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3 years ago

What is the pressure of 5.0 mol nitrogen (N2) gas in a 2.0 L container at 268

Chemistry

1 answer:

CaHeK987 [17]3 years ago

8 0

Answer: 54.94atm

Explanation: Please see attachment for explanation

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How many millimeters are equal to 0.399195L

WINSTONCH [101]

399.195 millimeters=0.399195 liters

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2 years ago

1) What is the electron configuration of the element with 27 protons?

Romashka [77]

1) cobalt u can tell bc on a periodic table there is a small number that on cobalt is 27.

2) think that would be 11 bc in the 4th shell there can be up to 18 electrons

6 0

3 years ago

What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?

natita [175]

This question provides us –

Weight of $\bf KCl$ is = 47 g
Volume, V = 375 mL

__________________________________________

Molar Mass of $\bf KCl$ –

$\qquad$ $\twoheadrightarrow\bf 39.0983 \times 35.453$

$\qquad$ $\twoheadrightarrow\bf 74.5513$

<u>Using formula</u> –

$\qquad$ $\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M$

$\qquad$ $\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M}$

Henceforth, Molarity of the solution is = 1.7M

___________________________________________

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2 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago

What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?

tensa zangetsu [6.8K]

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

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3 0

3 years ago

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