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s2008m [1.1K]
2 years ago
13

In a neutralization reaction, an aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to p

roduce:.
Chemistry
1 answer:
dem82 [27]2 years ago
6 0

An aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt.

<h3>What is a Salt?</h3>

This is a compound which is formed as a result of a neutralization reaction between acid and base.

Arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt due to increased concentration of H+ and OH- respectively.

Read more about Salt here brainly.com/question/13818836

#SPJ1

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Where in the eukaryotic cells does succinyl CoA take place​
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Calculate the % Oxygen in NaHCO3
Alexxx [7]

Answer:

Molar mass of NaHCO3 = 84.00661 g/mo

Explanation:

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Convert grams NaHCO3 to moles  or  moles NaHCO3 to grams

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3 years ago
A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC
inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K

Putting values in above equation, we get:

\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0
3 years ago
At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil
8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

\frac{0.725}{3.08}=\frac{S_{2}}{8}

On rearranging,

S_{2}=\frac{0.725\times 8}{3.08}=1.88

Therefore, solubility will be 1.88 mg of N_{2} gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0
2 years ago
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