The products will be magnesium phosphate and potassium chloride. You then have to watch a solubility chart to see which one of these is not soluable. In this case it is magnesium phosphate.
Answer:
<h2>0.15 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities.
From the question we have
We have the final answer as
<h3>0.15 moles</h3>
Hope this helps you
D. Chemical and Physical Properties
Correct question
The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)
Answer:
Explanation:
Given that;-
The density = 13.6 g/mL
Also, 1 kg = 2.205 lb
1 kg = 1000 g
So, 1000 g = 2.205 lb
1 g = 0.002205 lb
Also,
1 in = 2.54 cm
1 in³ = 16.39 cm³
1 cm³ = 1 mL
So, 1 in³ = 16.39 mL
1 mL = 0.061 in³
The expression for the calculation of density is shown below as:-
Thus,
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
