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Verdich [7]
2 years ago
11

What volume of hydrogen (in l) is produced

Chemistry
1 answer:
Zielflug [23.3K]2 years ago
3 0

Answer:

50.228L of H2.

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How many hydrogen atoms are in 3 molecules of ethyl alcohol, C2H3OH
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Answer: 3 hydrogen atoms

Explanation: H=hydrogen

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Given two temperatures, 36 °F and 72 °F, which of the following correctly compares the number of opportunities for particles to
Liono4ka [1.6K]

The correct answer is "Greater at 72 °F " hope it helps

5 0
3 years ago
What should you always do at the end of a lab
Murrr4er [49]

Answer:clean up the area

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3 0
3 years ago
which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4
olasank [31]

Answer:

Answer: CH₃ and C₂H₆ have same empirical formula.

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it just compares in that it its the same

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
Sophie [7]

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.

7 0
3 years ago
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