<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
Always. You never know what kind of chemical you’re dealing with and how powerful it is.
Answer:
No
Explanation:
One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.
When one gram phosphorus and 6 gram of iodine react they gives 8.234 g
ram of PI₃ .
Given data:
Mass of phosphorus = 1 g
Mass of iodine = 6 g
Mass of PI₃ = ?
Solution:
Chemical equation:
P₄ + 6I₂ → 4PI₃
Number of moles of P₄:
Number of moles = Mass /molar mass
Number of mole = 1 g / 123.9 g/mol
Number of moles = 0.01 mol
Number of moles of I₂:
Number of moles = Mass /molar mass
Number of moles = 6 g / 253.8 g/mol
Number of moles = 0.024 mol
Now we will compare the moles of PI₃ with I₂ and P₄.
I₂ : PI₃
6 : 4
0.024 :
4/6×0.024 = 0.02
P₄ : PI₃
1 : 4
0.01 : 4 × 0.01 = 0.04 mol
The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.
Mass of PI₃ = moles × molar mass
Mass of PI₃ = 0.02 mol × 411.7 g/mol
Mass of PI₃ = 8.234 g
Since you know the ratio of atoms, you can start to put a formula togeter. The formula might look like:<span>
X<span>H2.67
</span></span>but since atoms can't come in fractional amounts, we have to multiply the formula by some number in order to turn 2.67 into a whole #, while still maintaining the ratio. Multiplying 2.67 by 3 yields 8, so the most likely ratio in the molecule is
X3H8<span>so the ratio of 1:2.67 is still maintained. The mass percent tells you that out of every 100g of compound, 91.26g is element X, so the other 8.74g must be H. Dividing each mass by the number of moles in the formula gets us the molar mass of each element (approximately). DIviding 8.74g by 8 gets 1.09, roughly the molar mass of hydrogen. Dividing 91.26g by 3 gets us 30.4, roughly the molar mass of phosphorus. Element X is most likely phosphorus</span>
Iron(II) oxide also refers to a family of related non-stoichiometric compounds, which are typically iron deficient with compositions ranging from Fe0.84O to Fe0.95O.
