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Daniel [21]
3 years ago
9

What is the ph of a 0.135 m aqueous solution of potassium acetate, kch3coo? (ka for ch3cooh = 1.8×10-5)?

Chemistry
1 answer:
Gennadij [26K]3 years ago
6 0
[OH-] = √(Cs×Kw)/Ka
[OH-] = √(0,135×10^-14)/1,8×10^-5
[OH-] = √0,075×10^-9 = √75×10^-12 = 8,66×10^-6

pOH = -log[OH-]
pOH = -log[8,66×10^-6]
pOH ≈ 5,0625

pH + pOH = 14
pH = 14-5,0625
pH = 8,9375

:•)
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Shalnov [3]

Answer:

all good. tell me about your day

5 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1
jekas [21]

Answer:

y1 = 0.3162

y2 = 0.6838

Explanation:

ok let us begin,

first we would be defining the parameters;

at 25°C;

1-propanol P1° = 20.90 Torr

2-propanol P2° = 45.2 Torr

From Raoults law:

P(1-propanol) = P⁰ × X(1-propanol)

P(1-propanol) = 20.9 torr × 0.45 = 9.405

P(1-propanol) = 9.405 torr

Also P(2-propanol) = P⁰ × X(2-propanol)

P(2-propanol) = 45.2 torr × 0.45

P(2-propanol) = 20.34 torr

but the total pressure = sum of individual pressures

total pressure = 9.405 + 20.34

total pressure = 29.745 torr

given that y1 and y2 represent the mole fraction of each in the vapor phase

y1 = P1 / total pressure

y1 = 9.405/29.745

y1 = 0.3162

Since y1 + y2 = 1

y2 = 1 - y1

∴ y2 = 1 -  0.3162

y2 = 0.6838

cheers, i hope this helps.

7 0
3 years ago
What is the formula ofchromium(III) hydrogensulfate?
FromTheMoon [43]

Answer:

Cr (HSO4)3

Explanation:

its molecular weight is 343.20 g/mol

its molecular formula can also be written as CrH3O12S3

molar mass of Cr (HSO4)3 can be calculated by following method;

atomic mass of Cr = 51.9961 u

atomic mass of H = 1 u

atomic mass of S = 32.065 u

atomic mass of O = 16 u

molar mass of Cr(HSO4)3 =  51.9961+ 1.00784×3 + 32.065×3 + 15.999×12

molar mass of Cr(HSO4)3 =51.9961+3.02352+96.195+ 191.988

molar mass of Cr(HSO4)3 = 343.20 g/mol

3 0
3 years ago
There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a
Vanyuwa [196]
<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

  • The natural isotopic abundance of 10B is 19.60%.
  • The natural isotopic abundance of 11B is 80.40%.
  • The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages  we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10  B.

5 0
3 years ago
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