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3 years ago

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A fertilized egg receives 50% of its chromosomes from a sperm and 50% from the egg. Which process ensures that the sperm and egg

contain only 50% of the chromosomes of an organism?
A.fertilization
B.meiosis
C. mitosis
D.mutation

1 answer:

Helen [10]3 years ago

5 0

Answer:

B

Explanation:

Meiosis ensures that the sperm and egg contain only 50-50.10% of the chromosomes of an organism.

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What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?

insens350 [35]

First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm

6 0

3 years ago

An aqueous solution of barium hydroxide is standardized by titration with a 0.102 M solution of perchloric acid. If 10.3 mL of b

Vesna [10]

<u>Answer:</u> The molarity of barium hydroxide solution is 0.118 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

$n_1=1\\M_1=0.102M\\V_1=24.0mL\\n_2=2\\M_2=?M\\V_2=10.3mL$

Putting values in above equation, we get:

$1\times 0.102\times 24.0=2\times M_2\times 10.3\\\\M_2=0.118M$

Hence, the molarity of $Ba(OH)_2$ solution will be 0.118 M.

6 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago

A student observed a sample of water in three states of matter. The student should describe the liquid water as a state of matte

hodyreva [135]

Answer:

i say B

Explanation:

tell me if it is the right one

4 0

3 years ago

5 Examples of simple salts

kolbaska11 [484]

<h3>Five Examples of Salts </h3>

Sodium Chloride.
Sodium chloride (NaCl) is the most common type of salt in our lives.
Potassium Dichromate. •
Potassium dichromate (K2Cr2O7) is an orange-colored salt composed of potassium, chromium and oxygen.
Calcium Chloride.
Sodium Bisulfate.
Copper Sulfate.

3 0

1 year ago