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andreev551 [17]
2 years ago
10

How many moles are their in 62.0 grams of CO2

Chemistry
2 answers:
vladimir2022 [97]2 years ago
7 0
I attached my cross multiplication work below. Please let me now if you are still confused about how to get this answer. 

Answer: about 1.40877 mol CO_{2} 

Advocard [28]2 years ago
6 0
6.2 grams of CO2 = 1.408786739226764 moles
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PLEASE HELP ME!!! ASAP
shtirl [24]

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0
3 years ago
Can someone help me with number 1 and 2 plz!
GalinKa [24]

Answer:

1) 0 N

2) 8 N

Explanation:

The net force is the sum of all of the forces acting on the object.

For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left.  If we define the right to be positive and the left to be negative, then the net force equals:

Fnet = 5N - 5N = 0 N

Therefore, the net force in question 1 is 0 N.

For question 2, the process is very similar.  We want to find the sum of the forces acting on the object.  In this case, there are forces of 3 N and 5 N acting to the right.

Fnet = 3 N + 5 N = 8 N

Therefore, the net force in question 2 is 8 N.

Hope this helps!

3 0
2 years ago
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