Answer:
Explanation:
We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.
The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.
Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.
To convert to grams, find the molar mass of Ag from the periodic table.
Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.
(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)
Hands off, it's mine.
3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
Answer:
[H⁺] = 3.16 × 10⁻⁵ mol/L
Explanation:
Given data:
pH of solution = 4.5
Hydrogen ion concentration = ?
Solution;
pH = -log [H⁺]
we will rearrange this formula:
[H⁺] = 10∧-pH
[H⁺] = 10⁻⁴°⁵
[H⁺] = 3.16 × 10⁻⁵ mol/L
The liters in 3.25 g of ammonia 4.28 L
<u><em>calculation</em></u>
Step 1: find moles of ammonia
moles = mass÷ molar mass
From periodic table the molar mass of ammonia (NH₃) = 14 +(1×3 ) = 17 g/mol
3.25 g÷ 17 g/mol = 0.191 moles
Step 2: find the number of liters of ammonia
that is at STP 1 moles = 22.4 L
0.191 moles = ? L
<em>by cross multiplication</em>
={( 0.191 moles ×22.4 L) / 1 mole} = 4.28 L
D. 1-butyne.
The name of this molecule is 1-butyne.
