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3 years ago

What are the three isotopes of hydrogen and what is the significance of these isotopes?

Chemistry

2 answers:

antiseptic1488 [7]3 years ago

4 0

Answer:

21H deuterium (D), 31H is tritium (T), 11H is hydrogen

Explanation:

11H: it has the greatest abundance, it has only one proton (it does not have neutrons).

21H (D): it has a proton and a neutron in its nucleus.

31H (T): it has a proton and two neutrons in its nucleus.

Remember that isotopes have the same atomic number, but a different mass number (elements in the number of neutrons differed).

snow_lady [41]3 years ago

3 0

Answer:

See below.

Explanation:

They are Hydrogen 1 , Hydrogen 2 (also called deuterium) and Hydrogen 3 (tritium)

Hydrogen has no neutrons in the nucleus while deuterium has 1 and tritium has 2.

Deuterium has been used in the nuclear industry ( as heavy water , deuterium oxide) and as a compound lithium deuteride in the H bomb. Tritium is also used as a fuel in nuclear fusion reactions, and as a radioactive tracer.

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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.

4 0

3 years ago

URGENT!!!!!

Nikitich [7]

Answer:

buggy cyh gu hcgcggcfgggh5677

6 0

3 years ago

The chemical formula for strontium sulfite is SrSO3. TRUE FALSE

VLD [36.1K]

Valency is the number of electrons lost or gained by an atom to attain an stable configuration. Valency is important when writing the formula of chemical compounds in chemistry. Strontium has a valency of 2 while sulfite ion (radicle) has a valency of 2. Therefore, the chemical formula of strontium sulfite is written as SrSO3.

3 0

3 years ago

It refers to the cracked sections of the earth's crust.

poizon [28]

Answer:

Plates

Explanation:

The Earth's crust is broken up into sections called plates. Tectonic plates are on the mantle which allow them to move.

4 0

2 years ago

Read 2 more answers

How many liters of solvent are needed to make a 6.5M solution with 34 mol of solute?

Dominik [7]

Answer:

5.231 L.

Explanation:

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of KCl)/(Volume of the solution (L))

M = 6.5 M.

no. of moles of solute = 34.0 mol,

Volume of the solution = ??? L.

∴ (6.5 M) = (34.0 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (34.0 mol)/(6.5 M) = 5.231 L.

6 0

3 years ago