A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
1 answer:
Answer:- C. Hafnium.
Solution:- Mass of the sample is 46.0 g and it's volume is .
From mass and volume, we can calculate it's density using the formula:
On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.
The right choice is C) Hafnium.
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Answer:
ΔH = - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from 298.00 K to 302.16 K.
Then ;
Δ T = 302.16 K - 298.00 K
Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH = 56.576 kJ
The equation of the given reaction can be represented as :
Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ
For 1 mole of heat liberated now:
ΔH = 56.576 kJ/0.028 mol
ΔH = 2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH = - 2020.57 kJ/mol
Answer:
28.9%
Explanation:
Let's consider the following balanced equation.
2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂
We can establish the following relations:
- The molar mass of Fe₂O₃ is 159.6 g/mol
- 1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
- 1 mole of Fe is in 1 mole of FeS₂
- The molar mass of Fe is 55.84 g/mol
The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:
The percent of Fe in 1.25 g of the ore is: