8) 21.504 liters of gas
9) 122.5 g
10) 6.022*10^23
Explanation:
8)
1 mole= 22.4 liters
to calculate the volume of gas of 0.960 moles of CH4
22.4*0.960/1
= 21.504 liters of gas
9)
2.0 mole*18.02 g/ 1 mole
=122.5 g
10)
2.0 l * 1 mole/22.4 l
6.022*10^23
Answer:
- <em>The molar mass of an element is the mass of </em><u>one mole of atoms of the element.</u>
Explanation:
<em>The molar mass of an element </em>is its atomic mass, i.e. the mass in grams of one mole of atoms of the element.
Remember 1 mol is approximately 6.022 × 10²³.
So, 1 mol of atoms is 6.022 × 10²³ atoms.
The molar mass is an average: it is the weighted average mass of the natural isotopes of the element, taking into account their relative abundance.
For example, the molar mass or atomic mass of carbon is 12,0107 g/mol, instead of 12.0000, becasue carbon exists in several forms (isotopes), and so the weighted average is not a whole number.
Answer:
Explanation:
I burnt my ramen noodles today :) <3 help me <3 I need somewhere to live
1.95 or 2 is the molarity of a 45.3g sample of KNO3 (101g) dissolved in enough water to make a 0.225L solution.
The correct answer is option b
Explanation:
Data given:
mass of KN
= 45.3 grams
volume = 0.225 litre
molarity =?
atomic mass of KNO3 = 101 grams/mole
molarity is calculated by using the formula:
molarity = 
first the number of moles present in the given mass is calculated as:
number of moles = 
number of moles = 
0.44 moles of KNO3
Putting the values in the equation of molarity:
molarity = 
molarity = 1.95
It can be taken as 2.
The molarity of the potassium nitrate solution is 2.
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
