<span>Because </span>of the release of latent heat
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%
Answer:
1) harm life forms that rely on carbonate-based shells and skeletons, 2) harm organisms sensitive to acidity
Explanation:
The correct answer is (C. zirconium - 93.
Answer:
13.5 * 10^-2 g
Explanation:
What we know:
Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,
Grams of Ba: 1
Grams of Al2(SO4)3: 1.8g
Calculate the # of moles of Ba and Al2(SO4)3:
1g Ba/137.3 = 7.3 *10^-3 mol Ba
1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3
Find the limiting reactant:
Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3
Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3
2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.
finally, we just find the number of moles of Al
The ratio of Al to Ba is 2:3 so...
7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al
CONVERT TO GRAMS
5 *10^-3 mol Al * 27 = 13.5 * 10^-2 g
<u>Hope that was helpful! </u>