This
reaction is called the electrolysis of water. The balanced reaction is:
2H2O = 2H2 + O2
<span>
We are given the amount of water for the electrolysis reaction. This
will be the starting point of our calculation.
45.6 grams H2O (1 mol H2O / 18.02 g H2O) (1 mol O2 / 2 mol H2O) = 1.27 mol O2
V = nRT/P = </span><span>1.27 mol O2 (0.08206 atm L / mol K) (301 K) / 1.24 atm
V = 25.20 L O2</span>
Answer:
c) It describes the affinity an atom has for an electron.
Explanation:
C quite literally describes what electronegativity is, you can rule out most others by looking at periodic trends as attached below.
Answer:
_2Mg + _O2 --≥2MgO this is balanced
Answer:- 27.7 grams of 
are produced.
Solution:- The balanced equation is:
let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of ![Pb(NO_3)_2 = 207.2+2(14.01)+6(16) = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%20%3D%20207.2%2B2%2814.01%29%2B6%2816%29%20%20%3D%20331.22%20gram%20per%20mol%3C%2Fp%3E%3Cp%3Emolar%20mass%20of%20NaI%20%3D%2022.99%2B126.90%20%3D%20149.89%20gram%20per%20mol%3C%2Fp%3E%3Cp%3EMolar%20mass%20of%20%5Btex%5DPbI_2)
= 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:
= 
= 
From above calculations, NaI gives least amount of , so the answer is, 27.7 g of are produced.
... to be called elements<span>. This lesson shows </span>you how to<span> predict the </span>numbers<span> of </span>neutrons, electrons, andprotons<span> of the isotopes they are likely to find in nature. (</span><span>cont.) ... What </span>kind<span> of </span>generalization can you make<span> about how the </span>number<span> of </span>protons<span> and </span>neutrons<span> are </span>related<span> to </span>each other<span> in the </span>elements<span>? Unit 1 • Investigation IV</span>
