The number 6.022 × 1023 indicating the number of atoms or molecules in a mole of any substance
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Answer:
a) 1.61 mol
b) Al is limiting reactant
c) HBr is in excess
Explanation:
Given data:
Moles of Al = 3.22 mol
Moles of HBr = 4.96 mol
Moles of H₂ formed = ?
What is limiting reactant =
What is excess reactant = ?
Solution:
Chemical equation:
2Al + 2HBr → 2AlBr + H₂
Now we will compare the moles:
Al : H₂
2 : 1
3.22 : 1/2×3.22 = 1.61 mol
HBr : H₂
2 : 1
4.96 : 1/2×4.96 = 2.48 mol
The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.
Moles of H₂ :
Number of moles of H₂ = 1.61 mol
Answer:
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Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.
