The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.
Answer:
0.607mol
Explanation:
n(AR) = mass / molar máss
= 24.3 /40
=0.607
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
The correct answer is d!!
Answer:
677.39 g/mol
Explanation:
Step 1: Find molar masses of elements
Carbon (C) - 12.01 g/mol
Hydrogen (H) - 1.01 g/mol
Oxygen (O) - 16.00 g/mol
Step 2: Multiply the amount present
12.01 g/mol · 46 = 552.46 g/mol
1.01 g/mol · 92 = 92.92 g/mol
16.00 g/mol · 2 = 32.00 g/mol
Step 3: Add up all the molar masses to find compound molar mass
552.46 g/mol + 92.92 g/mol + 32.00 g/mol = 677.39 g/mol
