<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is
So, in 15.6 g of iron (III) oxide, mass of iron present will be =
Hence, the mass of iron in the ore is 10.9 g
Answer:
A. because the wood is stored chemical energy and it is converting into heat energy
Answer:
0.86 g
Explanation:
Given:
Pressure = 725 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 725 / 760 atm = 0.9539 atm
Volume = 225 mL = 0.225 L (1 mL = 0.001 L)
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25 + 273.15) K = 298.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9539 atm × 0.255 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.0099 moles
From the reaction,
1 mole of chlorine is formed from 1 mole of
Thus, moles of = 0.0099 moles
Molar mass of = 86.9368 g/mol
Mass = Moles*Molar mass = 0.0099 moles * 86.9368 g/mol = 0.86 g
Answer: 3
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