
or, [tex]V_{1}=\frac{(0.13M)\times (1.00L)}{(16.6M)}[/tex]
or, [tex]V_{1}=0.0078L[/tex]
Hence 0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]
A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. Calculate [H ], [OH -], and the pH for this solution.
Answer:
[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]
[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]
[tex]pH = 0.119[/tex]
Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
[tex]n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)[/tex]
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows
[tex]n_{H^{+} } from HCl = (5.00)(0.093)[/tex]
[tex]n_{H^{+} } from HCl = 0.465 mol[/tex]
[tex]n_{H^{+} } from HNO_{3} = (8.00)(0.037)[/tex]
[tex]n_{H^{+} } from HNO_{3} = 0.296 mol[/tex]
[tex]n_{H^{+}(total) } = 0.296 + 0.465[/tex]
[tex]n_{H^{+}(total) } = 0.761 mol[/tex]
For molar concentration of hydrogen ions:
[tex][H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}[/tex]
[tex][H^{+}] = \frac{0.761}{1.00}[/tex]
[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
[tex]K_{w} = [H^{+} ][OH^{-} ][/tex]
[tex][OH^{-}]=\frac{Kw}{[H^{+}] }[/tex]
[tex][OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }[/tex]
[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]
The pH of the solution can be measured by the following formula:
[tex]pH = -log[H^{+} ][/tex]
[tex]pH = -log(0.761)[/tex]
[tex]pH = 0.119[/tex]
a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how much water should be mixed?
Answer: The volume of acid and water that must be mixed will be 4.8 L and 11.2 L
Explanation:
We are given:
Volume of mixture = 16 L
Percent of acid present = 30 %
Calculating the percentage of acid present in the mixture:
[tex]\Rightarrow 16\times \frac{30}{100}=4.8L[/tex]
The mixture is made entirely of acid and water.
Volume of acid in the mixture = 4.8 L
Volume of water in the mixture = 16 - 4.8 = 11.2 L
Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L
Final answer:
The chemist needs 4.8 liters of pure acid and 11.2 liters of water to prepare 16 liters of a solution that is 30% acid.
Explanation:
To prepare 16 liters of a solution that is 30% acid, we start by setting up an equation to represent the mix of pure acid and water. Let the amount of pure acid needed be x liters, and therefore the amount of water would be 16 - x liters. Since the solution is 30% acid, we can write the equation as follows:
0.30 × 16 = x
When we solve for x, we get:
4.8 = x
This means we need 4.8 liters of pure acid and 16 - 4.8 = 11.2 liters of water to make the 30% acid solution.
Where would you expect to find many hydrophobic amino acids in the 3-D structure of a folded protein that functions in an aqueous solution?
Answer:
Hydrophobic amino cluster on the inside of the 3-D structure of the folded protein
Explanation:
Hydrophobic interactions, describe the process in which nonpolar, hydrophobic amino acids cluster or aggregate together in the internal part of the protein, to allow the hydrophilic amino acids interact with surrounding water molecules on the outside.
If you are involved in a vehicle accident, you should not drink any alcohol up to ___ hours after the accident or you could be legally charged with operating a vehicle while under the influence of alcohol.
Answer: 6
Explanation:
Alcohol is a psychoactive drug with a high number of side effects that can seriously affect our body. The amount and circumstances of consumption play an important role in determining the duration of intoxication. For example, consuming alcohol after a large meal is less likely to produce visible signs of intoxication than on an empty stomach.
Alcohol has a two-phase effect on the body, meaning that its effects change over time. Initially, it produces feelings of relaxation and joy, but later consumption can lead to blurred vision and coordination problems. Cell membranes are highly permeable to alcohol, so once alcohol is in the bloodstream, it can spread to almost all body tissues.
Unconsciousness can follow excessive consumption, and extreme levels of consumption can lead to alcohol poisoning and death. Death can also be caused by asphyxiation, if vomit - a frequent result of excessive drinking - blocks the windpipe and the individual is too drunk to respond. An appropriate first-aid response to an unconscious, drunk person is to place him or her in the recovery position.
