All categories

3 years ago

Why do we use balanced equations ?

Chemistry

1 answer:

polet [3.4K]3 years ago

8 0

Answer:

Balanced equations are used to allow chemists to calculate how much product they will produce from their reactants. The co-efficients in balanced equations represent the number of moles reacting and being produced, it is rare that this exact ratio will be used in a reaction.

Explanation:

You might be interested in

Ava’s best friend has an indoor cat. Ava notices that she sneezes every time that she visits her friend. Ava suspects that she i

zlopas [31]

Answer:

I say for her to test her suspicion by going to see an allergist to diagnose the allergy.

7 0

3 years ago

If the federal government wants to encourage businesses and consumers to spend more money, it would MOST LIKELY

ICE Princess25 [194]

If the federal government wants to encourage businesses and consumers to spend more money, it would most likely to reduce the price of the goods to attract more buyers and sellers

6 0

4 years ago

Read 2 more answers

Calculate the molar concentrations of H+ and OH− in solutions that have the following pH values:

Anastaziya [24]

Answer:

Explanation:

a )

pH = - log[ H⁺]

8.26 = - log[ H⁺]

[ H⁺] = 10⁻⁸°²⁶ mole / l

= 5.49 x 10⁻⁹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.49 x 10⁻⁹

= .182 x 10⁻⁵ moles / l

b )

10.25 = - log[ H⁺]

[ H⁺] = 10⁻¹⁰°²⁵ mole / l

= 5.62 x 10⁻¹¹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.62 x 10⁻¹¹

= .178 x 10⁻³ moles / l

c )

4.65 = - log[ H⁺]

[ H⁺] = 10⁻⁴°⁶⁵ mole / l

= 2.24 x 10⁻⁵ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.24 x 10⁻⁵

= .4464 x 10⁻⁹ moles / l

7 0

3 years ago

what acid-base properties can determine if a household chemical is an acid, base, or neutral substance.

damaskus [11]

if it is edible you can taste it if it taste bitter its a base and if it taste sour then it is a acid if it taste like nothing like water would it would be a neutral substance.you could also order a ph tester which is just a liquid that you drop in it.

yw:)

8 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)

Here,

$\left[{{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of also becomes 0.141 M.

Substitute 0.141 M for $\left[{{{\text{H}}^ + }}\right]$ in equation (3).

$\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}$

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

4 0

3 years ago

Read 2 more answers