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3 years ago

What type of chemical reaction is the following? BaCl2(aq) + Na2SO4(aq)

Chemistry

1 answer:

Basile [38]3 years ago

4 0

Answer:

it's a precipitation reaction.

Explanation:

since a solid is produced, one of the elements are insoluble with one another–making a precipitate.

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2CO + O2 --> 2CO2<br> What is the coefficient of carbon dioxide?

FrozenT [24]

Answer:

<h2> 2 is the coefficient of CO₂</h2>

Explanation:

The <u>coefficient</u> is the number in front of a substance that indicates how many molecules or atoms are. For example, when we see 5He, it shows that there are 5 atoms of helium. When we see 5H₂O, it indicates that there are 5 molecules of water, each having 2 atoms of hydrogen and 1 atom of oxygen, so a total of 10 hydrogen atoms and 5 oxygen atoms.

4 0

3 years ago

A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde

hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is $\boxed{{\text{0}}{\text{.85}}}$ .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}$

......(1)

Rearrange equation (1) to calculate the moles of solute.

${\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)$ ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}$

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}$

The reaction between HCl and NaOH occurs as follows:

${\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}$

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

$\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}$

The volume after the addition of NaOH can be calculated as follows:

$\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}$

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

$\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}$

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

${\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]$ ......(3)