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Answer:
Energy added to cold water, E1
E1 = m•Cp•ΔT = 6 x 4.187x(42-20) = 6x4.187x22
Energy lost by hot water, E2:
E2 = X(CpΔT) = X x 4.187 x (75-42) =X x 4.187 x 33
So, since E1 = E2
6x4.187x22 = X x 4.187 x 33
X = 6x22/33 = 4.0kg
So the mass of the hot water added to the 6kg of cold water is 4.0kg
Two identical copies formed by the replication of a single chromosomes
PP- purple homozygote
pp- white homozygote
Pp- purple hetrozygote
If these flowers cross, we obtain heterozygote offsprings with a genotype: Pp (100% of them will be like this)
Now, if those offsprings cross with each other: Pp x Pp
results:
1/4 would be purple homozygote -> PP
1/4 would be white -> pp
2/4 would be purple heterozygotes -> Pp
Genotipic ratio: 3:1 (3 are purple; 1 is white)
