Answer:- 
Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.
To calculate the heat absorbed or released we use the formula:
q = heat absorbed or released
m = mass of solution
s = specific heat capacity
and 
= change in temperature
mass of solution = mass of solute + mass of solvent
mass of solution = 5.00 g + 100.0 g = 105.0 g
(note:- density of pure water is 1 g per mL so the mass is same as its volume)
= 33.0 - 23.0 = 10.0 degree C
s = 
Let's plug in the values in the formula and calculate q.
q = 
q = 4389 J
To calculate the enthalpy of solution that is 
we convert q to kJ and divide by the moles of solute.
moles of solute = 
= 0.118 moles
q = 
= 4.389 kJ
= 
Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.
So,
Assume <span>the following: </span><span>Mass defect = 3.2008 x 10-27 kilograms</span><span>
</span><span>Use E = mc2, with c = 3 x 108 m/s
</span>
The nuclear binding energy for uranium-238 in joules is 2.8807 x 10-8 joules.
Answer:
0.012mol
Explanation:
1.
Given parameters:
Pressure in the balloon = 1.25atm
Volume of the balloon = 2.5L
Unknown:
Number of moles = ?
Solution:
We are going to used the combined gas law to solve this problem.
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
Insert the parameters and solve;
1.25 x 2.5 = n x 0.082 x 285
n = 0.13mol
The number of moles is 0.13mol
2.
Given parameters:
Pressure = 16torr;
to atm is ;
760torr = 1atm
16torr = 
= 0.02atm
Volume = 12L
Temperature = 253K
Using the ideal gas law;
PV = nRT
0.02x12 = n x 0.082 x 253
n = 0.012mol
Answer:
Some signs of a chemical change are a change in color and the formation of bubbles. The five conditions of chemical change: color change, formation of a precipitate, formation of a gas, odor change, temperature change.
We first can get the No of moles of H2S = 0.25 x 1 mol(H2S) / 34 g (H2S)
= 0.0074 Moles
and according to the ideal gas low:
we can get p for H2S
PV = nRT
when we have V = 5 L & n = 0.0074 & R (constant)= 0.082 & T= 25 + 273 = 298 K
By substitution:
P* (5L) = (0.0074)*(0.082)*(298)
∴ P = 0.036 atm
By assuming Kp (should be given, you just missed it) = 0.12 at 25 C°
By substitution: to get P for NH3
0.12 = X ( P + X)
0.12 = X ( 0.036 + X)
∴X^2 + 0.036 X -0.12 = 0
by solving this equation we get
X= 0.365 atm
So to get the no of moles of NH3:
PV = nRT
0.365 * 5 = n ( 0.082*298)
∴ n = 0.075 moles
and to get the mass on (g) =no.of moles * molar mass
0.075 * 51 = 3.825 g NH4Hs
