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lina2011 [118]
2 years ago
7

Ca(s) + O₂(g) + S(s) → CaSO4(s)

Chemistry
1 answer:
san4es73 [151]2 years ago
5 0

The mass of Calcium required to complete this reaction is 4.008 g.

  • Law of conservation of mass states that In a closed system, mass cannot be produced or destroyed, but it can be changed from one form to another.
  • The mass of the chemical constituents before a chemical reaction is equal to the mass of the constituents after the reaction.
  • In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.

In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.

So,

mass of Ca + mass of O₂ + mass of S = mass of CaSO4

Ca + 6.400 g + 3.206 g = 13.614 g

mass of Ca = 13.614 - 9.606 = 4.008 g

Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.

Learn more about mass conservation here:

brainly.com/question/2030891

#SPJ9

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How many valence electrons are in cbr2cohnh2
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Answer:

36 valence electrons

Explanation:

Given CBr₂COHNH₂ => Br₂C = C - O - H

                                                    |

                                             H - N - H

#Valence e⁻s  = 2Br + 2C + 3H + 1N + 1O = 2(7) + 2(4) + 3(1) + 1(5) + 1(6)

    = 14 + 8 + 3 + 5 + 6 = 36 valence electrons

Addendum ...

#Bonded e⁻s =  2Br + 2C + 3H + 1N + 1O = 2(8) + 2(8) + 3(2) + 1(8) + 1(8)

   = 16 + 16 + 6 + 8 + 8 = 54 bonded electrons

#Covalent Bonds = #Valence e⁻ - #Bonded e⁻ / 2 = (54 - 36) / 2 = 9 cov. bonds.

8 0
3 years ago
CHEMISTRY HELP (50 PTS!!!!!!) picture attached
vodomira [7]

Answer:

<h2>Na^+ Cl^- > NaCl</h2>

4 0
3 years ago
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How many molecules are 4.3 x 10^27 molecules of N2O5<br>​
horrorfan [7]

Answer:

<h2>7142.86 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4.3 \times  {10}^{27} }{6.02 \times  {10}^{23} }  \\  = 7142.857...

We have the final answer as

<h3>7142.86 moles</h3>

Hope this helps you

6 0
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Murrr4er [49]
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4 years ago
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Answer:

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Explanation:

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