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iogann1982 [59]
3 years ago
11

Pea plants have green seeds only if a plant receives an allele for green seeds from each of its parents. What kind of allele pro

duces green seeds in pea plants?
Your answer:


Hybrid

Dominant

Recessive

Mutated
Chemistry
1 answer:
Elena-2011 [213]3 years ago
6 0

Answer:If two heterozygous parents with a genotype of Yy are crossed, the possible combination of genotypes in their offspring would be YY, Yy, Yy, and yy. Therefore, the genotypic ratio of the offspring produced is 1:2:1.

Explanation:If two heterozygous parents with a genotype of Yy are crossed, the possible combination of genotypes in their offspring would be YY, Yy, Yy, and yy. Therefore, the genotypic ratio of the offspring produced is 1:2:1.

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Which parts (postulates) of Dalton's atomic theory has been modified in light of later discoveries
professor190 [17]

Explanation:

1) Atoms can not be subdivided. It has been changed as it is possible to divide an atom into Protons, Neutrons and Electrons as well as other smaller particles.

2) The regulations for the chemical mixture have always been altered to the rules for the chemical combination in the creation of organic compounds may be denied.

3)Atoms are of a particular element are identical in all respects i. e. they have same mass and similar properties.

It was modified. And according. to new search;

Atoms of the same element may not be always identical.

8 0
4 years ago
Using the graph below please draw a reaction potential energy diagram for a reaction with the
azamat

Answer:

12

Explanation:

because it is the sam

7 0
3 years ago
What is the Lewis Dot Diagram of gold?
ratelena [41]
Its structure is Au with one dot on top
6 0
3 years ago
When coke burns in air then evolved gas is​
Arturiano [62]

Answer:

Explanation:

Carbon (coke) burns in air to form carbon dioxide gas.

(i) C(s) + O2 ↑= CO2 ↑

7 0
4 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
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