A beaker and a microscope
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
Answer:
1(a) N = 3
(b) N = 0
(c) N = 5
(d) N = -2
(2) Molecular formula for benzene is C6H6
Explanation:
1(a) N02 1-
N + (2×-2) = -1
N-4 = -1
N = -1+4 = 3
(b) N2
2(N) = 0
N = 0/2 = 0
(c) NO2Cl
N + ( 2×-2) + (-1) = 0
N - 4 - 1 = 0
N - 5 = 0
N = 0+5 = 5
(d) N2H4
2(N) + (4×1) = 0
2N + 4 = 0
2N = 0 - 4 = -4
N = -4/2 = -2
(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole
Therefore, molecular formula for benzene is C6H6
Hey there!
The answer to this question would most likely be the 3rd choice, (option C)
Increasing the pressure on a gas decreases the volume
Good luck on your assignment and enjoy your day!
~
Answer:
In H2CO3(aq) + H2O(l) + CO2(g) there are:
4 hydrogen atoms
2 carbon atoms
6 oxygen atoms
