Answer:
oh it's easy
Explanation:
Take the hydrate
N
a
2
S
2
O
3
∙
5
H
2
O
. Are there ionic forces between the
N
a
+
and the
S
2
O
2
−
3
and ion-dipole forces between the cation/anions and the water?
Answer:
V₂ = 3227.46 L
Explanation:
Given data:
Initial volume of gas = 1000 L
Initial temperature = 50°C (50 +273 = 323 K)
Initial pressure = 101.3 KPa
Final pressure = 27.5 KPa
Final temperature = 10°C (10 +273 = 283 K)
Final volume = ?
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 101.3 KPa × 1000 L × 283 K / 323 K × 27.5 KPa
V₂ = 28667900 KPa .L. K /
8882.5 K.KPa
V₂ = 3227.46 L
Answer:
An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An ethothermic process absorbs heat and cools the surroundings. Hope this helped.
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
They are all stable and have eight valence electrons
