<span>MoO2
First, lookup the atomic weights of the elements involved
Atomic weight molybdenum = 95.94
Atomic weight oxygen = 15.999
Now calculate the molar mass of Mo2O3
2 * 95.94 + 3 * 15.999 = 239.877 g/mol
Now determine how many moles of the original Mo2O3 you had
10.63 g / 239.877 g/mol = 0.044314378 mol
Determine how much oxygen was added
11.340 g - 10.63 g = 0.71 g
How many moles of oxygen was added
0.71 g / 15.999 g/mol = 0.044377774 mol
Looking at the number of moles of oxygen added and the number of moles of the original compound, they're the same. So 1 oxygen atom was added to each molecule. Since the formula was Mo2O3, the new formula becomes Mo2O4. But since you're looking for the empirical formula, you need to reduce it. Both 2 and 4 are evenly divisible by 2, so the empirical formula becomes MoO2</span>
The Ions present in CaCl₂ are,
Ca²⁺ Cl⁻ Cl⁻
Means 1 formula unit contains 1 Ca²⁺ ion and 2 Cl⁻ ions.
Also, 1 mole of CaCl₂ contains 6.022 × 10²³ formula units.
So, 1 mole formula units of CaCl₂ contain,
2 × 6.022 × 10²³ = 1.20 × 10²⁴ Cl⁻ Ions
Now, Calculating number of moles contained by 220 g of CaCl₂,
As,
110.98 g of CaCl₂ = 1 mole
Then,
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g
X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol
X = 2.38 × 10²⁴ Cl⁻ Ions
Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.
Explanation:
1) Molarity of 0.250 L HCl solution : 0.0328 M
Moles of HCl in 0.250 L solution = 0.0082 moles
2) Molarity of 0.100 L NaOH solution : 0.0245 M
Moles of NaOH in 0.100 L solution = 0.00245 moles
3) Concentration of hydrochloric acid in the resulting solution.
0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.
Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L
Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles
Molarity of HCl left un-neutralized :
0.0164 molar concentration of hydrochloric acid in the resulting solution.
Answer:
C₂H₇F₂P
Explanation:
Given parameters:
Composition by mass:
C = 24%
H = 7%
F = 38%
P = 31%
Unknown:
Empirical formula of compound;
Solution :
The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;
C H F P
% composition
by mass 24 7 38 31
Molar mass 12 1 19 31
Number of
moles 24/12 7/1 38/19 31/31
2 7 2 1
Dividing
by the
smallest 2/1 7/1 2/1 1/1
2 7 2 1
Empirical formula C₂H₇F₂P
Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature
a. It is likely that more rock candy will be formed in batch A.
b. It is likely that less rock candy will be formed in batch A.
c. It is likely that no rock candy will be formed in either batch.
d. I need more information to predict which batch is more likely to form rock candy.
Answer: Option A
Explanation:
More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.
Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.
The sugar will be dissolved in water until the time all the space is filled sugar molecules.
Hence, the correct answer is Option A.
