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katrin2010 [14]
3 years ago
10

what is the realtionship between the kentic energy of molecules in an object and the objects tempeture

Chemistry
1 answer:
MrMuchimi3 years ago
6 0
Temperature is a measure of the average kinetic energy of the particles in an object .Temperatures also measure how kinetic energy is not how hot or cold it is. It’s measuring what the amount of Kinetic Energy there when you throw something in the air and it comes back down
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How much concentrated solution would it take to prepare 2.75 L of 0.400 M HCl upon dilution with water
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Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20g/mL. Calculate the molarity of the concd HCl.
1.20 g/mL x 1000 mL x 0.37 x (1/36.5) = about 12 M or so but you do it exactly.

Then mL x M = mL x M
mL x 12 M = 2800 mL x 0.475
Solve for mL of the concd HCl solution.
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3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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