Answer:
Relative and average atomic mass both describe properties of an element related to its different isotopes.
Explanation:However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.
Answer:
Four times the original amount if only one orange was used
Explanation:
We can assume that the oranges all have equal voltages. Connecting them in series will have an increasing effect on the voltage delivered. In our case, this will produce 4 times the voltage of the circuit when only one orange is used.
Whenever simple cells are connected in series, the voltages of the individual cells are added up to form the voltage of the whole circuit.
Let us assume that the voltage of each of the oranges is approximately 0.9 volts. The Voltage produced when the 4 oranges are joined in series is 0.9 + 0.9 + 0.9 + 0.9 = 3.6 volts
It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
75.0 mL in liters:
75.0 / 1000 => 0.075 L
1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L
moles Hg = 0.075 x 1 / 22.4
moles = 0.075 / 22.4
= 0.00334 moles of Hg
Hg => 200.59 u
1 mole Hg ----------------- 200.59 g
<span>0.00334 moles Hg ----- ( mass Hg )
</span>
mass Hg = 200.59 x 0.00334 / 1
mass Hg = 0.6699 / 1
= 0.6699 g of Hg
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
