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3 years ago

How could you use these solutions to determine the identities of each metal powder?

Chemistry

1 answer:

Anastaziya [24]3 years ago

5 0

Answer:

1.) The nitric acid solution will oxidize and thus dissolve _*(Zn and Pb)*_. This will allow to identify _**Pt**_.

2) To distinguish between _*(Zn and Pb)*_, we can use the nickel nitrate.

3) The nickel nitrate solution will oxidize and thus dissolve _**Zn**_ and will not oxidize or dissolve _**Pb**_.

Explanation:

1) Unlike Zinc and Lead, Platinum does not react with Nitric acid. So, it will be the only metal from step 1 that doesn't react. Pt is identified in this manner.

2) Nickel is higher than Lead in the activity series, but Zinc is higher than both of them in the activity series. This selectivity can be used to distinguish between Zinc and Lead metal powders.

3) Because Zinc is higher than Nickel in the activity series, it means that Zinc metal can and will displace Nickel from Nickel Nitrate solution. Therefore the Nickel Nitrate solution will oxidize and dissolve the Zinc metal.

But, there will be no reaction with the Lead metal powders sample as Pb is lower than Ni in the activity series, so, Nickel Nitrate solution will not oxidize or dissolve the Lead metal powders.

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Does anyone know how to make this a longer sentence?

Nadusha1986 [10]

Answer:

First, you have to analyze your problem or question. After you research and collect data about your topic, create a hypothesis to test to try and find the answer. After testing your hypothesis, come up with a conclusion based on the results.

7 0

3 years ago

Read 2 more answers

3.95 g of sugar (C6H12O6) is dissolved in water to make 158 mL of solution. Find the molarity.

12345 [234]

Answer:

[C₆H₁₂O₆] = 0.139 M

Explanation:

Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.

We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.

For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)

(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL

We determine moles, we convert them to mmoles, we divide by mL

M = 0.139 M

Moles = 3.95 g . 1mol / 180g → 0.0219 mol

We convert mL to L → 158 mL . 1L/1000mL = 0.158L

M = 0.0219 mol / 0.158L = 0.139 M

8 0

3 years ago

Instead of 6 M NaOH being added to the solution, 6 M HCl is added. How will this affect the test for the presence of ammonium io

Lera25 [3.4K]

Answer:

Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution

Explanation:

As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.

When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.

This gas turn red litmus paper blue.

This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).

If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.

3 0

3 years ago

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Darya [45]

Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

k = rate constant = $0.610M^{-1}s^{-1}$

$a_0$ = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

$\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}$

$t=3.120s$

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

8 0

3 years ago

MULTIPLE CHOICE

Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0

3 years ago