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taurus [48]
3 years ago
14

given a density of 350 grams/cm3 and a mass of 25 grams. What is the volume using the equation involving density?​

Chemistry
1 answer:
Eduardwww [97]3 years ago
6 0

0.07cm³

Explanation:

Given parameters:

Density of the body = 350g/cm³

Mass of the body = 25g

Unknown:

Volume of the body = ?

Solution:

Density is the amount of substance contained in a volume of body. It is expressed as:

    Density = \frac{mass}{volume}

Since the unknown is volume, we make it the subject of the expression;

   Volume = \frac{mass}{density}

 Volume = \frac{25}{350} = 0.07cm³

learn more:

Density brainly.com/question/2690299

#learnwithBrainly

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2 years ago
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

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

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