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o-na [289]
3 years ago
11

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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Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

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Now according to question, there are 6.022×1023 atoms of Na 

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Explanation:

When solid cadmium sulfide reacts with an aqueous solution of sulfuric acid then the reaction will be as follows.

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Hence, ionic equation for this reaction is as follows.

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Therefore, net ionic equation for this reaction is as follows.

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