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icang [17]
3 years ago
15

A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes

. At the end of meiosis I, each daughter cell would have ______ chromosomes. At the end of meiosis II, each daughter cell would have ______ chromosomes.
Chemistry
1 answer:
kozerog [31]3 years ago
7 0

Answer:

At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

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WHAT IS THE UNBALCNED EQUATION OF Methane burns in oxygen to produce carbon dioxide and water
deff fn [24]

Answer:

\boxed{CH_{4}+O_2 --> CO_2 + H_2O}

Explanation:

<h2>Part 1: Naming compound formulas given the names</h2>

<u>Step 1.</u> Methane's formula is CH_4.

<u>Step 2:</u> Oxygen is a diatomic molecule (it exists bonded to itself for stability purposes), so by itself in chemical equations, it is written as O_2.

<u>Step 3:</u> Carbon dioxide is the molecular compound of one atom of carbon and two atoms of oxygen → CO_2.

<u>Step 4:</u> Water is the common name of the compound of two hydrogen atoms and one oxygen atom → H_2O.

<h2>Part 2: Writing the skeleton equation</h2>

<u>Step 1:</u><em> </em>Use the determined formulas for the reactants and plug them into the equation. We are told that methane burns in oxygen -- hinting at a combustion reaction. Therefore, we may infer that these are the reactants that yield the products.

Skeleton equations are written with the reactant(s) on the left -- if there are several, they are separated by an addition symbol (+).

With this information, we may begin our equation: CH_4 + O_2, where CH_4 is methane and O_2 is the diatomic molecule of oxygen.

<u>Step 2:</u> Use the determined formulas for the products and plug them into the equation. We are told that the methane burns in oxygen to produce carbon dioxide and water. Hence, we can separate these two as we did with the reactants.

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This gives us our final equation, \boxed{CH_4 + O_2 --> CO_2 + H_2O}.

Because the problem asks for the unbalanced equation, we do not need to take any further steps of balancing the equation.

5 0
3 years ago
Read 2 more answers
A student recorded the following while completing an experiment. Color of substance: yellow, shiny powder Effect of magnet: yell
victus00 [196]

Answer:

The answer is letter b, mixture.

Explanation:

Let's define what a mixture is at first.

Mixture- In Chemistry, mixture is a material made up of two or more different substances that are physically combined. The identities of the substances are retained. There are two types of mixture: <em>homogenous and heterogenous.</em>

<u>Homogenous mixture</u> examples: saline solution and air.

<u>Heterogenous mixture</u> examples: sand and oil and water.

The student's record shows the color of the substance as yellow, shiny powder. It could mean that the substance is a mixture of different kinds of powders. This is already a hint, since powder is a form of mixture. The record also shows that the substance is attracted to magnet. Part of the powder mixture contains something that is attracted to magnet. It could be an iron powder, which is the shiny one. The substance can then be classified as letter b, mixture.  

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4 years ago
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Tomtit [17]
P= 1.50 atm
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we all this information, we can tell that we need to use ideal gas formula

PV= nRT

if we rearrange the formula for n, we get--> n= PV/ RT

let's plug in the values. 

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