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Assoli18 [71]
3 years ago
15

A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equa

tion
2NiS2+5O2=2NiO+4SO2
If the actual amount of SO2 made were (4.3090x10^2) g ehen the theoretical yield was (8.78x10^2) g what was the percent yield?
Enter in scientific notation
Chemistry
1 answer:
Ilya [14]3 years ago
5 0

Answer:

Percentage yield = 0.49 × 10² %

Explanation:

Given data:

Actual yield of SO₂ = 4.309 ×10² g

Theoretical yield of SO₂ = 8.78 ×10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 4.309 ×10² g  / 8.78 ×10² g  × 100

Percentage yield = 0.491 × 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 × 10² %

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Answer:

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Explanation:

You look at the type of atom and their electronegativity difference.

If ΔEN <1.6, covalent; if ΔEN >1.6, ionic

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N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.

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Explanation:

Credit goes to @znk

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