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qaws [65]
1 year ago
5

In the reaction of silver nitrate with copper metal, metallic silver comes out of solution, and the solution turns blue. This as

a ________ reaction. Group of answer choices combustion
Chemistry
1 answer:
nlexa [21]1 year ago
8 0

In the reaction of silver nitrate with copper metal, metallic silver comes out of solution, and the solution turns blue. This as a <u>single replacement</u> reaction.

<h3>What is single replacement reaction?</h3>

A single replacement reaction, also known as a single displacement reaction, occurs when one element in a molecule is swapped out for another. The starting materials are always pure elements, such as a pure zinc metal or hydrogen gas, plus an aqueous compound.

A + BC → B + AC

When A is more reactive than B or when the product AC is more stable than BC, single replacement reactions happen. A and B could either be two halogens or two metals (with hydrogen included) (C is a cation). C functions as a spectator ion when BC and AC are in aqueous solutions.

For example, 2HCl(aq)+Zn(s)→ZnCl₂(aq)+H₂(g)

Learn more about single replacement reactions here:

brainly.com/question/19068047

#SPJ4

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23) What is the volume of 2.454 x 1024 atoms of nitrogen gas at STP?
earnstyle [38]

Answer:

d) V =  91.3 L

Explanation:

Given data:

Volume of nitrogen = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Number of atoms of nitrogen = 2.454×10²⁴ atoms

Solution:

First of all we will calculate the number of moles of nitrogen by using Avogadro number.

1 mole = 6.022×10²³ atoms

2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms

0.407×10¹ mol

4.07 mol

Volume of nitrogen:

PV = nRT

1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K

V =  91.3 atm.L /1 atm

V =  91.3 L

3 0
3 years ago
Please help it’s due today
Romashka [77]
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8 0
3 years ago
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many
beks73 [17]

<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, A_0 = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}

where N is the no. of half life

Substituting the values,

\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}

\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0
3 years ago
A doctor is testing a new medication and he doesn’t tell the people in the study about the possible side effects of the medicine
balu736 [363]
Answer: He is being non-compliant
3 0
3 years ago
Calculate how many moles of nh3 form when each quantity of reactant completely reacts?
anygoal [31]

3.6 moles NH_{3} required when each quantity of reactant completely reacts.

The ratio of the mole quantities of any two compounds present in a balanced chemical reaction is known as the mole ratio. A reaction's product yield can be predicted using mole ratios, as can the amount of reactant required to produce a specific amount of result.

The balanced chemical equation is :

3N_{2} H_{4}(l) → 4NH_{3}(g) + N_{2}(g)

Given is = 2.6 mol N_{2} H_{4}

As we see, 3 moles of N_{2} H_{4} reacts with  4 moles of NH_{3}.

Therefore,

n( NH_{3}) = n(  N_{2} H_{4} ) × \frac{4}{3}

n( NH_{3}) = 2.6mol^{*} \frac{4mol}{3mol}

n( NH_{3}) = 3.6 moles.

Therefores, 3.6 moles NH_{3} required when each quantity of reactant completely reacts.

Learn more about moles here;

brainly.com/question/20486415

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6 0
2 years ago
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