Substances have different tendencies to donate or accept electrons. When a really good donor meets a great acceptor, the chemical reaction releases a lot of energy. Oxygen (O2) is the best electron acceptor and is used in many aerobic reactions (reactions with oxygen). Hydrogen gas (H2) is a good electron donor.
When O2 and H2 are combined, along with a catalyst, water (H2O) is formed. This example of a redox reaction can be written like this:

To answer the question above, multiply the given number of moles by the molar masses.
(A) (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B) (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C) (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D) (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O
The answer to the question above is letter D.
The answer is 492.8 g
1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample.
1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole
6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n
n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol
2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:
Ar(C) = 12 g/mol
Ar(H) = 1 g/mol
Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol
3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:
m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g
Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
Pentaarsenic decafluoride
Penta=5
Arsenic=As
Deca=10
Fluoride=F
Drop the -ine and add -ide
