Answer: Food production industry
Explanation:
Cooling is a process used mainly to reduce the temperature of food to either enter a different process or for storage purposes.
The temperature reached by the cooling process is usually between -1 and 8 degrees celcius.
Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8
1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))
COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm
Answer : The molarity of calcium ion on the original solution is, 0.131 M
Explanation :
The balanced chemical reaction is:

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.
First we have to calculate the moles of 

Given:
Mass of
= 0.524 g
Molar mass of
= 100 g/mol

Now we have to calculate the concentration of 

Now we have to calculate the concentration of calcium ion.
As, calcium carbonate dissociate to give calcium ion and carbonate ion.

So,
Concentration of calcium ion = Concentration of
= 0.131 M
Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M
Answer:
1188.0 mL.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁
</em>
V₁ = 900 mL, T₁ = 27.0°C + 273 = 300.0 K.
V₂ = ??? mL, T₂ = 123.0°C + 273 = 396.0 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (900 mL)(396 K)/(300.0 K) = <em>1188.0 mL.</em>
The answer is 0.975 L
Volume = mol/Molarity
We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:
Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol
So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles
Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L