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n200080 [17]
2 years ago
5

Calculate the molar mass of menthol, C10H20O

Chemistry
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

C10H200

Explanation:

The molecular formula C10H20O (molar mass : 156.27 g/mol) may refer to: Citronellol. Decanal.

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3

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr
Masja [62]

Answer: 116 g of copper

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds =  4.00 hr = 4.00\times 3600s=14400s  (1hr=3600s)

Q=24.5A\times 14400s=352800C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500C=193000C  of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = \frac{63.5}{193000}\times 352800=116g of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus  remaining in solution = (0.1-0.003)=0.097moles

8 0
2 years ago
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