Answer:
The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :
<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>
Explanation:
<u>Part I :</u>
n = 4.9983
n = 4.99 moles
(Note : You can also take n = 5 mole )
Molar mass of gold = 196.96 g/mole
This means, 1 mole of gold(Au) contain = 196.96 grams
So, 4.99 moles of gold contain = 
g
4.99 moles of gold contain = 984.8 g
Mass of 
atoms of gold = 984.5 g
<u>Part II :</u>
Density of Gold = 
Volume of the cuboid = 
Volume of the gold bar =
Volume of the gold bar = 51
Using formula,
Mass = 985.32 g
So, A gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>
<u>Difference</u> <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>
Friction <span>is most responsible for slowing a bicycle down when the brakes are applied</span>
Answer:
yes I think that they are correct
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
