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4 years ago

Fill in the coefficients that will balance the following reaction: a0Na + a1H2O → a2NaOH + a3H2

Chemistry

2 answers:

nexus9112 [7]4 years ago

7 0

Answer : The balanced chemical reaction will be,

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

$a_0Na+a_1H_2O\rightarrow a_2NaOH+a_3H_2$

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of hydrogen are not balanced while all the atoms are balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the $Na,H_2O\text{ and }NaOH$

Thus, the balanced chemical reaction will be,

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Alexandra [31]4 years ago

4 0

Hey there ! :

<span>2 Na + 2 H2O = 2 NaOH + 1 H<span>2

</span></span>Sodium<span> + </span>Dihydrogen Monoxide<span> = </span><span>Natriumhydroxid</span><span> + </span>Hydrogen

Coefficients:

Reagents : Na = 2
H2O = 2

Products : NaOH = 2
H2 = 1

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What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

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This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5% as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

$\% H=\frac{m_H*1}{M_{HF}}*100\%$

Where $m_H$ stands for the atomic mass of hydrogen and $M_{HF}$ for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

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2 years ago

A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is

DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

C = concentration of acid (HCl) = 0.0470 M,
V = volume of acid = 16.0 mL,
C' = concentration of base (ammonia solution) and
V' = volume of base = 26.0 mL.

<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

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C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] = where

[H⁺] = concentration of HCl at equivalence point,
[OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

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<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

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pH ≅ 12.46

So, the pH at equivalence point is 12.46

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