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Feliz [49]
3 years ago
7

Can anyone help me with this question?? No links please!

Chemistry
1 answer:
xeze [42]3 years ago
6 0

Answer: 6

Explanation: 2*3=6

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Norma-Jean [14]

Answer:

hi dont report me we are friends

Explanation:

6 0
3 years ago
CER sentence starter
jonny [76]

Answer:

CER all starts with a question asked by the teacher. This question is based on a phenomena or lab experience. The student's explanation or answer, as you may have guessed, will consist of three parts: a claim, the evidence, and the student's reasoning. A claim is a statement that answers the question. Start with a hook or attention getting sentence. Briefly summarize the texts • State your claim. Make sure you are restating the prompt. Include a topic sentence that restates your claim and your reason.

3 0
2 years ago
Draw the structure of the following atoms<br> (1) 19/9 F<br> (2)28/14 SI
Svet_ta [14]

Answer:

might help

Explanation:

8 0
3 years ago
Consider the decomposition of the compound C5H6O3 as follows: C5H6O3(g)  C2H6(g) + 3CO(g) A 5.63 g sample of pure C5H6O3(g) was
deff fn [24]

Answer:

K_{eq}=1.02x10^{-4}

Explanation:

Hello,

In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:

5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3

After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:

n_{total}^{eq}=\frac{P_{total}^{eq}V}{RT}=\frac{1.63atm*2.50L}{0.082\frac{atm*L}{mol*K}*473.15K} =0.105mol

Then, by knowing the moles at the equilibrium considering the change "x", which yields to:

\ \ \ \ \  C_5H_6O_3(g) \leftrightarrow  C_2H_6(g) + 3CO(g)\\I\ \ \ \ \ 0.0494mol\ \ \ \ \ \ 0mol \ \ \ \ \ \ \ \ 0mol\\C\ \ \ \ \ \ -x\ \ \ \ \ \ \ \ \ \ \  \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x\\E\ \ 0.0494mol-x\ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ 3x

The total moles at the equilibrium turn out:

n_{total}^{eq}=0.0494mol-x+x+3x

By solving for "x", we've got:

3x=0.105mol-0.0494mol\\x=\frac{0.0556mol}{3}\\x=0.0185mol

Finally, the equilibrium constant is:

K_{eq}=\frac{(x)(3x)^3}{0.0494-x}=\frac{(0.0185mol)(3*0.0185mol)^3}{0.0494mol-0.0185mol}=1.02x10^{-4}

Best regards.

4 0
4 years ago
In the chemical reaction: 2 h 2 (g) o2 (g) → 2 h2o (g), with 8 grams of h 2 (2 grams/mole) and 16 grams of o2 (32 grams/mole) an
Shtirlitz [24]
Use stoichiometric ratio to determine which is in excess and which is limiting

8 0
3 years ago
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