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RideAnS [48]
3 years ago
6

A football field is about 100 m long. If it takes a person 20 seconds to run its length, how fast were they running ?

Chemistry
1 answer:
madam [21]3 years ago
3 0
Hi Alex, thanks for asking a question here on Brainly!

We can figure how the fast the person is running by dividing the minutes by the seconds. 

\frac{5m}{sec}
= \frac{300m}{min}
= \frac{18km}{hr}

The person was running 18 km/hr. ✅ 

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or post another question and send the link to me. -UnicornFudge aka Nadia.
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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
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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

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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.

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I think this is correct? I hope it helps.        

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