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MrMuchimi
3 years ago
7

Which information is sufficient to differentiate a

Chemistry
2 answers:
blagie [28]3 years ago
8 0

Answer: (3) The reactivity of each sample with water

Explanation: Since sodium dissolves in water, you could place the two elements in water and see which of the two dissolve

Novay_Z [31]3 years ago
3 0

"The reactivity of each sample with water" is sufficient to differentiate the sample of sodium from a sample of silver.

<u>Explanation :</u>

Silver does not undergo any reaction with water. Also, silver is insoluble in water under normal conditions. Thus it remains in its metallic state when it comes in contact with water.

On the other hand, sodium is highly reactive in water. It forms a colorless solution in water, which is basic. This basic solution is sodium hydroxide and hydrogen gas. The reaction of sodium and water is exothermic.

2 N a(s)+2 H_{2} O \rightarrow 2 N a O H(a q)+H_{2}(g)

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the
Sergio [31]

Answer:

1.2\times 10^{-2}

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq)     K_3 = 1.89\times 10^{-10}

AgCl(aq)⇌Ag+(aq)+Cl−(aq)           K_4 = 1.25\times 10^{-4}

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

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4 0
3 years ago
A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius
Nonamiya [84]
Data Given:
                  Pressure  =  P  =  ?

                  Volume  =  V  =  3.0 L

                  Temperature  =  T  =  115 °C + 273  =  388 K

                  Mass  =  m  =  75.0 g

                  M.mass  =  M  =  44 g/mol

Solution:
              Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
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Solving for P,
                                      P  =  n R T / V      ------ (1)
Calculating Moles,
                                      n  =  m / M

                                      n  =  75.0 g / 44 g.mol⁻¹

                                      n  =  1.704 mol

Putting Values in Eq. 1,

                    P  =  (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

                    P  =  18.08 atm
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