Answer: The data are consistent with the law of conservation of mass because the mass of reactants equals the mass of products
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Given: mass of sodium = 12.8 g
mass of chlorine = 19.6 g
mass of reactants = mass of sodium + mass of chlorine = 12.8 g + 19.6 g = 32.4 g
Mass of product = Mass of sodium chloride = 32.4 g
As mass of reactant = mass of products
, Thus results are in accordance with the law of conservation of mass.
Answer:
1 mol SO2 contains 6.0213*10^23 molecules
6.023*10^24 molecules = 10 mol SO2
Equation
S(s) + O2(g) → SO2(g)
1 mol S reacts with 1 mol O2 to prepare 1 mol SO2
To prepare 10 mol SO2 you require : 10 mol S plus 10 mol O2
And that is the answer to the question
If you want a mass :
Molar mass S = 32 g/mol You require 10 mol = 320 g
Molar mass O2 = 32 g/mol :You require 10 mol = 320 g
Answer:
The answer to your question is: 16.7 g of KBr
Explanation:
Data
mass KBr = ? g
Volume = 0.400 L
Concentration = 0.350 M
Formula
Molarity = moles / volume
moles = molarity x volume
Process
moles = (0.350)(0.400)
= 0.14
MW KBr = 39 + 80 = 119 g
119 g of KBr -------------------- 1 mol
x -------------------- 0.14 mol
x = (0.14 x 119) / 1
x = 16.7 g of KBr
Lol yeah that’s what I’m doing I just want you know I got a lot of you guys to go and you don’t get
Oxidation-Reduction Reactions Suggested Reading Thus the oxidation number for oxygen in calcium oxide is -2. ... In effect, each calcium atom loses two electrons to form Ca2+ ions, and each O atom in O2 gains two electrons to form O2- ions. The net result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidation-reduction reaction.
+<u>O²</u><u>(</u><u>g</u><u>)</u><u>=</u><u>2</u><u>CaO</u><u>(</u><u>s</u><u>)</u>
Explanation:
we can conclude that in the reaction there is both reduction and oxidation.