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ZanzabumX [31]
3 years ago
11

What is the mass of 0.50 moles of Na? 11 grams 46 grams 3.0 × 1023 grams 6.0 × 1023 grams

Chemistry
2 answers:
Korolek [52]3 years ago
7 0

n= mass ÷ molar mass

mass= n × molar mass

=0.50 × 23

=11.5 grams

Nikitich [7]3 years ago
7 0

Answer : The mass of 0.50 moles of Na is, 11 grams

Explanation : Given,

Moles of Na = 0.50 mole

Molar mass of Na = 22.99 g/mole

Formula used :

n=\frac{w}{M}

where,

n = number of moles of Na

w = mass of Na

M = molar mass of Na

Now put all the given values in this formula, we get the mass of Na.

0.50mole=\frac{w}{22.99g/mole}

w=11.4g\approx 11g

Therefore, the mass of 0.50 moles of Na is, 11 grams

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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

Best regards.

3 0
3 years ago
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I believe it's answer #3. Logically, at least.

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The elements present in Ammonium Nitrate are Hydrogen, Nitrogen, and Oxygen at a ratio of 4:2:3, respectively. Hydrogen weighs in at 1.008 amu, Nitrogen at 14.007, and Oxygen at 15.999. This means that the molar mass would be:

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