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Karo-lina-s [1.5K]
2 years ago
8

What is the mass of 2.14 mol CaCl2?

Chemistry
1 answer:
balu736 [363]2 years ago
5 0

Answer:

237.5 grams CaCl2

Explanation:

Use the periodic table to calculate the mass of CaCl2

40.078+(35.45*2)=110.97800

Convert: 2.14 mol CaCl2 * 110.98g CaCl2/1 mol CaCl2 = 237.4972 g

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Chromium is electroplated on other metals. 2. Gold is electroplated on other cheap metals.

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10. The Proton is special! What does it tell us?<br> The proton tells me..
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These two numbers are fixed for an element. The mass number tells us the number the sum of nucleons of protons and neutrons in the nucleus of an atom. The atomic number also known as the proton number is the number of protons found in the nucleus of an atom. ... The atomic number uniquely identifies a chemical element.

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What is the solution of Ag + Cu2+ --&gt; ?
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2 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
2 years ago
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