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saveliy_v [14]
3 years ago
6

5.54 moles of AgNO3 are dissolved in 0.48 L of water. What is the molarity of the AgNO, solution?

Chemistry
2 answers:
expeople1 [14]3 years ago
7 0

Answer:

11.54 M

Explanation:

In this case, all we have to do is to apply the following expression:

M = n/V

M: molarity

n: moles

V: volume of solution in liters

In this case, we can assume that the volume of water will be the volume of solution. This is because the problem is not specifing if the moles of AgNO3 are liquid or solid, so we can make a safe assumption of the volume.

Using the expression above we have:

M = 5.54 / 0.48

M = 11.54 M

This is the molarity of solution

VashaNatasha [74]3 years ago
4 0

Answer:

the molarity of the AgNO₃ is 11.5417 mol/L

Explanation:

Given:

nAgNO₃ = 5.54 moles

V = 0.48 L

Question: What is the molarity of the AgNO₃, solution, M = ?

To calculate the molarity, the number of moles must be divided by the volume of the solution.

M=\frac{n}{V} =\frac{5.54}{0.48} =11.5417mol/L

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Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

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b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

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(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

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