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Zarrin [17]
3 years ago
14

You could add HCl(aq) to the solution to precipitate out AgCl(s). What volume of a 0.190 M HCl(aq) solution is needed to precipi

tate the silver ions from 12.0 mL of a 0.160 M AgNO3 solution? Express your answer with the appropriate units.
Chemistry
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer: 10.1mL

Explanation:

To solve this problem, let's generate the equation:

HCl + AgNO3 —> AgCl + HNO3

Next, we find the amount of AgNO3 in 12mL of the solution.

12mL = 0.012L

If 0.16mol of AgNO3 dissolves in 1L of solution,

Therefore Xmol will dissolve in 0.012L i.e

Xmol of AgNO3 = 0.16x0.012 = 0.00192mol

From the equation, 1mole of AgNO3 required 1mole HCl.

This implies that 0.00192mol of AgNO3 will also require 0.00192mol of HCl.

From this result, we can calculate the Volume of HCl that is needed for the reaction. This is done by:

Molarity = mole /Volume

Molarity = 0.19M

Mole = 0.00192mol

Volume =?

Volume = mole / Molarity

Volume = 0.00192 / 0.19

Volume = 0.0101L = 10.1mL

Therefore the volume of HCl needed is 10.1mL

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Why does the moon and the earth have the same mass
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Moon rocks contain few volatile substances (e.g. water), which implies extra baking of the lunar surface relative to that of Earth. The relative abundance of oxygen isotopes on Earth and on the Moon are identical, which suggests that the Earth and Moon formed at the same distance from the Sun.

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2 years ago
Read 2 more answers
Need help !!!!! ASAP
Julli [10]
<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

V_{2}=1L

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

P_{1}V_{1}=P_{2}V_{2}

So, we are given the information:

V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa

Then, isolating the new volume and substituting into the equation, we have:

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\frac{P_{1}V_{1}}{P_{2}}

V_{2}=\frac{50kPa*2L}{100kPa}=1L

Hence, the new volume will be 1 L.

V_{2}=1L

Have a nice day!

4 0
3 years ago
Find the mass of 5.82 mol of MgCl2. Round to the nearest tenth.
Oksana_A [137]

Answer:

554.13 grams

Explanation:

8 0
3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
2 years ago
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