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.
First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:
<span>9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed</span>
Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
5-Methyl-5-ethyldecane should be the answer .
I’m pretty sure it’s Nose
