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3 years ago

What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?

Chemistry

1 answer:

Zigmanuir [339]3 years ago

6 0

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

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molar concentration

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Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

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Therefore, the final temperature of the gas will be 207.03°C

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3 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

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(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

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Ganezh [65]

Answer:

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The total pressure is 3 Atm so all you have to do is subtract the other partial pressures from 3

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