Answer:
125 ml of HCl
Explanation:
The molarity of the stock solution to determine how many milliliters would contain 1.5 moles of HCl. Since a concentration of 12.0 mol/L means that you get 12.0 moles of hydrochloric acid per liter of solution,
Concentration of required HCl (C1) = 1.0M
Volume of required HCL (V1) = 1500 ml
Concentration of stock HCl (C2) = 12M
Volume of stock HCL (V2) = ?
C1V1 = C2V2
V2 = C1V1/C2 = 1*1500/12 = 125 ml
Answer:
1. It dissolves much more ice faster than sodium chloride
2. Calcium chloride is more effective in melting ice at lower temperatures.
Explanation:
Salts are used to melt ice on roadways and sidewalks because they help to lower the freezing point of water.
Sodium chloride and calcium chloride are both salts used for this purpose but calcium chloride is usually preferred for the following two reasons:
1. It dissolves much more ice faster than sodium chloride: Calcium chloride dissolves much more ice faster than sodium chloride because when it dissociates, it produces three ions instead of the two produced when sodium chloride. Therefore, the heat of hydration of its ions is greater than that of sodium chloride.
2. Calcium chloride is more effective in melting ice at lower temperatures. It lowers the freezing point of water more than sodium chloride. Calcium chloride is able to lower the freezing point of water to about -52°C while sodium chloride only lowers it to about -6°C.
Answer:
distance decreases and force increases
Explanation:
- Simple machine means less power
- Hence it will require more force.
- Hence distance will decrease with respect to force and force will decrease
Answer:
Whats the question here?
Explanation:
You didn't actually ask a question
Answer:
Explanation:
mass of the solution m = 1.6 + 75 = 76.6 g
fall in temperature = 25 - 23.34 = 1.66°C
heat absorbed = mass x specific heat x fall in temperature
= 76.6 x 1.66 x 4.18
= 531.5 J .
= .5315 kJ .
mol weight of ammonium nitrate = 80 g
heat absorbed by 1.6 g = .5315 kJ
heat absorbed by 80 g or one mole = 26.575 kJ
enthalpy change ΔH = +26.575 kJ
b )
enthalpy of hydration = 2630 kJ / mol
lattice energy = enthalpy of hydration + enthalpy change
= 2630 + 26.575
= 2656.575 kJ .
