Answer:
not sure what to say to this but the rays of the sun contain solar energy which is then converted into energy to use for electrical power
Mole of electron required by 
mole is 
- Faraday law expressed how the change that is been being produced by a current at an electrode-electrolyte interface is related and proportional to the quantity of electricity that is been used.
- There is one mole of electron required for 1 Faraday of electricity.
- Avogadro constant is
- Mole of electron can be calculated by dividing the number of electron by avogadro's constant.
=
= 
Therefore, it requires 
Faraday of electricity for the 
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A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask<span>The given are: </span>
<span><span>1. </span>Mass = ?</span><span><span /></span>
<span><span>2. </span>Density = 1298 g/L</span>
3. Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L
<span>Formula and derivation: </span><span><span>
1. </span>density = mass / volume</span> <span><span>
2. mass </span>= density / volume</span>
<span>Solution for the problem: </span><span><span>
1. mass = </span></span> <span> 1298 g/L / 0.125 L = 10384g
</span>
Answer:
Chemical reaction involves the breaking of bonds in the reactants and formation of bonds in the products. ... If a reaction is exothermic, more energy is released when the bonds of the products are formed than it takes to break the bonds of the reactants. This is the reason for temperature change during a reaction.
Explanation:
Here are just a few everyday demonstrations that temperature changes the rate of chemical reaction: Cookies bake faster at higher temperatures. Bread dough rises more quickly in a warm place than in a cool one.
The answer to this question would be: <span> 10 °K
Kelvin and Celcius scales are different by 273</span> degrees but their ratio is the same. One degree in Kelvin is equal to one degree in Celcius. That mean, 10 °C change in Celcius would be same as <span> 10 °K changes in Kelvin too. </span>
