a 1.1 mole sample of Kr has a volume of 371mL. How many moles of Kr are in a 2.85L sample at the same temperature and pressure

The value of Kp at 200°C for the given reaction is 2.2 x 10^3. The equilibrium partial pressure of CO in the reaction H2(g) + CO2(g) ? H2O(g) + CO(g) is 0.200 atm. Increasing the temperature in an exothermic equilibrium reaction favors the formation of products, not reactants. The equilibrium-constant expressions for a reaction and its reverse are related by the reciprocal of each other.

To determine the value of Kp at 200°C for the reaction 3 Fe (s) + 4 H2O (g) ? Fe3O4 (s) + 4 H2 (g), we can use the relationship between Kc and Kp. Since the reaction involves gases, we can use the equation Kp = Kc(RT)Δn, where Δn is the change in the number of moles of gas. In this case, there is no change in the number of moles of gas, so Δn = 0. Therefore, the value of Kp is equal to the value of Kc, which is given as 2.2 x 10^3 at 50°C. Thus, the value of Kp at 200°C is also 2.2 x 10^3.

To calculate the equilibrium partial pressure of CO in the reaction H2(g) + CO2(g) ? H2O(g) + CO(g), first, we need to determine the initial partial pressures of H2 and CO2. Both gases have a partial pressure of 0.200 atm when admitted into the rigid container. Since the reaction is at equilibrium, the equilibrium partial pressure of CO is the same as the initial partial pressure of CO2, which is 0.200 atm.

The statement that increasing the temperature favors the formation of reactants in an exothermic equilibrium reaction is False. In an exothermic reaction, the formation of products is favored with an increase in temperature. The reaction releases heat, so increasing the temperature shifts the equilibrium position towards the products.

The statement that the equilibrium-constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written for the reverse direction is False. The equilibrium-constant expression for a reaction and its reverse are related by the reciprocal of each other, but they are not exactly the same.

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Using a starch solution instead of a protein in an experiment would yield different results because proteins and starch have different biochemical properties.

If you used a starch solution instead of a protein, different biochemical reactions would occur, as they differ fundamentally in their nature. Proteins are made up of amino acids, while starch is a polysaccharide composed of glucose units. As such, they interact differently with various substances. For instance, if you perform a Iodine test, the results would vary: iodine reacts with starch to give a blue-black colour, but it shows no color change with proteins. Likewise with the Biuret's test, you would obtain a negative result with starch but a positive (purple color) with proteins.

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Explanation :

Zwitter ion : An ion or molecule having separate negatively and positively charged groups.

Methionine is an amino acid with sulfur atom in its molecule .The proton from carboxylic group with get attached to amino group present in the methionine and negative charge will be generated on oxygen atom of a carboxylic group which will get in conjugation with an another oxygen atom of the carboxylic group.

The positive charge will generated on the nitrogen atom due  attachment of the proton from the carboxylic group.

As shown in the image attached.

To form the zwitterion of methionine, the amino group gains a proton to become -NH₃⁺, while the carboxyl group loses a proton to become CO²⁻, resulting in a molecule with no overall electrical charge.

To convert methionine to its zwitterion form, you need to recognize the conditions under which amino acids typically exist as zwitterions. The zwitterion form of an amino acid has both a positive charge and a negative charge, but the overall molecule is electrically neutral. Methionine, which has the chemical formula CH₃SCH₂CH₂CH(NH₂)CO₂H, will have its amino group (-NH₂) gain a proton to become -NH₃⁺, and the carboxyl group (CO₂H) lose a proton to become CO²⁻. This results in a molecule with a positively charged ammonium (NH₃⁺) at one end and a negatively charged carboxylate (CO²⁻) at the other, thus forming the zwitterion.

At the biological pH of approximately 7.4, or the isoelectric point of methionine, the molecule will adopt this zwitterionic form naturally. It's important to note that methionine is the only common amino acid with sulfur in its side chain (R group). As with alanine and other amino acids, this zwitterionic form is more soluble in water, has a high melting point, and is less soluble in nonpolar solvents.

Final answer:

To find the new pressure of an ideal gas after heating, we use the combined gas law. The initial temperature and pressure are converted to Kelvins and atmospheres respectively, and then the formula is applied to give a new pressure of approximately 3.038 atm after heating the gas to 820.0 °C.

Explanation:

To determine the new pressure of an ideal gas after heating, we can use the combined gas law, which states that (P1 x V1) / T1 = (P2 x V2) / T2, where P is pressure, V is volume, and T is temperature in Kelvins. Given that volume remains constant (as the gas remains in the same cylinder) and considering the temperature change from 30.0 °C to 820.0 °C, we just need to convert these temperatures into Kelvins by adding 273.15 and calculate the new pressure.

Convert the initial temperature to Kelvins: T1 = 30.0 °C + 273.15 = 303.15 K
Convert the final temperature to Kelvins: T2 = 820.0 °C + 273.15 = 1093.15 K

Convert the initial pressure to atmospheres: P1 = 710 torr × (1 atm / 760 torr) ≈ 0.934 atm

Now apply the combined gas law: P2 = (P1 × T2) / T1
P2 = (0.934 atm × 1093.15 K) / 303.15 K ≈ 3.038 atm

Therefore, the new pressure of the gas after heating to 820.0 °C is approximately 3.038 atm.

Answer:

[tex]M=39.1g/mol[/tex]

[tex]\rho=1.70g/L[/tex]

Explanation:

Hello,

In this case, one uses the ideal gas equation:

[tex]PV=nRT[/tex]

By solving for moles in terms of mass and molar mass one obtains:

[tex]PV=\frac{m}{M}RT\\M=\frac{mRT}{PV} =\frac{3.15g*0.082 \frac{atm*L}{mol*K} *(35+273)K}{1.10atm*1.85L} \\M=39.1g/mol[/tex]

On the other hand, the density is easily computed as shown below:

[tex]\rho=\frac{m}{V}=\frac{3.15g}{1.85L}=1.70g/L[/tex]

Best regards.

