Answer:
4.67M
Explanation:
The concentration of methanol (CH3OH) can be calculated using the following:
Molarity (M) = number of moles(n)/volume(v)
However, mole is not given. It can be obtained by using:
Mole = mass / molar mass
Where; mass = 34.4g
Molar mass (MM) of CH3OH is:
= 12 + 1(3) + 16 + 1
= 12 + 3 + 17
= 32g/mol
mole = 34.4/32
mole = 1.075mol
The volume needs to be converted to L by dividing by 1000
230mL = 230/1000
= 0.230L
Molarity = mol/volume
Molarity = 1.075/0.230
Molarity = 4.6739
Molarity = 4.67M
The concentration of CH3OH in solution is 4.67M
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Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
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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
