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2 years ago

Meiosis makes four blank cells A) diploid B) haploid C) metabolic D) prokaryotic

Chemistry

2 answers:

mel-nik [20]2 years ago

7 0

Answer:

haploid

Explanation:

Meiosis is a type of cell division that occurs during the process of formation of gametes particularly sexually reproducing organisms.

Gametes are reproductive, sex cells that are haploid (having half the number of chromosomes as their parent cell).

Thus, during meiosis a diploid parent cell gives 4 haploid daughter cells, called gametes (such as eggs and sperms).

kap26 [50]2 years ago

6 0

Answer:

the answer is haploid

Explanation:

i took the test

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A mineral that has _______ can be split fairly easily along planes with weak atomic attraction. A. hardness B. luster C. streak

Tomtit [17]

A mineral that has CLEAVAGE can be split fairly easily along the planes with a weak atomic attraction. Technically a mineral may have a luster but it isn't the cause of it's split. Additionally, hardened minerals is the opposite of what is being discussed and I cannot truly remember what "streak" meant in regards to minerals.

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What does a plant produce during photosynthesis? A. Oxygen and glucose B. Carbohydrates and carbon C. Carbon, oxygen, and hydrog

Nataliya [291]

Answer:

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Explanation:

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2 years ago

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

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2 years ago

How much silver can be produced from 125g of Ag2S

rosijanka [135]

Answer:

108.9g of Silver can be produced from 125g of Ag2S

Explanation:

The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1

•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.

•Molar mass of Silver Ag= 107.9g/mol

•Molar mass of Sulphur S= 32g/mol

•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol

•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol

•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S

If 247.8g Ag2S produced 215.8g Ag

125g Ag2S will produce xg Ag

cross multiplying we have

xg= 215.8g × 125g / 247.8g

xg= 26975g/247.8

xg= 108.85g

Therefore, 108.9g of Silver can be produced from 125g of Ag2S

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2 years ago

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Why are bottles made from amorphous solid's such as plastic and glass

notsponge [240]

They will mold into different shapes

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2 years ago