Half-life is the time it takes for the radioactive material's mass to decrease by half.
So, half of 96 is 48
half of 48 is 24
half of 24 is 12
half of 12 is 6
96->48->24->12->6
You count how many arrows (->) there is.
Which is 4
Then you use the 12 minutes the question gave you to do:
12 divided by 4
= 3
So, one half-life (Or the time it takes for the radioactive material's mass to decrease in half) is 3 minutes
and every 3 minutes it will decrease by half.
ΔG = -nEF
ΔG = Gibbs free energy change
n = moles of electrones participated in
E = Electrode potential
F = Faraday constant
By substituting,
ΔG = -(3 mol) x 96485 A S/ mol x (-0.46) V
= + 133149.3 J
= + 133 kJ
Hence the answer is "b".
Since ΔG is a positive value, the reaction is non spontaneous reaction.
Answer:
C
Explanation:
∆H°f means the enthalpy change of formation of one mole of substance by its constituent elements under standard conditions.
So in an equation for ∆H°f, we must see 2 or more elements as reactants combining to form a compound.
In the 4 answers, only C represents elements forming a substance (Al(s) + 3/2O2(g) + 3/2H2(g) -> Al(OH)3(s)),
while the others include compounds as one of their reactants.
Answer:
Neutral.
Explanation:
Protons have a positive charge. In this problem there are 12.
Electrons have a negative charge. In this problem there are 12.
Neutrons have a neutral charge that isn't relevant tot the overall charge, it can be ignored.
#protons - #e- = charge
12 - 12 = 0
The protons and electrons balance each other out, resulting in a 0 net charge. This means the atom has a neutral charge.
Answer:
<h2>
</h2>
Explanation:
According to hooks law "provide the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force F"
where F= applied force
k= spring constant
e= extension
Given data
length of string l = 16cm
extension e = 20-16= 4cm
applied force = 5N
we need t o first calculate the spring constant k
apply the formula
we can now calculate the extension of the string when supporting a 6N weight
The length of the string when supporting a 6N weight is
<em>COMMENT </em><em>:According to the analysis an extra weight of 1N will cause add 0.8cm to the length of the string</em>
