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3 years ago

Step 1: H2(g) + → ICI(9) HI(g) + HCl(g) (slow)

Chemistry

1 answer:

sineoko [7]3 years ago

6 0

Answer:

Explanation:

Step 1 occurs at a slower rate than Step 2.

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3 years ago

Find density of nitrogen dioxide at 75*C and 0.805 atm.

Eva8 [605]

Answer:

1 (348) (D2) = 273 (2.05) (0.805) D2= 1.29 g/L

Explanation:

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4 years ago

112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.

dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

$\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol$

For the given chemical reaction:

$2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)$

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = $\frac{12}{2}\times 0.778=4.668 mol$ of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

$\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol$

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

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3 years ago

Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha

nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

$35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L$

The moles of air are:

$3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol$

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

$1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J$

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

$5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}$

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3 years ago

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Explanation:

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3 years ago

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