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
The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
To learn more about relative abundance with the given link
brainly.com/question/1594226
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Answer: Butane will effuse more quickly because it has a smaller molar mass
Explanation:
Molar mass of C4H10 = 58.123 g/mole
Molar mass of I2 = 253.808 g/mole
Answer:
400 mL
Explanation:
Given data:
Mass of barium = 2.17 g
Pressure = 748 mmHg (748/760 = 0.98 atm)
Temperature = 21 °C ( 273+ 21 = 294k)
Milliliters of H₂ evolved = ?
Solution:
chemical equation:
Ba + 2H₂O → Ba(OH)₂ + H₂
Number of moles of barium:
Number of moles = mass/ molar mass
Number of moles = 2.17 g / 137.327 g/mol
Number of moles = 0.016 mol
Now we will compare the moles of barium with H₂.
Ba : H₂
1 : 1
0.016 : 0.016
Milliliters of H₂:
PV = nRT
V = nRT/P
V = 0.016 mol × 0.0821 atm. mol⁻¹.k⁻¹.L×294 k/0.98 atm
V = 0.39 atm. L/0.98 atm
V = 0.4 L
L to mL
0.4 × 1000 = 400 mL