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Iteru [2.4K]
3 years ago
9

A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa a

nd the new volume is 55.0 L. What was the initial pressure of the sample? Show your work, using the G.U.E.S.S. method.
Chemistry
1 answer:
Damm [24]3 years ago
4 0

Answer:

71.92 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

T1 = 50°C = 50 + 273 = 323K

V1 = 105L

T2 = -25°C = -25 + 273 = 248K

P2 = 105.4 kPa

P1 = ?

V2 = 55.0 L

Using P1V1/T1 = P2V2/T2

P1 × 105/323 = 105.4 × 55/248

105P1/323 = 5797/248

0.325P1 = 23.375

P1 = 23.375 ÷ 0.325

P1 = 71.92 kPa

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3 years ago
Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p
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1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g 
molar mass of <span>N2O4 = 92.02 g/mol
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</span>mass of N2H4:45.0 g

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</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
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3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
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moles= \frac{grams}{molar mass}             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

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