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worty [1.4K]
3 years ago
13

How many molecules are in 67.0 grams of potassium chloride KCl

Chemistry
1 answer:
valentinak56 [21]3 years ago
8 0
67g KCl(1 mol KCl/74.55g)=0.8987256875 mol KCl
0.8987256875 mol KCl(6.02*10^23 molecules/ 1 mol)
=5.410328638*10^23 molecules KCl
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"Chilling" is most commonly practiced by:
zimovet [89]
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.
4 0
3 years ago
The Kp for the reaction below is 1.49 × 108 at 100.0°C:
Anna007 [38]
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
5 0
3 years ago
If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t
tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3

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 CaCO_3

\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}

Given:

Mass of CaCO_3 = 0.524 g

Molar mass of CaCO_3 = 100 g/mol

\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol

Now we have to calculate the concentration of CaCO_3

\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}

So,

Concentration of calcium ion = Concentration of CaCO_3 = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0
3 years ago
A gas occupies 900.0 ML at temperatures of 27.0 Celsius. What is the volume at 123.0 celsius
Artemon [7]

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>

4 0
3 years ago
Read 2 more answers
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Law Incorporation [45]
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
5 0
3 years ago
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