The answer is , C. Both the atomic mass and the atomic number increase from left to right .
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
Answer:
N₃O₆
Step-by-step explanation:
Data:
EF = NO₂
MM = 138.02 g/mol
Calculations:
EF Mass = (14.01 + 32.00) u
EF Mass = 46.01 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ
n = MF Mass/EF Mass
n = 138.02 u/46.01 u
n = 3.000 ≈3
MF = (NO₂)₃
MF = N₃O₆
Magnesium Peroxydisulphate with formula of MgS₂O₈.
<h3><u>Explanation:</u></h3>
The question clearly stands on the concept of molarity and atomic weights.
The atomic weight of magnesium = 24.
The atomic weight of sulphur = 32.
The atomic weight of oxygen = 16.
The amount of magnesium present = 0.248 g
The amount of sulphur present = 0.665 g.
The amount of oxygen present = 1.31g.
So, moles of magnesium present = 0.01 moles.
Moles of sulphur present = 0.02 moles.
Moles of oxygen present = 0.08 moles.
So, mole ratio of the compound as magnesium : sulphur : oxygen = 1:2:8.
So the compound is Magnesium Peroxydisulphate with formula of MgS₂O₈.
Answer:
C
Explanation:
If you think about it treated sewage would have a certain place that it is put therefore C is the answer
