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katrin [286]
4 years ago
15

What is the density of a mineral with a mass of 41.2 g and a volume 8.2 cm3?

Chemistry
2 answers:
Tanya [424]4 years ago
7 0

Answer : The density of a mineral is, 5.024 g/ml

Solution : Given,

Mass of mineral = 41.2 g

Volume of mineral = 8.2cm^3=8.2ml      (1cm^3=1ml)

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get the density of mineral.

Density=\frac{41.2g}{8.2ml}=5.02g/ml

Therefore, the density of a mineral is, 5.024 g/ml

dezoksy [38]4 years ago
4 0
Hey there!:

mass = 41.2 g

Volume = 8.2 cm³

Therefore:

D = m / V

D = 41.2 / 8.2

D = 5.02 g/cm³
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Chemical Formula:

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Explanation:

• Phosphorus trichloride (PCl3) contains three chlorine atoms and one phosphorus atoms.

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Which would most easily form negative one ions.
irinina [24]

Answer:

Cl

Explanation:

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8 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}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

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

To know more about neutralization visit:

brainly.com/question/12498769

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4 0
2 years ago
4.Calculate the volume of 40g of Helium (He) at rtp.
Oksana_A [137]

Answer:

2.24dm³

Explanation:

Given parameters:

Mass of He = 40g

Unknown:

Volume of Helium  = ?

Solution:

To solve this problem, we convert the given mass to number of moles.

   Number of moles  = \frac{mass}{molar mass}  

   molar mass of He  = 4g/mol

  Number of moles  = \frac{4}{40}   = 0.1mole

So;

                  1 mole of gas at rtp occupies a volume of 22.4dm³

                0.1 mole of He will occupy a volume of 0.1 x 22.4 = 2.24dm³

4 0
3 years ago
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