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
"Increase Pressure " is the right answer. if you need help , let me know
At higher temperature, and lower pressure.
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
