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2 years ago

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How many moles of nitrogen gas would be produced if 4.92 moles of copper(II) oxide were reacted with excess ammonia in the follo

wing chemical reaction? 2 NH₃(g) + 3 CuO (s) → 3 Cu(s) + N₂(g) + 3 H₂O(g)\

1 answer:

WITCHER [35]2 years ago

3 0

Answer:

4.92/3

Explanation:

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Help me!!!<br>will give the brainliest!!<br>plz answer correctly<br>Urgent!!!

Jet001 [13]

1.a) All except copper can react with dilute sulphuric acid.
REACTION OF DILUTE SULPHURIC ACID WITH COPPER (II) CARBONATE
CuCO3 (s) + H2S04 (aq)———> CuSO4(aq) +CO2(g)+ H2O (l)

REACTION OF COPPER (II) OXIDE WITH DILUTE SULPHURIC ACID
CuO (s) + H2SO4 (aq)———> CuSO4 (aq) + 2H2O (l)

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3 years ago

Read 2 more answers

A certain reaction is endothermic in the forward direction. The reaction has less moles of gas on the product side. Which of the

xxMikexx [17]

It would have to increase pressure... but I don’t see that option here..?

6 0

3 years ago

A sample of aspirin weighing 5.945 g became contaminated with 2.134 g of sodium sulfate. The resulting

ira [324]

Answer:

The correct answer to the following question will be "62.9 %".

Explanation:

The given values are:

The aspirin's initial amount = 5.945 g.

and is polluted containing 2,134 g of sodium sulfate.

After extraction we provided 3,739 g of pure aspirin.

Now,

$Percentage \ of \ recovery \ aspirin =\frac{Aspirin \ amount \ isolated \ after \ extraction}{initial \ amount \ of \ aspirin}\times 100$

On putting the values in the above formula, we get

⇒ $=\frac{3.739}{5.945}\times 100$

⇒ $=62.9 \ percent$

Note: percent = %

7 0

3 years ago

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

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3 years ago

I need help solving this problem !!

Tatiana [17]

Answer:

A:Boyle's Law or B:Charles's Law

Explanation:

5 0

4 years ago