All categories

3 years ago

12) Which of the above solutions could be used to remove lead from lead (II) nitrate?

Chemistry

1 answer:

maw [93]3 years ago

7 0

Ag can be used to remove lead from lead (II) nitrate.

Option C

Explanation:

In order to separate the lead from lead (II) nitrate, we have to react it with another solution having same oxidation state and which can replace lead and make it as a precipitate.

So among the given options, Ag has high tendency to react with nitrate easily. Since, the oxidation state of Ag is also similar to the oxidation state of Pb, it can easily substitute Pb. Thus, Ag can be used to remove lead from lead nitrate.

You might be interested in

Technetium-99m, a “metastable” nuclide of

Bond [772]

In beta decay, a neutron is transformed into a proton via emission of an electron. This means that the total nucleon number remains the same; however, the proton number is increased by one (and the neutron number decreases by one). Therefore, the parent atom must also have mass number 99 but be of one less atomic number.
The atomic number of Tc is 43 and proton number 42 is Mo. Therefore, the answer is:
99Mo

4 0

2 years ago

Compound a reacts with one equivalent of h2 in the presence of a catalyst to give methylcyclohexane. Compound a can be formed up

V125BC [204]

Answer:

The compound a is 1-methyl cyclohexene (see attachment for structure).

Explanation:

The reaction of 1-Bromo-1-methylcyclohexane with sodium methoxide is a second-order reaction since the methoxide ion is a strong base and also a strong nucleophile. This ion attacks the alkyl halide faster than the alkyl halide can ionize to produce a first-order reaction. However, we can not see the product of nucleophilic substitution. The SN₂ mechanism is blocked due to the impediment of the 1-Bromo-1-methylcyclohexane. The main product, according to the Zaitsev rule, is the 1-methyl cyclohexene, thus forming a double bond.

Then, this cyclohexene is hydrogenated to form the cyclohexane.

6 0

3 years ago

the combustion of a sample of butane C4H10 (lighter fluid ),produced 2.16 grams of water. 2C4H10 +13O2 ------->8CO2+10H2O how

ratelena [41]

I think this is learned in chemistry do you have any notes that can help

7 0

3 years ago

HELP. Answer as much as possible. Thnx.

blondinia [14]

Answer:4

Explanation:

6 0

3 years ago

The lab just ran out of 1 M HCl that you need to complete the Benzillic Acid lab. The TA tells you there is 12 M HCl in the fume

Viktor [21]

Answer:

0.83 mL

Explanation:

Given data

Initial concentration (C₁): 12 M

Initial volume (V₁): ?

Final concentration (C₂): 1.0 M

Final volume (V₂): 10.0 mL

We can calculate the initial volume of HCl using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.0 M × 10.0 mL / 12 M

V₁ = 0.83 mL

The required volume of the initial solution is 0.83 mL.

7 0

3 years ago