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3 years ago

The Ka of carbonic acid is 4.3 x 10-7.

Chemistry

1 answer:

ratelena [41]3 years ago

3 0

Answer:

poor hydrogen-ion donor

Explanation:

Acid dissociation constant constant chemistry is the equilibrium constant of the dissociation reaction of an acid, it is denoted by Ka. This equilibrium constant is a measure of the strength of an acid in a solution.

Note these as a rule of thumb:

When Ka is large, the dissociation of the acid is favored.

When Ka is small, the acid does not dissociate to a large extent.

Hence, a Ka of 4.3 x 10-7 shows a weak acid. A weak acid is a poor hydrogen ion donor because it does not dissociate to a large extent in solution.

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Aqueous sulfuric acid H2SO4 reacts with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2

Pie

Answer:

109.34 g

Explanation:

2NaOH(aq) + H2SO4(aq) ------> Na2SO4(aq) + 2H2O(l)

Number of moles of NaOH = 105g/40g/mol = 2.6 moles

From the reaction equation;

2 moles of NaOH yields 1 mole of sodium sulphate

2.6 moles of NaOH yields = 2.6 × 1/2 = 1.3 moles of sodium sulphate

Number of moles of H2SO4= 75.5g/98 g/mol = 0.77 moles

From the reaction equation;

1 mole of H2SO4 yields 1 mole of sodium sulphate

Hence, 0.77 moles of H2SO4 yields 0.77 moles of sodium sulphate

So H2SO4 is the limiting reactant.

Theoretical yield = number of moles × molar mass

= 0.77 mol ×142 g/mol

= 109.34 g

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3 years ago

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Answer:

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Explanation:

1) C

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3) A

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3 years ago

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3 years ago

Read 2 more answers

The average lung capacity of a human is 6.0L.

Darya [45]

Answer:

(a) 0.25 mol

(b) 0.11 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol$

(b)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol$

(c)

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

n = number of moles = ?

Putting values in above equation, we get:

$30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol$

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4 years ago