Explanation:
The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.
Total pressure = 2atm
Mole Fraction = number of moles / total number of moles
Neon
Mole Fraction = 4.46 / 7.35 = 0.607
Partial Pressure = 0.607 * 2 = 1.214 atm
Argon
Mole Fraction = 0.74 / 7.35 = 0.101
Partial Pressure = 0.101 * 2 = 0.202 atm
Xenon
Mole Fraction = 2.15 / 7.35 = 0.293
Partial Pressure = 0.293 * 2 = 0.586 atm
Answer:
73.88 g/mol
Explanation:
For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:
In this case we will have:
P= 1 atm
V= 3.16 L
T = 32 ªC = 305.15 ºK
R= 0.082 
n= ?
So, we can <u>solve for "n"</u> (moles):
Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:
The mass percent of hydrogen in CH₄O is 12.5%.
<h3>What is the mass percent?</h3>
Mass percent is the mass of the element divided by the mass of the compound or solute.
- Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
- Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
- Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
Learn more about mass percent here:brainly.com/question/4336659
1 m = 0,001 km
1m³ = 0,000000001 km³
278 m³ = 0,000000278 km³ = 2,78×10^(-7) km³
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