6. What are two uses for nonmetals?

1 answer:

5 0

Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.

Nonmetals used in fertilizers: Fertilizers contain nitrogen.

Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.

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Which of the following will form a solution? (Select all that apply.)

SSSSS [86.1K]

Answer:

sand and water

Explanation:

when u mix these both it will become like a paste

4 0

2 years ago

One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in

Dennis_Churaev [7]

Answer:

48.32 g of anhydrous MnSO4.

Explanation:

Equation of dehydration reaction:

MnSO4 •4H2O --> MnSO4 + 4H2O

Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)

= 223 g/mol

Mass of MnSO4 • 4H2O = 71.6 g

Number of moles = mass/molar mass

= 71.6/223

= 0.32 mol.

By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

Number of moles of MnSO4 = 0.32 mol.

Molar mass = 55 + 32 + (4*16)

= 151 g/mol.

Mass = 151 * 0.32

= 48.32 g of anhydrous MnSO4.

3 0

3 years ago

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$