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Grace [21]
3 years ago
14

An unknown gas with a formula of HX is contained in a 25.0 L tank at 1.2 atm and 25.0 oC. What is the chemical formula of the ga

s if the mass of the gas is 44.7 g?
Chemistry
1 answer:
ozzi3 years ago
8 0

Answer:

HCl

Explanation:

Volume = 25L

Temperature = 25.0 oC + 273 = 298 K (Converting to kelvin units)

Pressure = 1.2 atm

Mass of gas = 44.7g

Formular of gas = HX

To solve this, we have to know the identity of X. One qay to do that is by obtaining the molecular mass of the compound.

To get the molar mass, we need the number of mole.

Using the ideal gas equation;

PV = nRT

where R = gas constant = 0.082 057 L atm K−1  mol−1

n = PV / RT

n = 1.2 * 25  / 0.082 057 * 298

n = 30  /  24.45

n = 1.227 mol

The relationship between number of moles and molar mass is given as;

Nummber f moles = mass  / molar mass

Molar mass = mass  / Number of moles

Molar mass = 44.7 / 1.227

Molar mass =  36.43 g/mol

Molar mass of gas X = Molar mass of H + Molar mass of X

36.43 = 1 + X

X = 36.43 - 1 = 35.43

Chlorine is the only element with molar mass approximate to 35.43

Hence X = Cl

The chemical formular = HCl

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What is the molarity of a solution made by dissolving 8.60 g of a solid with a
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Answer:

1.43 M

Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

Mass of solid = 8.60 g

Molar mass of solid = 21.50 g/mol

Mole of solid =?

Mole = mass / molar mass

Mole of solid = 8.60 / 21.50

Mole of solid = 0.4 mole

Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

280 mL = 280 mL × 1 L / 1000 mL

280 mL = 0.28 L

Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

Mole of solid = 0.4 mole

Volume = 0.28 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.4 / 0.28

Molarity = 1.43 M

Thus, the molarity of the solution is 1.43 M.

8 0
3 years ago
What is it called when evaporation takes place beneath the surface of a liquid
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5 0
4 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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