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Alexxandr [17]
3 years ago
15

In a certain acidic solution at 25 ∘c, [h+] is 100 times greater than [oh −]. what is the value for [oh −] for the solution? in

a certain acidic solution at 25 , [] is 100 times greater than [ ]. what is the value for [ ] for the solution? 1.0×10−8 m 1.0×10−7 m 1.0×10−6 m 1.0×10−2 m 1.0×10−9 m
Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Answer:

1.0 x 10⁻⁸ M.

Explanation:

<em>∵ [H⁺][OH⁻] = 10⁻¹⁴. </em>

∵ [H⁺] = 100 [OH⁻].

∴ 100 [OH⁻][OH⁻] = 10⁻¹⁴.

∴ 100 [OH⁻]² = 10⁻¹⁴.

[OH⁻]² = 10⁻¹⁴/ 100 = 1.0 x 10⁻¹⁶.

<em>∴ [OH⁻] = √(1.0 x 10⁻¹⁶) = 1.0 x 10⁻⁸ M. </em>

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Explanation:

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Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]
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<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq) + Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>) = 0.021 M.
Ka(HCN) =  4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻] / [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] = x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M - x).
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Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --&gt; C2H2(g) + Ca(OH)2(aq
nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of CaC_2 = 2.54\ g

Molar mass of CaC_2 = 64.099 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{2.54\ g}{64.099\ g/mol}

Moles_{CaC_2}= 0.0396\ mol

According to the given reaction:-

CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}

1 mole of CaC_2 on reaction forms 1 mole of C_2H_2

0.0396 mole of CaC_2 on reaction forms 0.0396 mole of C_2H_2

Moles of C_2H_2 = 0.0396 moles

Considering ideal gas equation as:-

PV=nRT

where,

P = pressure of the gas = 742 mmHg  

V = Volume of the gas = ?

T = Temperature of the gas = 26^oC=[26+273]K=299K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L

<u>1.0 L of acetylene  can be produced from 2.54 g CaC_2.</u>

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