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pantera1 [17]
3 years ago
5

In the experiment, you added water to the reaction vessel after the reaction was complete, but we did not discuss why. Based on

what happened after adding the water, which of the following is the reason for the addition of water?
A. lower the solubility of the product in solution so it precipitated out
B. increase the solubility of the product in solution so it precipitated out
C. lower the solubility of the product in solution so it stayed in solution
D. increase the solubility of the product in solution so it stayed in solution
Chemistry
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

A.

Explanation:

Water was added to the reaction after the completion of the reaction so as to lower the solubility if the product in the solution therefore, the product can be precipitated out. On adding water the reaction moves in forward direction and more product is formed. (By Le Chatelier's principle). Thus, the precipitation occurs. Hence, option A is correct.

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zepelin [54]

Answer:

B?

Explanation:

In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles. Use the mass of the hydrogen gas to calculate the gas moles directly; divide the hydrogen weight by its molar mass of 2 g/mole. For example, 250 grams (g) of the hydrogen gas corresponds to 250 g / 2 g/mole = 125 moles.

7 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
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Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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3 years ago
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sasho [114]

Answer:a leech

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Solution attached.

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3 years ago
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