A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
<h3>What is a decomposition reaction?</h3>
A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.
- Step 1: Write the balanced equation for the decomposition of KHCO₃.
2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)
The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.
- Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.
In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.
0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃
- Step 3: Calculate the mass percentage of KHCO₃ in the sample.
There are 1.25 g of KHCO₃ in the 1.775 g sample.
%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%
A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
Learn more about decomposition reactions here: brainly.com/question/14219426
Answer:
Explanation:
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Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Solving:
1,82 = <em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!
