Explain why we must balance all chemical equations

Chemistry

2 answers:

Illusion [34]3 years ago

4 0

Explanation:

the chemical equation needs to be balance so that it follows the law of conservation of mass. a balance chemical equation occurs when the number of the different atoms of elements to the reactants side is equal to that of the product side. balancing chemical equation is a process of trial and error .

enyata [817]3 years ago

3 0

Answer:

The chemical equation needs to be balanced so that it follows the law of conservation of mass.

Explanation:

You might be interested in

What is the mole fraction of solute in a 3.19 m aqueous solution?

Bumek [7]

Molality= mol/ Kg

if we assume that we have 1 kg of water, we have 3.19 moles of solute.

the formula for mole fraction --> mole fraction= mol of solule/ mol of solution

1) if we have 1 kg of water which is same as 1000 grams of water.

2) we need to convert grams to moles using the molar mass of water

molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol

1000 g (1 mol/ 18.02 grams)= 55.5 mol

3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution

4) mole fraction= 3.19 / 58.7= 0.0543

4 0

2 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.