Answer:
517.3 K
Explanation:
Initial volume of gas V1= 114 L
Initial temp. T1= 273 K
Final volume V2= 216 L
Final temp. T2= ?
From Charles law, Volume is directly proportional to temperature provided pressure is kept constant
V1/T1 = V2/T2
T2 = V2T1/V1
T2 = (216×273)/114
T2 = 517.3 K
Answer:
2.28 m
Explanation:
Use the relationship
λ=vf
to solve for wavelength λ. Substituting the known quantities yields:
λ=vf
λ=2.99 x 108 m/s1.31 x 108 Hz
λ=2.28 m
Answer:
(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ
(b) 4.898 MeV
Explanation:
In the nuclear reaction, it was stated that plutonium-242 decayed firstly to uranium-238 and alpha, and lastly to a stable uranium-238 by emitting a gamma ray. The balanced equation for the nuclear reactions is shown below:
(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ
(b) The energy emitted by releasing a gamma ray is calculated using:
E = hc/λ
where
h = 4.136*10^-15 eV.s
c = 299792458 m/s
λ = 0.02757 nm = 0.02757*10^-9 m
Therefore:
E = (4.136*10^-15)*(299792458)/0.02757*10^-9 = 44974.31 eV
The total energy if the stable 238U was produced directly would be
4.853*10^6 + 44974.31 = 4.898 MeV
4 atoms of Oxygen are required to make 2 atoms of hydrogen. The compound H2O is 1 Hydrogen and 2 Oxygen.
Answer:The correct answer is ;
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
Explanation:
In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.
In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.
In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in 
molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent
In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in 
molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.
