Temperature and Pressure One way to increase the solubility of a gas is to decrease the temperature of the liquid. The solubility of a gas in a liquid is usually temperature dependent, although it depends on the particular combination of which gas and which liquid. Usually the solubility of a gas goes down with increasing temperature (think of warm carbonated beverages going flat).
<span>The other way to increase the solubility is to increase the pressure of the gas. The higher the pressure of the gas above the liquid, the more will dissolve. Again, think of a carbonated beverage: when it is sealed it doesn't go flat because it is under pressure, but when open to air, it will go flat. </span>
The mass of 1.72 mol of magnesium fluoride is 107 grams.
To determine the mass of 1.72 mol of magnesium fluoride, we first need the chemical formula of magnesium fluoride. Magnesium forms a +2 ion (Mg+2) and fluoride forms a -1 ion (F-1). Since all compounds formed from ions have to be electrically neutral, we need 2 fluoride ions and 1 magnesium ion. Therefore, the formula for magnesium fluoride is MgF2.
Now we need to determine the molar mass of the compound from the molar mass values from the periodic table. Let's use a table to calculate this molar mass.
Molar mass of MgF2
Element Molar Mass (g/mol) Quantity Total (g/mol)
Mg 24.31 1 24.31
F 19.00 2 38.00
Total molar mass of MgF2 = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol
This is the mass of one mole of the substance. If we have 1.72 mols of it, we multiply 1.72 by 62.31.
1.72 mol (62.31 g/mol) = 107 grams
We rounded to 107 to keep the correct number of significant digits in our answer.
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
