Question One
It's more a question of unscrambling what is there. The answer is correct.
If you look at your periodic table, there is such an element as Na which I suspect might be the problem.
It has an atomic mass of 23. It's atomic number is 11 which means it is the 11th member (by mass) on the periodic table. The other 2 are oxygen and hydrogen. So Oxygen has an atomic mass of 16 and Hydrogen has an atomic mass of 1.
Oxygen is number 8 on the periodic table and hydrogen is number 1.
Molecular Mass
Na = 23
O = 16
H = 1
Total 40, just as the answer key says. I've probably given you more than you wanted to know, but if you still have trouble, leave a note.
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
The question is: 10 g of carbonic acid H2CO3 are dissolved in 150 g of water. Determine the% m / m concentration of that solution?
Answer: The% m / m concentration of that solution is 6.66%.
Explanation:
Given: Mass of solute = 10 g
Mass of solvent = 150 g
Formula used to calculate the %m/m is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the% m / m concentration of that solution is 6.66%.
Cl-35 occurs in greater abundance.
The <em>weighted atomic mass</em> lies <em>closer to 35 u</em> than to 37 u, so the Cl-35 isotope must be more abundant.
