Answer: 0.0450 moles of
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number of particles.
To calculate the moles, we use the equation:
Thus there are 0.0450 moles of
Answer:
Standard boiling point
Explanation:
Note that there are 2 major units of pressure except Pa .
- bar
- atm
At 1 atm pressure the boiling temperature is called normal boiling point.
At 1 bar pressure the boiling temperature is called standard boiling point
Answer:
(a) 0.2518 g; (b) 6
Explanation:
The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.
That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.
(a) Methane and ethane
In methane, the mass ratio of H:C is r₁ = 0.3357:1.00.
In ethane, the mass ratio of H:C is r₂
The ratio of the ratios r = ⁴/₃
r = r₁/r₂
⁴/₃ = 0.3357/r₂
(⁴/₃)r₂ = 0.3357
r₂ = 0.3357 × ¾ = 0.2518:1
Ethane contains 0.2518 g of H for every 1 g of C.
(b) Xenon fluorides
In XeF₂, the mass ratio of F:Xe is r₁ = 0.2894:1
In XeFₙ, the mass ratio of F:Xe is r₂ = 0.8682:1
r = r₂/r₁ = 0.8682/0.2894 = 3.000:1 ≈ 3:1
XeFₙ contains three times as many F atoms as XeF₂.
n = 6
Fomented means encouraged/excited state.
Therefore, our answer is encouraged.
So there are three formulas that you must be aware of hen attempting these questions (by technicality, only two because the third is a transposition of the second).
These formulas are :
mole x L = # of atoms (where L = Avogadro's Constant)
mass / Mr = mole (where Mr = Molar Mass)
Mole x Mr = mass
So no we can easily ace these questions
1) a. 98.3 g of Hg
mass of Hg = 98.3g
Mr " " = 201 g / mol
mole =
= 0.489 mol
∴ # of atoms = 0.489 mol * (6.03 * 10²³)
= 2.95 * 10²³ atoms
c. 10.7g of Li
mass of Li = 10.7 g
Mr " " = 7 g / mol
thus mole =
= 1.529 mol
# of atoms = 1.529 mol * (6.03 * 10²³)
= 9.22 * 10²³
2) a. 6.84 g of C₁₂H₂₂O₁₁
mass of sucrose = 6.84 g
Mr " " = ∑ (Mr of each element * amount of each element's atom)
= ∑ [(12 * 12)+(1 * 12)+(16 * 11)] g / mol
= 342 g / mol
∴ mole of sucrose =
= 0.02 mol
c) 68.0g of NH<span>₃</span>
<span> mass of NH₃ = 68.0 g</span>
<span> Mr " " = [(14 * 1) + ( 1 * 3)] g / mol</span>
<span> ∴ mole of NH₃ = </span>
= 4 mol
3) a. 3.52 mol of Si
mole of Si = 3.52 mol
Mr " " = 28 g / mol
Mass = 3.52 mol * 28 g / mol
= 98.56 g
c. 0.550 mol of F₂
mol of F₂ = 0.55 mol
Mr " " = (19 * 2 ) g / mol
Mass = 0.55 mol * 38 g / mol
= 20.9 g
Hope I as clear in my answers and that they helped