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Mamont248 [21]
3 years ago
15

Can methene exist? Explain yes or no

Chemistry
1 answer:
finlep [7]3 years ago
7 0
Methene (non existent) would be an alkene and therefore would need a carbon to carbon double bond, since it only has one carbon it can't bond to another, this is why the lowest alkene is ethene (C2H4).         i hope this helps  
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Round off each number to the indicated number of significant figures (sf)
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Answer:

The answer is B

Explanation:

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Which statements are true of solutions?
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Answer: they contain solutes and solvents, their particles must be evenly distributed, they may contain solid liquid and gas simultaneously, and they are homogeneous matter.

Explanation:

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Balance the following H2 SO4 + NaOH ==== NaSO + H2O​
katrin [286]

Answer:

H2SO4 + 2NaOH --->   Na2SO4 + 2H2O

Explanation:

H2SO4 + NaOH --->   Na2SO4 + H2O

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3 years ago
What is the oxidation number for as in as2o3?
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8 0
3 years ago
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You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0
3 years ago
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