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3 years ago

What is formed in the neutralization reaction between a strong acid and a strong base? (2 points)

Chemistry

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer: A neutral salt and water is formed as products.

Explanation:

Neutralization reaction is defined as the reaction in which an acid reacts with a base to produce salt and water molecule.

$BOH+HX\rightarrow BX+H_2O$

Salts are formed when an acid reacts with a base during a neutralization reaction.

When a strong acid and a weak base reacts, it leads to the formation of acidic salt.
When a strong base and weak acid reacts, it leads to the formation of basic salt.
When a strong acid and strong base or weak acid and weak base reacts, it leads to the formation of neutral salts.

Hence, a neutral salt and water is formed as products.

Alborosie3 years ago

5 0

A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.

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What mass of Iron (III) acetate will be produced when 1.50 grams of iron reacts in acetic acid?

densk [106]

Answer:

Fe + H(C2H3O2) ........

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3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

Harman [31]

Answer:

$\boxed{\text{10 L}}$

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

$\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}$

Calculation:

$\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}$

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3 years ago

What’s the empirical formula of taurine

zhenek [66]

Answer: The formula is C2H7NO3S or NH2CH2CH2SO3H

Hope this helps!

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Which statement best describes a physical change?

yulyashka [42]

The answer is the third one .

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