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Marina86 [1]
3 years ago
15

Two of the seven different pea traits examined by Mendel involved genes that we now know are linked. Knowing this, can you expla

in why Mendel was still able to use results from his crossing experiments to develop the principle of independent assortment?
Chemistry
1 answer:
ser-zykov [4K]3 years ago
4 0

Answer:: Mendel studied how traits are been passed from parents to offspring using seven features in peas, including height, flower color, seed color, and seed shape. To do this he divided the pea plant into short height and tall height. From this experiment he proposed a principle called independent assortment, which describes how different genes independently separate from one another when reproductive cells develop. Though this experiment was studied using gene formation in prokaryotic cell.

This principle of independent assortment is also seen in eukaryotic cells during meiosis.

Mendel proposed this principle because during cell formation of the offspring, each individual Gene from the parents will first separate to stand on its own before cross linking up together, which made the offspring look different from the parents. The principle of independent assortment does not criticize gene linkage, it only highlight how gene in the garments of the parents forms offspring, by sperating to assort independently.

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Chemists recognize many different elements, such as gold, or oxygen, or carbon. Suppose you got some carbon, and started splitti
Otrada [13]

Answer:

An Atom.

Explanation:

An atom is the smallest particle of a chemical element that can exist. It the fundamental piece of matter. Therefore, splitting carbon into smaller pieces , the smallest piece that would still be called "carbon" would be an Atom of carbon. However, atom is also made of three subatomic particles called electrons, protons and neutrons.

6 0
3 years ago
1. If we have a reaction of Zn3PO4 and HCl, what would the products of this
Vladimir79 [104]

Answer:

1. The products of this reaction are ZnCl₂ and H₃PO₄.

2. 14.57 g.

Explanation:

<em>1. What would the products of this  reaction be?</em>

  • The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:

<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>

It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.

So, the products of this reaction are ZnCl₂ and H₃PO₄.

<em>2. If we produced 13.05 g of H₃PO₄, how many grams of  hydrochloric acid would be need to start with?​</em>

  • Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:

n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.

<u><em>Using cross-multiplication:</em></u>

6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.

??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.

∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.

∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.

<em>So, the grams of  hydrochloric acid would be need to start with = 14.57 g.</em>

3 0
3 years ago
PLEASE HELP!
svetoff [14.1K]

Answer:

period 3 and group 3

Explanation:

I'm saying group 3 because that is how I learnt it at school, but if you count it then it's in group 13.

8 0
2 years ago
1 times 2900 inches for a car
Delicious77 [7]

One times anything is the same number. 1 x 2900= 2900

5 0
3 years ago
How many grams of ammonium chloride (gram formula mass= 53.5 g) are contained in .500 L of a 2.00 M solution?
Elza [17]

Answer:

53.5g of NH4Cl

Explanation:

First, we need to obtain the number of mole of NH4Cl. This is illustrated below:

Volume = 0.5L

Molarity = 2M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 2 x 0.5

Mole = 1mole

Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:

Molar Mass of NH4Cl = 53.5g/mol

Number of mole = 1

Mass =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass = 1 x 53.5

Mass = 53.5g

Therefore, 53.5g of NH4Cl is contained in the solution.

8 0
2 years ago
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