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luda_lava [24]
3 years ago
5

How many grams of calcium hydride are needed to form 4.550 g of hydrogen gas

Chemistry
1 answer:
stealth61 [152]3 years ago
4 0
Molar mass 

CaH₂ = 42.0 g/mol

H₂ = 2.0 g/mol

<span>Balanced chemical equation :

</span>CaH2 + 2 H₂O = Ca(OH)₂ + 2 H₂

42.0 g CaH₂ ---------------------  4.0 g H2
    ? g CaH₂ ---------------------- 4.550 g H2

mass = 4.550 * 42.0 / 4

mass = 191,1 / 4

mass =<span> 47.775 g of CaH₂</span>

<span>hope this helps!</span>
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Answer:

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Looking at the nuclear reaction at hand, the total mass on the left hand side has to be 246 while the total charge on the left hand side of the equation will be 98. Coming over to the right hand side, the total masses are also found to be the same as a neutron with no charge and a mass of 1 is produced. Hence the nuclear reaction equation is balanced when an alpha particle (helium nucleus) is substituted for X in the equation.

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What coefficient is assumed if no coefficient <br> written before a formula in a chemical equation
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Answer:

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For which blocks of elements are outer electrons the same as valence electrons? For which are d electrons often included among v
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3 years ago
Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+
DIA [1.3K]

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
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  • X = +1 + 4
  • X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

6 0
3 years ago
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