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Gnoma [55]
3 years ago
6

PLEASE HELP!!!!!!!!!!! WILL AWARD 50 POINTS!!!!!!!!!111

Chemistry
2 answers:
Novay_Z [31]3 years ago
7 0
  • B

(Might be wrong).................

Vinil7 [7]3 years ago
3 0

I am pretty sure the answer is . But I might be wrong.

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Another word for oxidation is<br> Decaying<br> Rusting<br> Cutting<br> Burning
anyanavicka [17]
I believe that the answer is A. Decaying
7 0
3 years ago
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What is the chemical formula of iron (II) sulfide would be: (EO S = +2 and EO Fe = +2). please
Anni [7]

Answer:

FeS

Explanation:

The reason why their is no subscripts is because as put in the question, the charges are equivalent. As you can see, +2 and -2 equal each other out.

Note: S has a charge of -2, not +2. (Look at oxidation rules.)

7 0
4 years ago
Explain why flourine forms a negative ion more easily than chlorine.​
Svetllana [295]

Answer:

the electron is outermost

Explanation:

shell of a flourine are closer together .

5 0
3 years ago
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Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2​
lorasvet [3.4K]

Al

Explanation:

The limiting reactant will be Al:

        4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

     Given parameters:

  Mass of Al = 30g  Molar mass = 27g/mol

         Number of moles = \frac{mass}{molar mass} =  \frac{30}{27}

         Number of moles of Al = 1.111 mole

  Mass of O₂ = 30g, molar mass = 32g/mol

   Number of moles =  \frac{30}{32} = 0.94mol

In the reaction:

          4 moles of Al reacted with 3 moles of O₂

         1.11moles of Al will require \frac{1.11 x3}{4} = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0
3 years ago
How many grams of na2co3 would be needed to produce 1000g of nahco3
Ivenika [448]

Answer:

630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃

Explanation:

The balanced reaction is:

Na₂CO₃ + CO₂+ H₂O → 2 NaHCO₃

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 moles
  • CO₂: 1 mole
  • H₂O: 1 mole
  • NaHCO₃: 2 moles

Being the molar mass:

  • Na₂CO₃: 106 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole
  • NaHCO₃: 84 g/mole

Then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • CO₂: 1 mole* 44 g/mole= 44 g
  • H₂O: 1 mole* 18 g/mole= 18 g
  • NaHCO₃: 2 moles* 84 g/mole= 168 g

You can apply the following rule of three: if 106 grams of Na₂CO₃ are needed to produce 168 grams of NaHCO₃, how much mass of Na₂CO₃ is necessary to produce 1000 grams of NaHCO₃?

mass of Na_{2} CO_{3}=\frac{1000grams ofNaHCO_{3} *106gramsofNa_{2} CO_{3} }{168grams ofNaHCO_{3}}

mass of Na₂CO₃= 630.95 grams

<u><em>630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃</em></u>

6 0
3 years ago
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