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gogolik [260]
3 years ago
13

A sample of nitrogen gas had a volume of 500. mL, a pressure in its closed container of 740 torr, and a temperature of 25 °C. Wh

at was the new volume of the gas when the temperature was changed to 50 °C and the new pressure was 760 torr, if the amount of gas does not change?
A) 530 mL
B) 450 ml
C) 970 mL
D) 240 ml
E) 400 mL
Chemistry
1 answer:
iren2701 [21]3 years ago
6 0

Answer:  Thus the new volume of the gas is 530  ml

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas =  740 torr

P_2 = final pressure of gas = 760 torr

V_1 = initial volume of gas = 500 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 25^oC=273+25=298K

T_2 = final temperature of gas = 50^oC=273+50=323K

Now put all the given values in the above equation, we get:

\frac{740\times 500}{298}=\frac{760\times V_2}{323}

V_2=530ml

Thus the new volume of the gas is 530 ml

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Density of iron is 7.8 g/cc. How much would 10 cc of iron weigh?
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Answer:

<h3>The answer is 78 g</h3>

Explanation:

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<h3>mass = Density × volume</h3>

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Hope this helps you

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Read 2 more answers
Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

(12*1)+(16*2)=44~g/mol

In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

8 0
3 years ago
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