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Lina20 [59]
3 years ago
11

Determine the pH of a 0.461 M C6H5CO2H M solution if the Ka of C6H5CO2H is 6.5 x 10-5. Determine the pH of a 0.461 M C6H5CO2H M

solution if the Ka of C6H5CO2H is 6.5 x 10-5. 11.74 9.48 5.48 4.52 2.26

Chemistry
1 answer:
valkas [14]3 years ago
3 0

Answer:

the answer

Explanation:

the answer

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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
What is the valency of hydroxide​
Simora [160]

Answer:

Its valancy is 1.

As it has the formula OH-.

It has valancy 1.

<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

4 0
3 years ago
What explains osmosis
Reika [66]
A process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one thus equalizing the concentrations on each side of the membrane. I hope this helps :D

8 0
3 years ago
3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa
Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

7 0
3 years ago
KMnO₄ solution
kicyunya [14]
"B. purple" is the correct answer.
7 0
3 years ago
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