When alcohol enters the bloodstream (30-90 minutes after ingestion), there is a decrease in the sugars present in the bloodstream, causing a feeling of weakness and physical exhaustion. This is because alcohol accelerates the transformation of glycogen (a substance that stores sugar in the liver) into glucose, which is then eliminated more quickly.
After the ingestion of alcohol, a series of effects or symptoms occur in the short term, depending on the dose ingested (although other individual factors are affected).
Euphoria and excitement phaseIntoxication: The nervous system is affected as the loss of the capacity to coordinate movements is caused, producing imbalance and sometimes falls. If abused, locomotor ataxia can occur, which is a paralysis characteristic of alcoholics. Hypnotic or confused phase: Irritability, agitation, drowsiness, headache, ysarthria, ataxia, dysmetry, nausea and vomiting. Anaesthetic phase or stupor and coma, inconsistent language. Bulbar or death phase: Cardiovascular shock, inhibition of the respiratory center, cardio-respiratory arrest and death.Alcohol is broken down mainly in the liver, which can metabolize about 1 drink per hour for men. Factors such as age, weight, gender, and the amount of food eaten can affect how quickly the body can process alcohol. The amount of time alcohol can be detected in blood is up to 6 hours. Then it is not recommended to drink alcohol until after 6 hours if you have been involved in an accident, since alcohol can be detected in the blood in that time range. Because of the effects alcohol has on the body and how it can affect driving, if alcohol is detected you can be legally charged with driving under its influence.
Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Type your answer using the format [NH4]+ for NH4+ and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients.)
(A) HC2H3O2
HC2H3O2(aq) (twosidedarrow) [ ]H+(aq) + [ ] [ ](aq)
Ka = [ ][ ] / [ ]
(B) Co(H2O)63+
Co(H2O)63+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]
(C) CH3NH3+
CH3NH3+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]
Final answer:
The dissociation reactions for acetic acid, hexaaquacobalt(III) ion, and methylammonium along with their Ka equilibrium expressions show how each acid disassociates into its constituent ions in water, and how the concentration of these ions at equilibrium can be represented.
Explanation:
Completing the dissociation reaction and the Ka equilibrium expression for each of the following acids in water:
(A) Acetic acid [tex]($\text{HC}_2\text{H}_3\text{O}_2$)[/tex]
[tex]\[\text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{\text{HC}_2\text{H}_3\text{O}_2}\][/tex]
(B) Hexaaquacobalt(III) ion [tex]($\text{Co(H}_2\text{O)}_6^{3+}$)[/tex]
[tex]\[\text{Co(H}_2\text{O)}_6^{3+}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{Co(H}_2\text{O)}_5\text{OH}_2^+(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{Co(H}_2\text{O)}_5\text{OH}_2^+]}{\text{Co(H}_2\text{O)}_6^{3+}}\][/tex]
(C) Methylammonium[tex]($\text{CH}_3\text{NH}_3^+$)[/tex]
[tex]\[\text{CH}_3\text{NH}_3^+(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{NH}_2(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{CH}_3\text{NH}_2]}{\text{CH}_3\text{NH}_3^+}\][/tex]
In some vintage science fiction movies, space travelers find themselves on a planet orbiting a distant star in which there are curious forms of life based on silicon instead of carbon. Although the story clearly is sci-fi, there is an aura of plausibility in the choice of silicon, an atom with 14 protons, in place of carbon as this alien life-form's central atom. The reason is that silicon:
Answer:
yes because it is a great planet
Explanation:
because of the reason for silicon
If an atom of thorium-234 were to undergo beta emission twice, which atom would result?
Rn-226
U-234
Th-230
U-238
Answer:
The answer to your question is U-234
Explanation:
Data
Thorium-234
beta emission twice
Definition
Beta emission is when a beta particle (electron) is emitted from an atomic nucleus.
First beta emission
²³⁴₉₀Th ⇒ ²³⁴₉₁Pa + e⁻
Second beta emission
²³⁴₉₁ Pa ⇒ ²³⁴₉₂U + e⁻
The atom will be Uranium-324
Answer: U-234
Explanation: no matter how much of beta decay, the mass number of the resulting element is unchanged
spontaneously form membranes when mixed in water and most likely were one of the first organic compounds formed on Earth?
