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borishaifa [10]
2 years ago
15

CH4 + 202 → CO2 + 2H2O

Chemistry
1 answer:
ohaa [14]2 years ago
6 0
It’s a combustion reaction but what are you trying to find?
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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
PLEASE HELP
AfilCa [17]
I think it’s salt water
7 0
3 years ago
Read 2 more answers
In this picture below one object is attractive to another by-
Ne4ueva [31]
I think it'd be gravity. 
5 0
3 years ago
Read 2 more answers
If you have 400 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days? To help
alisha [4.7K]

Answer:

25 grams

Explanation:

You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.

3 0
3 years ago
Use the data in the table above to calculate an approximate ΔH∘ in kilojoules for the synthesis of hydrazine from ammonia:
ankoles [38]
DeltaH formation = deltaH of broken bonds - deltaH of formed bonds

Broken bonds: tiple bond N-N and H-H bond
Formed bonds: N-H and N-N bonds

You also have to take note of the molar coefficients

deltaH formation = <span> [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)] 
                           = (945 + 2*436) - (4*390 + 240)
                            = 17 kJ/mol

The answer is 17 kJ/mol.</span>
8 0
3 years ago
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