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VashaNatasha [74]

3 years ago

9

A mixture of coarse sand and sugar is 45.0 percent sand by mass. 120.0 grams (g) of the mixture is placed in a fine-mesh cloth b

ag and dunked repeatedly in Lake Michigan. After drying, the mass of the contents of the bag equals: ________.
A. 66.0 g
B. 120.0 g
C. 65.0 g
D. 72.00 g
E. 54.0 g

1 answer:

8090 [49]3 years ago

8 0

Answer:

Option E

Explanation:

From the question we are told that:

Amount of sand in percentage $s_p=45%$

Sample size $n=120g$

Note:After being dumped in the river repeatedly the sugar melts away leaving behind the insoluble sand

Generally the equation for Amount of sand content is mathematically given by

$X=n*s_p$

$X=120*\frac{45}{100}$

$X=54g$

Therefore

After drying, the mass of the contents of the bag equals

$X=54g$

Option E

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This is an incomplete question, here is a complete question.

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Explanation :

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This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the $HN_3$ and the coefficient '3' put before the $N_2$ then we get the balanced chemical equation.

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$\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25$

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$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 6.0 atm

$X_i$ = mole fraction of gas

$p_{N_2}=X_{N_2}\times p_T$

$p_{N_2}=0.75\times 6.0atm=4.5atm$

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$p_{H_2}=X_{H_2}\times p_T$

$p_{H_2}=0.25\times 6.0atm=1.5atm$

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