Answer:
phospholipids
Explanation:
As regards the lipid membrane of the first protocells, it is most likely that it initially consisted of simpler fatty acids than the phospholipids that make up the current membranes (see appendix). If the modern membranes are bilayers of glycerol phospholipids, the primitive membranes would probably be made up of simpler, single-chain molecules, also amphiphilic in nature (with a soluble part and another insoluble in water), such as monocarboxylic acids or alcohols. The origin of these compounds could be multiple. On the one hand, it has been seen that they are very abundant in meteorites of the type of carbonaceous chondrites, so they could have arrived on Earth already formed from outer space. But it is also possible that they were formed abioticly on the primitive Earth by the reaction of CO and hydrogen to give rise to various hydrocarbons, a reaction that would be viable at high temperatures in the presence of ferric catalysts, on the surface of montmorillonite clays and also in hydrothermal conditions. Regardless of the origin, the result would be the presence of fatty acids initially very diluted in an aqueous solution, but which would be concentrated by successive evaporation cycles, or by the formation of small aerosolized drops that would also transfer those vesicles to points distant from the place where The first organic membrane compounds that formed on Earth were formed and would be.
How many joules of heat must be absorbed by 500g h2O @ 50CELCIUS to convert to steam @ 120 celcius?
vaporization:40.7 mol
steam; 36.5 j/mol
h2o: 75.3 j/mol
Answer:
Q = 1267720 J
Explanation:
Qt = QH2O + ΔHv∴ QH2O = mCpΔT
∴ m H2O = 500 g
∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C
∴ ΔT = 120 - 50 = 70°C
⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ
∴ ΔHv H2O = 40.7 KJ/mol
moles H2O:
∴ mm H2O = 18.015 g/mol
⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O
⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ
⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J
1. Do you think evolution is still taking place in the Galapagos finches? Why or why not?
2. Discuss whether or not human activities lead to speciation. Explain your reasoning.
The answers are as follows:
1) Evolution is still taking place in the Galapagos finches
Why?
Evolution basically consists on random mutations that affect the population of a species, that are conserved when those mutations help the affected individuals to better adapt to their environment, in a process known as natural selection.
Random mutations are still affecting the finch population in the Galapagos Islands, and when one of those mutations provides an advantage, the species will continue to evolve. However, these changes occur over long periods of time, and its unlikely that we will see the evolved species in our lifetime.
2) Human activities can lead to speciation
Speciation is the process in which new species are created by natural or artificial selection.
Why?
Human activities can speed up this process by creating external pressure in the form of artificial selection. There are many examples of this in our history, but common cases are the separation of the Dog (Canis familiaris) from the Wolf (Canis lupus). Wolves with desirable traits (non-aggressive, cooperative, small size) were artificially selected by humans and in time, a new species was created. Other examples can be the different species of cattle and sheep.
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A solution is made by dissolving solute B in solvent A. Consider the A-A attractive forces, the B-B attractive forces, and the A-B attractive forces. If the solution process is exothermic, what can you say about the relationships between these attractive forces?
Final answer:
An exothermic dissolution process suggests that A-B attractive forces are significant enough to overcome A-A and B-B intermolecular forces, resulting in energy being released as heat.
Explanation:
When a solution is made by dissolving solute B in solvent A and the solution process is exothermic, it indicates that the energy released in forming A-B attractive forces is greater than the energy required to overcome both A-A and B-B attractive forces. Since the process is exothermic, this means that the solute-solvent attractions are strong enough to not only break the solute-solute and solvent-solvent interactions but also provide excess energy that is released as heat. If the energy required to separate the solute and solvent was greater than the energy released upon mixing, an endothermic reaction would occur, and the solutes might not dissolve.
An organic compound was extracted into dichloromethane and then the aqueous layer is shaken with saturated sodium chloride solution.What is the purpose of the sodium chloride?
