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3 years ago

What inspired Mendeleev to create his organization for the elements

1 answer:

const2013 [10]3 years ago

7 0

Dimitri Mendeleev was inspired primarily by the work of Antoine Lavoisier and his work on writing the first extensive list of known elements. Lavoisier also collaborated in the construction of the metric system and worked to develop a better nomenclature for chemical compounds which parts are still used today.

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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

B. Determine the percent composition of each element in NaCl. (1 point) Must show work to

andrey2020 [161]

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433. 22.99/58.433= 39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433. 35.453/58.433= 61%

Explanation:

8 0

2 years ago

The following equation is balanced correctly. 4C2H6 + 7O2 --> 6H2O + 4CO2 True False

Ostrovityanka [42]

Answer: False

Explanation:

4C2H6 + 7O2 --> 6H2O + 4CO2

8 Carbons on the reactant side, but 4 Carbons on the Product

24 Hydrogens on Reactant, 12 H on product

But Oxygen is balanced, 14 on each side

4 0

3 years ago

Read 2 more answers

By replacing hardware safety features with software controls in the therac-25 machine, aecl was able to?

Ipatiy [6.2K]

Therac-25, AECL was able to reduce costs by replacing hardware safety features with software controls.

the Therac-25 was a laptop-controlled radiation therapy machine produced by Atomic energy state Canada limited in 1982 after the Therac-6 and Therac-20 units. It was involved in at least six accidents between 1985 and 1987, in which patients were given massive overdoses state radiation.

The Therac-25 machine was a state-of-the-art linear accelerator developed by using the company Atomic strength Canada restricted (AECL) and a French company CGR to provide radiation treatment to cancer sufferers. The Therac-25 become the most computerized and complicated radiation therapy device state its time.

learn more about cancer here brainly.com/question/11710623

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5 0

2 years ago

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

kozerog [31]

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 20 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 20 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.

7 0

3 years ago