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weqwewe [10]
3 years ago
10

4. You have a very concentrated solution (12 M) of potassium chloride (KCI). You need it to be at the least concentration possib

le for the experiment you are
about to conduct. The problem is you forgot to order lab supplies, so you only have 2 L of distilled water left. What would be the final concentration if you
added the two 2 L of distilled water to the 0.5 L of 12 M KCI?
a 3.0 M
b 24 M
C 2.4M
d 48 M
Chemistry
1 answer:
Alenkinab [10]3 years ago
4 0

Answer:

C. 2.4 M

Explanation:

Because you started with 12M solution of KCl, that means that there was a concentration of 12 mols of KCl per Liter of solution. (12mol/1L =12 M). Since there was only 0.5L of solution, there was only 6 mols of KCl because there is only 12 mols per 1 Liter, so half of that volume would have half the amount of solute to keep that true. (6mol/.5L = 12 M). With the new water added, the formula becomes (6mols KCl/ 2.5L of water =2.4 M)

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5. Calcium-40 and Calcium-42 both share the same
gayaneshka [121]
They share the same number or protons
6 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane is
DanielleElmas [232]

Answer:

6g

Explanation:

Step 1:

The balanced equation for the reaction between gaseous ethane and gaseous oxygen. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Step 2:

Determination of the masses of C2H6 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol

Mass of C2H6 from the balanced equation = 2 x 30 = 60g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 7 x 32 = 224g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 4 x 44 = 176g

From the balanced equation above,

60g of C2H6 reacted with 224g of O2 to produce 176g of CO2.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting reactant as it will be needed to obtain the maximum mass of CO2.

From the balanced equation above,

60g of C2H6 reacted with 224g of O2.

Therefore, 2.71g of C2H6 will react with = (2.71 x 224)/60 = 10.12g of O2.

From the above calculation, we can see that a higher mass of O2 is needed to react with 2.71g of C2H6, therefore, O2 is the limiting reactant.

Step 4:

Determination of the maximum mass of CO2 produced when 2.71 g of ethane is mixed with 7.6 g of oxygen.

The limiting reactant is used to determine the maximum mass.

From the balanced equation above,

224g of O2 produce 176g of CO2.

Therefore, 7.6g of O2 will produce = (7.6 x 176)/224 = 5.97g ≈ 6g of CO2

From the calculations made above, the maximum mass of CO2 produced is 5.97 ≈ 6g

5 0
3 years ago
How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
lisov135 [29]
Moles of glucose = Molarity x volume solution 
                              = 4.5 x 1.5
                              = 6.75 moles.

Hope this helps, have a great day ahead!
3 0
3 years ago
Which of the following is not part of the proper protocol for using acids and bases?
Anni [7]

Answer:

B.Add acid to water,not water to acid

Explanation:

they should not be mixed

8 0
3 years ago
Two or more assistance
Goryan [66]
I believe it would be a compound.
4 0
3 years ago
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