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3 years ago

Help me please!!!!!!

Chemistry

1 answer:

Bingel [31]3 years ago

7 0

Answer:

CH₄ + 2O₂ –> CO₂ + 2H₂O

The coefficients are: 1, 2, 1, 2

Explanation:

CH₄ + O₂ –> CO₂ + H₂O

The above equation can be balance by as follow:

CH₄ + O₂ –> CO₂ + H₂O

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by writing 2 before H₂O as shown below:

CH₄ + O₂ –> CO₂ + 2H₂O

There are 2 atoms of O on the left side and a total of 4 atoms on the right side. It can be balance by writing 2 before O₂ as shown below:

CH₄ + 2O₂ –> CO₂ + 2H₂O

Now, the equation is balanced.

The coefficients are: 1, 2, 1, 2

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Chemical Reactions

irakobra [83]

Answer:

It works by separating the fuel from the oxygen. The oxygen comes from the air. It is the same oxygen we breathe. Since the oxygen has to be in contact with the fuel, if you can coat the fuel with something that keeps the oxygen away, the fire will go out which is what the fire extinguisher does.

6 0

2 years ago

Give 3 example of a solid that sublimes.

tino4ka555 [31]

Here are a few examples :)

iodine (I2)

naphthalene

aresenic (As)

ferrocene

water (H2O)

carbon dioxide (CO2)

Hope this helps :)

5 0

3 years ago

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

Becker did an experiment. Separately, Tracy did the exact same experiment. Becker and Tracy found they each got very different r

maxonik [38]

Answer:

I don't know

Explanation:

Maybe they shouldn't copy each other

5 0

3 years ago

Read 2 more answers