S=50.05/32 O=49.95/16, S=1.5640625/1.5640625 O=3.121875/1.5640625, S=1 O=1.99, S=1 O=2 Empirical Formula= SO2
The final volume of the gas is 73.359 mL
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Given :
A sample gas has an initial volume of 72.0 mL
The work done = 141.2 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm; we have
External Pressure P:
= 783 torr ×
= 1.03 atm
The work done W =
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 0.001359 L
ΔV = 1.359 mL
The initial volume = 72.0 mL
The change in volume V is ΔV = V₂ - V₁
- V₂ = - ΔV - V₁
multiply both sides by (-), we have:
V₂ = ΔV + V₁
= 1.359 mL + 72.0 mL
= 73.359 mL
Therefore, the final volume of the gas is 73.359 mL .
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Answer:
The correct answer is : It is a salt because it is formed by the reaction of an acid and a base.
Explanation:
Sodium chloride is formed from the neutralization reaction between a strong acid (hydrochloric acid) and a strong base (sodium hydroxide):
HCl + NaOH ---> NaCl + H20
By reacting strong acid and base, they neutralize each other resulting in a salt (NaCl) and water.
Corresponds to an exothermic reaction (heat is released).
Answer:
Mass = 51 g
Explanation:
Given data:
Mass of ammonia formed = ?
Number of moles of hydrogen = 4.50 mol
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Now we will compare the moles of hydrogen and ammonia.
H₂ : NH₃
3 : 2
4.50 : 2/3×4.50 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
Answer:-
[Kr] 5s1
Explanation:-
Atomic number of rubidium = 37
Atomic number of Kr = 36
So only 1 electron needs to be assigned an orbital.
Kr is also the last element of the fourth period
So the shell number for the 1 electron will be 5 since it is after Kr
Since s subshell is filled first the electron will go into 5s.
Hence the electron configuration for rubidium (Rb) in noble-gas notation is
[Kr] 5s1