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monitta
4 years ago
6

During a lab activity, students prepared two solutions separately at the same temperature. Later they mixed the solutions and th

e temperature dropped. Why?
A:there was no energy input during the chemical reaction
B:energy was absorbed during the chemical reaction
C:energy was released during the chemical reaction
Chemistry
1 answer:
Yuri [45]4 years ago
6 0
The temperature dropped because B. energy was absorbed during the chemical reaction.
If energy was released, the temperature would rise. If there was no energy input, the temperature would stay the same. Since temperature dropped, it means that energy was absorbed.
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Nitrogen and oxygen are the most prevalent in the atmosphere.
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1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
How many grams of carbon dioxide are produced from the combustion of 250.G of ethane, C3H8 in the reaction C3H8 +5O2 - 3CO2 +4H2
ludmilkaskok [199]

Answer:

750 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

C₃H₈ + 5O₂ —> 3CO₂ + 4H₂O

Next, we shall determine the mass of C₃H₈ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C₃H₈ = (12×3) + (8×1)

= 36 + 8 = 44 g/mol

Mass of C₃H₈ from the balanced equation = 1 × 44 = 44 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ = 3 × 44 = 132 g

SUMMARY:

From the balanced equation above,

44 g of C₃H₈ reacted to produce 132 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 250 g of C₃H₈. This can be obtained as follow:

From the balanced equation above,

44 g of C₃H₈ reacted to produce 132 g of CO₂.

Therefore, 250 g of C₃H₈ will react to produce = (250 × 132)/44 = 750 g of CO₂.

Thus, 750 g of CO₂ were obtained from the reaction.

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