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4 years ago

How did I get the sand out the mixture ?

Chemistry

2 answers:

svetoff [14.1K]4 years ago

4 0

Answer:

They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.

Oliga [24]4 years ago

3 0

Answer:

Sifter

Explanation: They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.

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What is the net charge of the ionic compound calcium fluoride

yKpoI14uk [10]

Answer:

The net charge of the ionic compound calcium fluoride is zero (0).

Explanation:

Ionic compounds, such as covalent ones, have zero net charge; this is, they are neutral.

Substances with net positive charge are cations and substances with net negative charge are anions.

The charges in the ionic compound calcium flouride are distributed in this way:

Compound formula: CaF₂

Calcium charge: Ca²⁺: this is, each calcium ion has a 2 positive charge

Fluoride charge: F⁻: each fluoride ion has a 1 negative charge.

Then, the net charge is: 1 × (2+) + 2 × (1-) = +2 - 2 = 0.

So, a two positve charge, from one calcium ion, is equal to two negative charges, from two fluoride tions, yielding a zero net charge.

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3 years ago

What does the symbol "E3" represent?

Anna71 [15]

Answer:

your answer to your question is B

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3 years ago

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A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A

Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

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3 years ago

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Charra [1.4K]

C. all the other choices are positive impacts to the environment unlike C

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3 years ago

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If a 1.50 L gas sample is held at a constant temperature, and its pressure is changed from 2.30 atm to 3.60 atm, what will the f

likoan [24]

Answer:

the new temperature is 11.49pa

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3 years ago

How did I get the sand out the mixture ￼?

How did I get the sand out the mixture ?