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3 years ago

3. The equation for the decomposition of sodium hydrogencarbonate is :

Chemistry

1 answer:

Nastasia [14]3 years ago

5 0

a. 13.25 g

b. 3 L

<h3>Further explanation</h3>

Given

21 g of Sodium hydrogencarbonate

Reaction

2 NaHCO3 (s) → Na2CO3 (s) + H2O(l) + CO2(g)

Required

mass of residue

volume of CO2

Solution

mol NaHCO₃ (MW=84 g/mol) :

$\tt mol=\dfrac{21}{84}=0.25$

a. the residue = Na₂CO₃

From the equation, mol ratio NaHCO₃ : Na₂CO₃ = 2 : 1, so mol Na₂CO₃ :

$\tt \dfrac{1}{2}\times 0.25=0.125$

mass Na₂CO₃ (MW=106 g/mol) :

$\tt mass=0.125\times 106=13.25~g$

b. From equation, mol CO₂ = 0.5 x mol NaHCO₃ = 0.125

RTP= 25 C, 1 atm ⇒ 1 mol =24 L

So volume CO₂ :

$\tt 0.125\times 24=3~L$

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Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation

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Number of moles of A = Number of moles of HA = Number of moles of NaOH

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Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

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a = 9.673 * 10^-7

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If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?

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14.434 r.a.m.

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The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

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Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

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No we have to calculate the concentration of hydronium ion or hydrogen ion.

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$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

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pH + pOH = 14

pOH =14 -2.1 = 11.9

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