Answer:
molarity = moles of solution/liters of solution
molarity = 1 mole/2 liters
molarity = 0.5 M
molarity = 50 moles/200 kg
molarity = 0.25 M
Explanation:
Answer:
Gas pressure is caused when the gas particles hit the walls of their container. The more often the particles hit the walls, and the faster they are moving when they do this, the higher the pressure. This is why the pressure in a tyre or balloon goes up when more air is pumped in
Explanation:
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answer:
2.75 mol
Explanation:
Given data:
Mass of Nitrogen = 38.5 g
Moles of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 38.5 g/ 28 g/mol
Number of moles = 1.375 mol
Now we will compare the moles of ammonia and nitrogen from balance chemical equation.
N₂ : NH₃
1 : 2
1.375 : 2×1.375 = 2.75 mol
Thus 2.75 moles of ammonia are produced from 38.5 g of nitrogen.
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g