First, we need to determine the half reaction of magnesium. It would be expressed as:
Mg2+ + 2e- = Mg
Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:
4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C
We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.
35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
The common substance among the product(s) of the first equation and among the reactant(s) in the second equation is H2O(g). We can eliminate that as an intermediate. The overall chemical equation will thus be:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l),
which is the first answer choice.
In essence, all you’re doing here is swapping water vapor for liquid water.
If your in a city, go inside a building immediately. most buildings have lightening rod, so the lightening voltage won't hit on you.
if you're in an open area, don't go under a tree, or it'll hit you and the tree, even if it doesn't hit you, you still cannot take the impact by the exploding tree. leave the water or metal bicycle or stuff that conducts electricity.
Answer: (C)
The frequency increases as the wavelength decreases
Explanation:
The relation between the frequency and wavelength of a wave is
Frequency = 1 / Wavelength
The Frequency of electromagnetic wave is inversely proportional to the wavelength. So, as the frequency increases, the wavelength of the wave decreases and vise-versa.
The frequency of a wave is number of complete cycles passing a particular point per second. Its S.I unit is Hertz whereas the wavelength of a wave is the distance between two consecutive crest and trough in meters.
So, on increasing the frequency of a wave, there will be more number of the cycles of wave per second which will decrease the distance between the consecutive crest and trough i.e wavelength.
