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Mrac [35]
3 years ago
12

An 8.0 L sample of neon gas at 300. K exerts a pressure of 900. kPa. If the gas is compressed to 2.0 L and the temperature is ra

ised to 600. K, what will be the new pressure?
Chemistry
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

7200 kPa

Explanation:

Applying,

PV/T = P'V'/T'................ Equation 1

Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas

Make P' the subject of the equation

P' = PVT'/V'T.............. Equation 2

Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K

Substitute these values into equation 2

P' = (900×8×600)/(2×300)

P' = 7200 kPa

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Answer:

44.23 L

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Write 10,847,100 in Scientific Notation with 4 significant figures.
masha68 [24]

Answer:

The number 10,847,100 in Scientific Notation is 1.0847x10^{7}

Explanation:

Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.

5 0
3 years ago
Match each term with the correct location on a trophic pyramid.
zubka84 [21]

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6 0
2 years ago
How many total atoms are represented by this formula? Na2H3IO6<br> A 1<br> B 6<br> C 11<br> D 12
Sindrei [870]

Answer:

D 12

Explanation:

Assume that you have one formula unit of Na₂H₃IO₆. Then, you have

2 atoms Na

3 atoms   H

1 atom       I

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3 0
3 years ago
Read 2 more answers
A mixture of 50 wt% methane, 35 wt% ethane, and 15 wt% propane. Determeine the mole fraction of methane.
dmitriy555 [2]

Answer:

a) Molar fraction:

Methane: 67,5%

Ethane: 25,1%

Propane: 7,4%

b) Molecular weight of the mixture: 25,16 g/mol

Explanation:

with a basis of 100 kg:

Moles of methane:

500 g ×\frac{1mol}{16,04 g} = 31,2 moles

Mass of ethane

350 g ×\frac{1mol}{30,07 g} = 11,6 moles

Mass of propane

150 g ×\frac{1mol}{44,1 g} = 3,4 moles

Total moles: 31,2 moles + 11,6 moles + 3,4 moles = <em>46,2 moles</em>

Molar fraction of n-methane:

\frac{31,2moles}{46,2moles} =67,5%

Molar fraction of ethane:

\frac{11,6moles}{46,2moles} = 25,1%

Molar fraction of propane:

\frac{3,4moles}{46,2moles} = 7,4%

b) Average molecular weight:

0,5*16,04g/mol + 0,35*30,07g/mol+0,15*44,1g/mol} = <em>25,16g/mol</em>

I hope it helps!

5 0
3 years ago
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