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vfiekz [6]
3 years ago
14

Each equation is incorrect. find the errors, then rewrite and balance each equation. cl2 + nai → nacl2 + i nh3 → n + h2 na + o2

→ nao2
Chemistry
1 answer:
VikaD [51]3 years ago
6 0
1) Balanced chemical reaction: Cl₂ + 2NaI → 2NaCl + I₂.
Chlorine and iodine are diatomic molecules.
2) Balanced chemical reaction: 2NH₃ → N₂ + 3H₂.
Nitrogen and hydrogen are diatomic molecules.
3) Balanced chemical reaction: 4Na + O₂ → 2Na₂O.
Sodium in compounds has oxidation number +1 and oxygen -2.
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What is the molecular formula of a compound containing 89% cesium (Cs) and 11% oxygen (O) with a molar mass = 298 g/mol?
Maurinko [17]
M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11

CsxOy

x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2

y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2

Cs₂O₂  cesium peroxide
6 0
3 years ago
Read 2 more answers
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 <u>half-reactions</u>:

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

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

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

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 <u>flip</u> the reduction potential value:

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

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

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
What is the main difference between extensive and intensive physical properties?
mr_godi [17]
Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present.
8 0
3 years ago
Read 2 more answers
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
2 years ago
1/4 checkpoint
DochEvi [55]

Answer:

here's the answers

Explanation:

7 0
2 years ago
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