To solve this problem we will use the following equation:
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
<span>w = mass CaCl2/(mass of water + mass of CaCl2)
</span>
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".
This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate
Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters
Answer: 
for this reaction at this temperature is 0.029
Explanation:
Moles of 
= 2.00 mole
Volume of solution = 4.00 L
Initial concentration of 
The given balanced equilibrium reaction is,
Initial conc. 0.500 M 0 M 0 M
At eqm. conc. (0.500-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5Ctimes%20%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D)
Equilibrium concentration of ![[Br_2]](https://tex.z-dn.net/?f=%5BBr_2%5D)
= x = 0.0955 M
Now put all the given values in this expression, we get :
Thus for this reaction at this temperature is 0.029
Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
Answer:
-154KJ/mol
Explanation:
mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter
= 0.2 × 100 / 1000 ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.
Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K
where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK
heat gain by calorimeter = (4.184 × 200 + 480) × 2.34 = 3081.3 J
ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol
