Answer:
The correct option is;
24 moles
Explanation:
Here, we have the reaction as follows;
Sn(s) + 2HF(g) → SnF₂ (s) + H₂ (g)
Therefore, one mole of Sn reacts with 2 moles HF to form one mole of SnF₂ and one mole of H₂
Molar mass of H₂ = 2.01588 g/mol
Therefore, the number of moles of H₂ in 48 grams of H₂ is given by the relation;
Since one mole each of SnF₂ and H₂ are produced, the number of moles of SnF₂ produced = 24 moles.
The number of moles of SnF₂ that will be produced is 24 moles.
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
In 1st orbit 2
2nd 8
3rd 10
f orbital has 16
Answer:
- <em>The maximum amount of copper allowed in 100 g of water is </em><u><em>0.00013 g</em></u>
Explanation:
To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg) and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):
First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.
Now, set the proportion:
- 1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O
Solve for x:
- x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu
Convert mg to grams:
- 0.13 mg × 1 g / 1,000 mg = 0.00013 g
Answer: 0.00013 g of copper.
