Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
A. easily combines chemically with other substances is your answer.
Answer: The average atomic mass of the element = 88.242amu
Explanation:
The abundance of the first isotope is =35.5%
Atomic mass of first isotope = 68.9257
The average atomic mass of the first isotope =86.95amu X 35.5% =86.95amu X 0.355 =30.8725 amu
The abundance of the second isotope =64.5%
Atomic mass of the second isotope =88.95amu
The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu
Now the average atomic mass =30.8725 +57.37275 = 88.242amu
OR using the formulae
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100
8,824/100
=88.24amu