Explanation:
Heat is a form of thermal energy.
Heat is the sum of all the energy of the molecular motion in an object.
Temperature measures the average heat possessed by each molecule in a given substance.
Molecules at a higher temperature possess more kinetic energy and they will move faster. This kinetic energy form is the heat variant of thermal energy.
Temperature is the measure of this heat energy of molecules.
Answer:
The molality of the glycerol solution is 2.960×10^-2 mol/kg
Explanation:
Number of moles of glycerol = Molarity × volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles
Mass of water = density × volume = 0.9982 g/mL × 998.7 mL = 996.90 g = 996.90/1000 = 0.9969 kg
Molality = number of moles of glycerol/mass of water in kg = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg
Answer:
B?
Explanation:
In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles. Use the mass of the hydrogen gas to calculate the gas moles directly; divide the hydrogen weight by its molar mass of 2 g/mole. For example, 250 grams (g) of the hydrogen gas corresponds to 250 g / 2 g/mole = 125 moles.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
<u>Answer:</u> The mass of calcium chloride present in given amount of solution is 87.5 g
<u>Explanation:</u>
We are given:
Mass of solution = 277.8 grams
Also, 31.5 % (m/m) of calcium chloride in water. This means that 31.5 g of calcium chloride is present in 100 g of solution.
To calculate the mass of calcium chloride in the given amount of solution, we use unitary method:
in 100 g of solution, the mass of calcium chloride present is 31.5 g
So, 277.8 g of solution, the mass of calcium chloride present is
Hence, the mass of calcium chloride present in given amount of solution is 87.5 g