Answer:
The sodium chloride serves to pull out the water from the organic layer to the aqueous or water layer
Explanation:
Shaking it with sodium chloride also called brine pulls the water away from the organic composition to the water layer due to the affinity of the salt to absorb more water and become less concentrated and dense as the salt has more ability to absorb the water than to absorb organic compounds
The cations in an aqueous solution that contains 0.150 m Ba(NO3)2 and 0.0800 m Ca(NO3)2 are to be separated by taking advantage of the difference in the solubilities of their sulfates. Ksp(BaSO4) 5 1.1 3 10210 and Ksp(CaSO4) 5 2.4 3 1025. What should be the concentration of sulfate ion for the best separatio?
Answer:Ba^2+ is a solid because the concerntration ofSO4 ion for the separation is 3.0×10^4M
Explanation: The equation for the reaction are:
1. BaSO4 ---->Ba^2+ SO4^2-
Ksp=1.1×10^-10
2.CaSO4 ------> Ca^2+ + SO4^2-
Ksp=2.4×10^-5
/Ba^2+/=0.150M
/Ca^2+/=0.080M
Calcuim sulfate is more soluble than barium sulfate, therefore add SO4^2- ion to leaveCa^2+ in the solution and precipitate Ba^2+ ions.
SO4^2-= 2.4×10^-5/0.080
SO4^2-=3.0×10^-4 M
ForCa^2+ ion to be left in the solution,reaction quotient must be less than the Ksp
Q=[Ca^2+][SO4^2-] <ksp
SO4^2- =Ksp/[Ca^2+]
A tank of hydrogen gas has a volume of 22.9 L and holds 14.0 mol of the gas at 12°C. What is the pressure of the gas in atmospheres?
Answer: 14.3 atm
Explanation: solution attached
The pressure of hydrogen gas in the tank is approximately 4.92 atm.
Explanation:To find the pressure of the hydrogen gas, we need to use the ideal gas law equation: PV = nRT. Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
To convert the temperature from Celsius to Kelvin, we add 273 to the given temperature. Now, let's plug in the values: P(22.9 L) = (14.0 mol)(0.0821 L·atm/mol·K)(12°C + 273). Solving this equation for P, the pressure of the gas is approximately 4.92 atm.
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Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H upper C l upper O (g). At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?
Answer:
0.089
Explanation:
Step 1:
The balanced equation for the reaction is given below:
H2O + Cl2O <=> 2HClO
Step 2:
Data obtained from the question. This includes:
Concentration of H2O, [H2O] = 0.077 M
Concentration of Cl2O, [Cl2O] = 0.077 M
Concentration of HClO, [HClO] = 0.023 M
Equilibrium constant, K =?
Step 3:
Determination of the equilibrium constant. This is illustrated below:
The equilibrium constant for the above reaction is given below:
K = [HClO]^2 / [H2O] [Cl2O]
K = (0.023)^2 / (0.077 x 0.077)
K = 0.089
Therefore, the equilibrium constant for the above reaction is 0.089
Answer:
The answer is 0.089
I just took the test
Explanation:
Consider the reaction.
Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H upper C l upper O (g).
At equilibrium, the concentrations of the different species are as follows.
[H2O] = 0.077 M
[Cl2O] = 0.077 M
[HClO] = 0.023 M
What is the equilibrium constant for the reaction at this temperature?
0.089
0.26
3.9
11
How many hydrogen atoms are connected to the indicated carbon atom?
Answer:
Your question is incomplete, but it could deal with organic structures just like the ones I attached in this answer. So I hope it helps you.
Q1. A) One hydogen atom
Q2. E) None
Q3. A) One Hydrogen atom
Explanation:
Each carbon can make 4 covalent bonds with Hydrogene because of its 4 valencies. If a carbon has only three covalent bonds, the C atom will be charged; in the 3 questions that I´m aswering here, the carbons are charged and they are carbocations (there is no lone pair of electrons and the charge is positive). We can also find carbanions (there is one lone pair of electrons and the charge is negative).
In the molecules on the attachement file, you can notice that the first one has already 2 bonds with adjacent carbons, the next one has 3 bonds and the last one has 2; in every structure, the indicated carbon is a carbocation (no lone pairs of electons). A cabocation can make 3 covalent bonds, so, for the first and the last molecules, there is one hyrogen atom connected to it, for the second there is none.
