Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
= 
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
Learn more about calculation at STP here:
brainly.com/question/9509278
#SPJ4
Answer:
yes it can λ =265 nm
Explanation:
Here we will use the relationship
E = h c/λ ∴ λ = E/ hc where
h= Plank's constant
c= Speed of light
λ = Wavelength = ?
Substituting
note need E in J ,
E = 4.7 eV x 1.602 x 10⁻¹⁹ J/eV = 7.5 x 10⁻¹⁹ J)
λ = 7.5 x 10 ⁻¹⁹ J / ( 6.626 x 10⁻³⁴ Js x 3 x 10^8) = 2.65 x 10⁻⁷ m = 2.65
= 2.65 x 10⁻⁷ m x 1 x 10⁹ nm/m = 265 nm
In viewing a chemical equation,we <span>determine the ratios that relate reactants and products </span>from the coefficients of the reactants and products .
Answer:
222.2 grams of CaCl2 in 2 moles
Explanation:
Mole=given mass/gram mass formula
2moles=x/Gram Mass Formula
(Gram Mass Formula of CaCl2):
Ca=40.1
Cl= 35.5 x 2= 71
-----------------------
GFM=111.1
2 moles = x/111.1 g
x=222.2
