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Nezavi [6.7K]
3 years ago
14

Write balanced molecular equation for the reaction between nitric acid and calcium hydroxide. Express your answer as a chemical

equation. Identify all of the phases in your answer.
Chemistry
1 answer:
eimsori [14]3 years ago
6 0

<u>Answer:</u> The balanced chemical equation for the reaction of nitric acid and calcium hydroxide is written below.

<u>Explanation:</u>

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The balanced chemical equation for the reaction of nitric acid and calcium hydroxide follows:

2HNO_3(aq.)+Ca(OH)_2(aq.)\rightarrow Ca(NO_3)_2(aq.)+2H_2O(l)

By Stoichiometry of the reaction:

2 moles of nitric acid reacts with 1 mole of calcium hydroxide to produce 1 mole of calcium nitrate and 2 moles of water molecule.

Hence, the balanced chemical equation for the reaction of nitric acid and calcium hydroxide is written below.

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Answer:

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Explanation:

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3 0
3 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

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3 years ago
Why the ph of glycine increases when 0.1 M NaOH is added dropwise​
shtirl [24]

Answer:

The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

Explanation:

The glycine is an amino acid with the following chemical formula:

NH₂CH₂COOH  

The COOH functional group is what gives the acid properties in the molecule.      

Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:

NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O

The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.  

Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.  

I hope it helps you!  

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Explanation:

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