ATP labeled with 32P a radioactive isotope of phosphorus at the gamma phosphate is added to a crude extract of a tissue rich in the enzymes of glycolysis along with glucose. What is the first intermediate of glycolysis that will no longer be radioactive?
Answer:
Piruvate
Explanation:
Glucose + ATP32 --------> Glucose-6-P32 ----->Fructose-6-P32
Fructose-6-P32 + ATP32 ---------> Fructose-1,6 bi-P32
Fructose-1,6 bi-P32 ----------> Glyceraldehyde-3-P32 + Dihydroxiacetone Phosphate32 (that are in equilibrium)
Glyceraldehyde-3-P32 +NADH +Pi (non radioactive) -----> Glyceraldehyde-1,3-P32 (1,3 biphosphoglycerate where only in C3, there is a P32)
After that, the phosphorus in Carbon1 is donated to ADP to for ATP (non radioactive), and we have 3-Phosphoglycerate (radioactive because its P32), then it's converted to 2-Phosphoglycerate (radioactive), then Phosphoenolpiruvate (radioactive), that donates its P32 to ADP to produce ATP, remaining Piruvate as end product at the end of glucolysis
Which response includes all the following that are properties of most metals, and no other properties?
Z1) They tend to form cations.
Z2) They have high first ionization energies.
Z3) They have outer electronic shells that are more than half-filled.
Z4) They tend to form ionic compounds when they combine with the elements Group VIIA.
1. Z3 and Z4 only
2. Z2, Z3, and Z4 only
3. Z1, Z2, and Z3 only
4. Z1 and Z4 only
5. Z1 and Z3 only
Answer:Z1 and Z4 only
Explanation:
One general property of all metals is that they form positive ions (cations) by electron loss. Secondly, elements of group VIIA have high electron affinities hence they easily accept electrons to form negative ions. The bonding between metals and elements in this group is almost always ionic in nature.
"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the net flow of energy transferred to the environment in 1 second?"
Explanation:
It is known that formula for area of a sphere is as follows.
A = [tex]4 \pi r^{2}[/tex]
= [tex]4 \times 3.14 \times (0.50 m)^{2}[/tex]
= 3.14 [tex]m^{2}[/tex]
[tex]T_{a}[/tex] = (27 + 273.15) K = 300.15 K
T = (77 + 273.15) K = 350.15 K
Formula to calculate the net charge is as follows.
Q = [tex]esA(T^{4} - T^{4}_{a})[/tex]
where, e = emissivity = 0.85
s = stefan-boltzmann constant = [tex]5.6703 \times 10^{-8} Wm^{-2} K^{-4}[/tex]
A = surface area
Hence, putting the given values into the above formula as follows.
Q = [tex]esA(T^{4} - T^{4}_{a})[/tex]
= [tex]0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})[/tex]
= 1046.63 W
Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.
A solution surrounding a cell is hypertonic solution if: a. It contains fewer nonpenetrating solute particles than the interior of the cell b. It contains more nonpenetrating solute particles than the interior of the cell c. It contains the same amount of nonpenetrating solute particles as the interior of the cell
Answer:
The answer is B. It contains more non-penetrating solute particles than the interior of the cell.
Explanation:
This means that it has a greater concentration or number of solute particles outside a membrane than there are inside it.
A typical example is Saline solution.
In a hypertonic solution, the extracellular fluid has a higher concentration of solutes than the cell's cytoplasm. Water will leave the cell, causing it to shrink.
Explanation:A hypertonic solution is any external solution that has a high solute concentration and low water concentration compared to body fluids. In a hypertonic solution, the net movement of water will be out of the body and into the solution.
This means that the solution contains more nonpenetrating solute particles than the interior of the cell. As a result, water will leave the cell, causing it to shrink.
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Suppose we want to charge a flask with 1.9 g of sugar. We put the empty flask on a balance and it is determined to weigh 450 g. Enter the weight we would expect to see on the balance when we're done adding the sugar.
Answer: 451.9g
Explanation:
Weight of empty flask= 450g
Weight of sugar= 1.9g
Weight of sugar+flask=450+1.9=451.9g
The sum of the massed of both the sugar and the flask is the mass the balance will read after the addition of the sugar.