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Westkost [7]
3 years ago
5

How many atoms are in 5.5 moles of carbon dioxide

Chemistry
2 answers:
Veronika [31]3 years ago
4 0

Answer: I think the answer is, 3.312177825e+24 atoms

Explanation: I had a problem similar to this, Hope this helps!

Rasek [7]3 years ago
4 0

Answer:

So if you have 5 mole, you have: 5 x (6.022 x 10^23) = 3.011 x 10^24 atoms.

Explanation:

Basically the answer is 44.0095 :)) have a great day!!

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Which is an example of flow of heat through conduction
JulsSmile [24]

Answer:

a metal spoon left in boiling water

Explanation:

5 0
3 years ago
How many grams of NH3 can be produced from 12.0g of H2?
RSB [31]

Answer:

Balanced reaction:

3 H2 (g)  + N2 (g)  → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3.  The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0
2 years ago
The second stage of photosynthesis produces _____ and _____.
lukranit [14]
The second stage of photosynthesis also called Calvin stage produces glucose

5 0
3 years ago
Read 2 more answers
How is alkyne converted to acetone?
cricket20 [7]

Answer:

Propyne can be converted to the acetone when it is made to undergo the reaction with mercuric sulphate followed by hydrolysis and thus, resultant product, thus formed is acetone.

Explanation:

3 0
3 years ago
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
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