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Daniel [21]
3 years ago
15

When a certain powder is mixed into a solution of silver nitrate, a white precipitate forms. When the powder is added to a solut

ion of sodium hydroxide, bubbles form, which produce a noxious odor. These bubbles, when bubbled into a solution of nickel (II) hydroxide, produce a drastic color change from green to dark blue. What is the identity of this powder?
Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer:

Ammonium chloride

Explanation:

The powder is:- Ammonium chloride

When mixed with silver nitrate, white prescipitate of silver chloride is formed as:-

NH_4Cl+AgNO_3\rightarrow NH_4NO_3+AgCl

When mixed with sodium hydroxide, ammonia gas is formed which has noxious order.

NH_4Cl +NaOH\rightarrow NH_3 + H_2O + NaCl

Ammonia gas on reaction with nickel (II) hydroxide forms deep blue colored complex as shown below:-

Ni(OH)_2(s) + 6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}(aq) + 2OH^{-}(aq)

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Calculate the molarity of an hcl solution if a 10.00 ml sample requires 25.24 ml of a 1.600 m naoh solution to be neutralized.
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Answer:

The molarity of the HCl solution should be 4.04 M

Explanation:

<u>Step 1:</u> Data given

volume of HCl solution = 10.00 mL = 0.01 L

volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L

<u>Step 2:</u> The balanced equation

HCl + NaOH → NaCL + H2O

Step 3: Calculate molarity of HCl

n1*C1*V1 = n2*C2*V2

Since the mole ratio for HCl and NaOH is 1:1 we can just write:

C1*V1 =C2*V2

⇒ with C1 : the molarity of HCl = TO BE DETERMINED

⇒ with V1 = the volume og HCl = 10 mL = 0.01 L

⇒ with C2 = The molarity of NaOH = 1.6 M

⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L

C1 * 0.01 = 1.6 * 0.02524

C1 = (1.6*0.02524)/0.01

C1 = 4.04M

The molarity of the HCl solution should be 4.04 M

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