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gogolik [260]
3 years ago
15

What is the mass of 3.2×10²³ molecules of CH4?

Chemistry
1 answer:
Nikitich [7]3 years ago
8 0

Answer:

Mass? Molar mass? Grams? Or Mols?

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Which component is required for the activation of some enzymes?
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The best answer is letter A. 
The main factors that change the speed of enzymatic reactions are temperature, pH and substrate concentration (quantity). <span>There are enzymes that need other associated molecules to work. These molecules are called enzyme co-factors. They can be organic ions like mineral salts.</span>
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A mixture of C3H8 and C2H2 has a mass of 2.8 g. It is burned in excess O2 to form a mixture of water and carbon dioxide that con
astraxan [27]

Answer:

The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.

Explanation:

The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.

From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.

The individual balanced equation of reaction is:

C3H3 +5O2 >> 3CO2 + 4H2O

and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.

The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.

4 0
3 years ago
What organelle is thin and flexible
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Cell Membrane

The thin, flexible outer covering of a cell. It controls what enters and leaves a cell.

Explanation:

7 0
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3) Which of the following transformations creates a figure that is similar (but not congruent) to the
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For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

4 0
4 years ago
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