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Tanzania [10]
2 years ago
8

Write the formula of the conjugate acid of HCO₂⁻.

Chemistry
1 answer:
Alex2 years ago
7 0

Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions.

<h3>Conjugate base and conjugate acid</h3>

Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

<h3>Conjugate acid of HCO₂⁻</h3>

Like a conjugate acid is an ion or molecule resulting from the base that gains the proton, the conjugate acid of HCO₂⁻ is H₂CO₂.

Learn more about the Brønsted-Lowry acid-base theory:

<u>brainly.com/question/12916250?referrer=searchResults</u>

<u>brainly.com/question/1191429?referrer=searchResults</u>

<u>brainly.com/question/4000152?referrer=searchResults</u>

<u>brainly.com/question/12808135?referrer=searchResults</u>

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A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect
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Answer:

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Explanation:

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27 g of fat on burning gives 9*27 = 243 kilocalories of energy

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Total energy = 515 kilocalories

Using,

Q=m_{water}\times C_{water}\times (T_f-T_i)

Given: Volume of water = 23 L = 23×10⁻³ m³

Density=\frac{Mass}{Volume}  

Density of water= 1000 kg/m³

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}  

Mass of water  = 23 kg

Initial temperature = 16°C  

Specific heat of water = 0.9998 kcal/kg°C  

515=23\times 0.9998\times (T_f-16)

Solving for final temperature as:

<u>Final temperature = 38.3958 °C  </u>

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