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seraphim [82]
2 years ago
7

If a student calculated 0.03256 moles of copper metal (Cu0), how many grams of copper metal (Cu0) should be recovered at the end

of the experiment?
Chemistry
1 answer:
Liula [17]2 years ago
7 0

Answer:

Mass of copper metal will be 2.06756 gram

Explanation:

We have given number of moles of copper n = 0.03256

Molar mass of copper = 63.5 u

We have to calculate mass of copper metals in gram

We know that number of moles is given by number\ of\ moles=\frac{mass\ in\ gram}{molar\ mass}

0.03256=\frac{mass\ in\ gram}{63.5}

Mass in gram = 2.06756 gram

So mass of copper metal will be 2.06756 gram

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Identify which statement describes the following process. 2Ce + 3Cu2+ → 3Cu + 2Ce3+
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Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.

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Which of the following items is a salt? (1 point)
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Learn more at ; brainly.com/question/14091731?referrer=searchResults

5 0
2 years ago
Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

3 0
3 years ago
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