B. 11,540
<h3>Further explanation
</h3>
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
General formulas used in decay:
T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Nt=25 g
No=100 g
t1/2=5770 years
Answer:
18. E. 0%
19. D. 25%
Explanation:
Question 18:
Let's use "B" to represent the dominant allele of "light blue skin", and
"b" for recessive "light green skin".
Squidward => BB - light blue skin
Squidward's bride => bb - light green skin
When they cross, they will have the following offsprings:
(BB) × (bb) - Parent
(Bb) (Bb) (Bb) (Bb) - Offspring
All the offspring would be light green skin. The dominant allele of light green skin will express itself over the recessive allele.
Therefore, the chances of Squidward and his bride having light green skin is 0%
Question 19:
Squidward's son => Bb - light blue skin
Squidward's son's bride => Bb - light blue skin
(Bb) × (Bb) - parent
(BB) (Bb) (Bb) (bb) - offspring
They will have the following offspring:
BB and Bb - light blue - 75%
bb - light green - 25%
Chance of having offspring with light green skin is 25%
What they have in common is that they both have the same number of atoms.
Answer:
7.5 g of AlCl3
Explanation:
The given equation is;
NaOH + AlCl3 --> Al(OH)3 + NaCI.
By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.
Thus, let's make them equal by balancing the equation.
Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:
3NaOH + AlCl3 --> Al(OH)3 + 3NaCI
We can see that;
NaOH has 3 moles
While AlCl3 has 1 mole
Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;
2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3
