Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
V₂ = 4.0L
Explanation:
Decreasing temperature => Decreasing Volume (Charles Law)
For a given volume, use a temperature ratio that will give a smaller volume.
Volume at lower temp = 4.6L(70K/82K) = 4.0L ... Using (82K/70K) would give a larger volume => contrary to temperature effects on gas volumes when pressure and mass are kept constant.
Pressure effects on Gas Volumes:
Note: The same idea is applied to pressure effects on gas volumes also except that changes in pressure affect gas volumes indirectly. That is, an increase in pressure => decrease in volume, or a decrease in pressure => increase in volume. Boyles Law => V ∝ 1/P.
Given a gas volume of 4.60L at 760mmHg, what is volume at 848mmHg?
Increasing pressure => Decreases Volume (Boyles Law)
For the given volume use a pressure ratio that will give a smaller volume.
Volume at higher pressure = 4.6L(760mm/848mm) =4.1L. Using (848mm/760mm) would give a larger volume => contrary to pressure effects on gas volume when temperature and mass of gas are kept constant.
