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shtirl [24]
3 years ago
13

Vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvoops wrong

Chemistry
2 answers:
Andrei [34K]3 years ago
8 0

Answer:

do you need help? feel free to ask

shepuryov [24]3 years ago
7 0
Hi, what’s your question?
You might be interested in
Help plz ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!
ddd [48]
Honestly I don’t even know
7 0
3 years ago
2. Some nitrogen at a pressure of 35.75 p.s.i is in a 100 L container. If the container's volume is reduced to 2250 ml then what
elena-s [515]

Answer:

1455.6

Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values

35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6

4 0
3 years ago
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
Which of the following models a synthesis reaction?
Nuetrik [128]

Answer: D

Explanation: i’m right

5 0
3 years ago
How would one convert 6 pounds to ounces (16 oz = 1b)?
NNADVOKAT [17]

Answer:

D.

Explanation:

I just got it right in the test :)

6 0
3 years ago
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