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rodikova [14]
3 years ago
5

What is the chemical formula for the compound formed between chromium(iii) and the sulfate ion

Chemistry
2 answers:
Anastasy [175]3 years ago
7 0
Cr2(SO4)3 is the formula
zlopas [31]3 years ago
7 0

<u>Answer:</u> The chemical formula of the given compound is Cr_2(SO_4)_3

<u>Explanation:</u>

The combination of chromium and sulfate ions leads to the formation of an ionic compound having chemical name as chromium (III) sulfate. The electrons get transferred from chromium atom to sulfate ion.

Chromium is the 24th element of the periodic table. It is a metal and it looses its electrons. The electronic configuration for this element is 1s^22s^22p^63s^23p^64s^23d^{4}

This element will loose 3 electrons and forms Cr^{3+} ion.

Sulfate ion is a polyatomic ion having chemical formula of SO_4^{2-}

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for chromium (III) sulfate is Cr_2(SO_4)_3

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Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0
3 years ago
Hhheppppp fffaaast
umka21 [38]

Answer:

False

Explanation:

Lead sulfate is insoluble in water and sinks in water

(sorry if it's wrong)

4 0
2 years ago
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