The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.
Answer:
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The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M
calculation
molarity =moles of solute/ Kg of the solvent
mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol
moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles
calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol
mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg
molarity is therefore = 0.06/0.09842 = 0.6 M
<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: 50. 4g
Explanation:
First calculate number of moles of aluminium in 38.8g
Moles = 38.8g/ 26.982mol/g
= 1.44mol
By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.
4 = 2
1.4 = x
Find the value of x
x= (1.4×2)/4= 0.72 mol
0.72 moles of aluminium oxide are produced from 38.8g of aluminium
Now find the mass of aluminium produced.
Mass = moles × molar mass
= 0.72mol × 69.93 mol/g
= 50.4g
