Answer:
I think it should be Carbon.
Nitrogen has 5 valence electrons (ve-), so a diatomic nitrogen molecule will have twice as many, 10 valence electrons. Then, just draw electrons in pairs of 2 until you both get ride of all of them (reach 0) and you fill every atom (eight electrons each). It can be drawn either way, the important thing is that there are 3 electron pairs shared between the two atoms.
Answer:
The frequency of this light is
.
Explanation:
Wavelength of the light = 
Speed of the light = c = 
Frequency of the light = 


The frequency of this light is
.
Answer: 404.04 kJ.
Explanation:
To calculate the moles, we use the equation:
moles of

According to stoichiometry :
2 moles of
on burning produces = 1036 kJ
Thus 0.78 moles of
on burning produces =
Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.
The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.
First-order reactions are very commonplace. we have already encountered examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.
Rate constant K₁ = 0. 543 /s
T₁ = 25°C
Activation energy Eₐ = 75. 9 k j/mol.
T₂ = 32. 3 °C.
K₂ =?
formula;
log K₂/K₁= Eₐ /2.303 R [1/T₁ - 1/T₂]
putting the value in the equation
K₂ = 0.343 /s
Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s
The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.
Learn more about activation energy here:- brainly.com/question/26724488
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