Answer:
[K₂CrO₄] → 8.1×10⁻⁵ M
Explanation:
First of all, you may know that if you dilute, molarity must decrease.
In the first solution we need to calculate the mmoles:
M = mmol/mL
mL . M = mmol
0.0027 mmol/mL . 3mL = 0.0081 mmoles
These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.
New molarity is:
0.0081 mmoles / 100mL = 8.1×10⁻⁵ M
Answer:
A)
<u>4, 7, 4, 6</u>
B)
<u>12 moles</u>
Explanation:

__↑______↑
8.00 mol | 14.00 mol
________________

You can turn this into a system of variables which are solvable.
To do this, create variables for the coefficients of each compound in the reaction respectively.

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.
a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.
(Reactant = Product)
Reactant: 1a [N] Product: 1c.
Reactant: 3a [H] Product: 2d.
Reactant: 2b [O] Product: 2c + 1d.
Thus the system is:
1a = 1c
3a = 2d
2b = 2c + 1d.
Then just use the substitution methods to solve.
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
1 mol ---- 40g
X --------- 15g
X = 15/40 = 0,375 moles
:)