Ask question

Ask question

All categories

3 years ago

10

How can balancing equations support the law of conservation

1 answer:

amm18123 years ago

4 0

Answer:

detail is given below.

Explanation:

This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

For example:

In given photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

The given equation is balanced chemical equation of photosynthesis. There are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

If equation is not balanced,

CO₂ + H₂O → C₆H₁₂O₆ + O₂

It can not follow the law of conservation of mass because mass is not equal on both side of equation.

You might be interested in

Help me find the word elliptical

maria [59]

Omg where is it?? i’ve been looking for a long time

7 0

2 years ago

WILL MARK BRAINLIEST!

jarptica [38.1K]

An atomic mass unit is defined as precisely 1/12 the mass of an atom of carbon-12.

7 0

3 years ago

My science homework is due tomorrow and I am stuck on this last question.

stealth61 [152]

I think 25m hopes dis helps

3 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

<u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

<u>Initial:</u> 0.1372 0.04514

<u>Final:</u> 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Please help me with this friends.

Radda [10]

Where is the picture to this?

3 0

3 years ago