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:
178.35g
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles ÷ volume
Based on the information provided in this question, molarity (M) of the solution = 1.50 M, volume = 725 mL = 725/1000 = 0.725L, n = ?
1.50 = n / 0.725
n = 1.50 × 0.725
n = 1.0875mol
Molar mass of Na3PO4
23(3) + 31 + 16(4)
= 69 + 31 + 64
= 164g/mol
Mole = mass ÷ molar mass
1.0875 = mass/164
mass = 178.35g
Answer: Enthalpy of combustion (per mole) of is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
We are given:
Putting values in above equation, we get:
Enthalpy of combustion (per mole) of is -2657.5 kJ
There are 100 degrees between the freezing (0°) and boiling points (100°) of water on the Celsius scale and 180 degrees between the similar points (32° and 212°) on the Fahrenheit scale.
Problem: Two scientists are doing an experiment designed to identify the boiling point
Answer: 250°F is the higher temperature by 2°F