Answer:

d) bromothymol blue

Explanation:

Hello,

For that pH change between 5.3 and 6.3, we need an indicator which changes the color within that rank, in such a way, among the given options, one recalls the bromothymol blue because it turns yellow below about a pH of 6 and starts getting green above 6, therefore, one states that it is the most appropriated indicator.

Best regards.

Answer:

answer b

Explanation:

According to Boyle's law and as 1 atmosphere= 760 torr the new volume when pressure is increased to 1.5 atmospheres is 3.37 L.

Boyle's law is an experimental gas law which describes how the pressure of the gas decreases as the volume increases. It's statement can be stated as, the absolute pressure which is exerted by a given mass of an ideal gas is inversely proportional to its volume provided temperature and amount of gas remains unchanged.

Mathematically, it can be stated as,

P∝1/V or PV=K. The equation states that the product of of pressure and volume is constant for a given mass of gas and the equation holds true as long as temperature is maintained constant.

According to the equation the unknown pressure and volume of any one gas can be determined if two gases are to be considered.That is,

P₁V₁=P₂V₂

Substitution in above equation gives V₂=0.989×5.12/1.5=3.37 L.

Thus, the new volume of gas is 3.37 L.

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Final answer:

To calculate the equilibrium partial pressure of H2S, use the equilibrium constant expression and rearrange to solve for PH2S.

Explanation:

To calculate the equilibrium partial pressure of H2S, we need to use the equilibrium constant, Kp. The equilibrium constant expression for the given reaction is:

Kp = [H2S] / [NH4HS]

Given that Kp = 0.110, we can establish the relationship:

Kp = (PH2S) / (PNH4HS)

Since the initial moles of NH4HS is 0.371 moles and the volume of the vessel is 1.00 L, we can calculate the initial partial pressure of NH4HS:

PNH4HS = (0.371 mol) / (1.00 L) = 0.371 atm

Now, we can rearrange the equilibrium constant expression and solve for PH2S:

PH2S = Kp * PNH4HS = (0.110)(0.371 atm) = 0.0408 atm

The equilibrium partial pressure of H₂S is 0.332 atm. The equilibrium partial pressures for PCl₅, PCl₃, and Cl₂ are 0.83 atm, 0.64 atm, and 0.64 atm, respectively.

A.) To determine the equilibrium partial pressure of H₂S for the reaction NH₄HS(s) ⇌ NH₃(g) + H₂S(g) with Kp = 0.110 at 298 K, follow these steps:

Therefore, the equilibrium partial pressure of H₂S is 0.332 atm.

B.) For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kp = 0.497 at 500 K:

Therefore, the equilibrium partial pressures are PPCl₅ = 0.83 atm, PPCl₃ = 0.64 atm, and PCl₂ = 0.64 atm.

Answer:

Answer has been given below

Explanation:

Answer:

[tex]M=1.02M[/tex]

Explanation:

Hello,

In this case, we compute the concentration as the molarity of the nitrate ion as the ratio between the nitrate ion moles and the volume of the solution as shown below:

[tex]M_{NO_3^-}=\frac{n_{NO_3^-}}{V_{solution}}[/tex]

Now, the nitrate ion moles are computed considering that in mole of magnesium nitrate, there are two moles of the nitrate ion:

[tex]n_{NO_3^-}=32.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3gMg(NO_3)_2}*\frac{2molNO_3^-}{1molMg(NO_3)_2}\\n_{NO_3^-}=0.432molNO_3^-[/tex]

Finally, the requested molarity turns out into:

[tex]M_{NO_3^-}=\frac{0.432molNO_3^-}{0.425L}\\M=1.02M[/tex]

Best regards.

Final answer:

To calculate the nitrate ion concentration, convert the mass of magnesium nitrate to moles, convert volume to liters, determine the molarity of magnesium nitrate, and then multiply by 2, since each formula unit yields two nitrate ions. The concentration of nitrate ion is 1.0146 M.

Explanation:

The student is asking how to calculate the concentration of nitrate ion in a solution with a known volume and mass of magnesium nitrate. We can start by finding the molarity of magnesium nitrate and then use stoichiometry to determine the concentration of nitrate ions.

First, convert the mass of Mg(NO3)2 to moles:

Moles of Mg(NO3)2 = mass (g) / molar mass (g/mol)

32.0g / 148.3g/mol = 0.2156 mol

Next, convert the volume from mL to L:

425 mL * (1L / 1000 mL) = 0.425 L

Now calculate the molarity of the Mg(NO3)2 solution:

Molarity (M) = moles / volume (L)

0.2156 mol / 0.425 L = 0.5073 M

Each unit of Mg(NO3)2 gives two nitrate ions, so the concentration of nitrate ion is:

Concentration of NO3- = 2 × molarity of Mg(NO3)2

2 × 0.5073 M = 1.0146 M

Therefore, the concentration of nitrate ion in the solution is 1.0146 M.

The rate constant of the reaction is 0.0002 s-1.

Using the formula;

lnC = lnCo - kt

Where;

C = concentration at time t

Co = initial concentration

k = rate constant

t = time taken

ln(335) = ln(502) - k(2000)

k = ln(335)  - ln(502)/(-2000)

k = 5.8 - 6.2/(-2000)

k = 0.0002 s-1

For a first order reaction;

t1/2 = 0.693/k

t1/2 = 0.693/ 0.0002 s-1

t1/2 =3465 s

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