Answer:
B: Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)
Explanation:
We are looking for enthalpy of formation, so we want to see reactance in their natural standard form.
Thus, we want to see the reactance of Na, Cl2 and O2.
The only option that has the correct form of Na, Cl2 and O2 is B.
Na(s) + Cl2(g) + 3O2(g) = 2NaClO3(s)
Answer:- 100
Explanations:- The given number is 124.683 and it has 6 significant figures. We are asked to round it to one significant digit number. To round it as one significant digit we need to make two second and third digits zero and all the three digits next to the decimal are dropped since we want only one significant digit. Second and third digits left to the decimal could not be dropped as they hold the place values. So, 124.683 is round to 100 that has just one significant digit.
Phosphorus can be prepared from calcium phosphate by the following reaction:
![2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)](https://tex.z-dn.net/?f=2Ca_3%28PO_4%29_2%28s%29%2B6SiO_2%28s%29%2B10C%28s%29%5Crightarrow%206CaSiO_3%28s%29%2BP_4%28s%29%2B%2010CO%28g%29)
Phosphorite is a mineral that contains
plus other non-phosphorus-containing compounds. What is the maximum amount of
that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75%
by mass? Assume an excess of the other reactants.
Answer: Thus the maximum amount of
that can be produced is 0.345 kg
Explanation:
Given mass of phosphorite
= 2.3 kg
As given percentage of phosphorite
is = ![\frac{75}{100}\times 2.3kg=1.725kg=1725g](https://tex.z-dn.net/?f=%5Cfrac%7B75%7D%7B100%7D%5Ctimes%202.3kg%3D1.725kg%3D1725g)
![moles=\frac{\text {given mass}}{\text {Molar mass}}](https://tex.z-dn.net/?f=moles%3D%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D)
![{\text {moles of}Ca_3(PO_4)_2=\frac{1725g}{310g/mol}=5.56moles](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bmoles%20of%7DCa_3%28PO_4%29_2%3D%5Cfrac%7B1725g%7D%7B310g%2Fmol%7D%3D5.56moles)
![2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)](https://tex.z-dn.net/?f=2Ca_3%28PO_4%29_2%28s%29%2B6SiO_2%28s%29%2B10C%28s%29%5Crightarrow%206CaSiO_3%28s%29%2BP_4%28s%29%2B%2010CO%28g%29)
According to stoichiometry:
2 moles of phosphorite gives = 1 mole of ![P_4](https://tex.z-dn.net/?f=P_4)
Thus 5.56 moles of phosphorite give=
of ![P_4](https://tex.z-dn.net/?f=P_4)
Mass of ![P_4=moles\times {\text {Molar mass}}=2.78mol\times 124g/mol=345g=0.345kg](https://tex.z-dn.net/?f=P_4%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D2.78mol%5Ctimes%20124g%2Fmol%3D345g%3D0.345kg)
Thus the maximum amount of
that can be produced is 0.345 kg
Answer:
nickel, cobalt would decrease in size when it became an iron