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3 years ago

Elements in Group 4A

Chemistry

1 answer:

steposvetlana [31]3 years ago

6 0

Answer:

helium hydrogen

Explanation:

lithium beryllium bottom carbon

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Can you identify 2 characteristics of combustion reactions?

monitta

An element or compound will react with oxygen and will produce carbon dioxide, water, and sometimes carbon (if it is an incomplete combustion).

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3 years ago

In the reaction Fe2O3 + 3CO a 2Fe + 3CO2, 10 moles of solid iron and 15 moles of carbon dioxide are produced from 5 moles of iro

Schach [20]

Answer:

Ratio is 3:2

3CO = 2Fe or 1.5 CO = 1 Fe

Explanation:

Fe2O3 + 3CO = 2Fe + 3CO2

Fe2O3 = Iron (|||) oxide

CO = Carbon monoxide

Fe = Solid Iron

CO2 = Carbon dioxide

Excellent is already balanced.

10 Moles Fe and 15 Moles of CO2

5 Moles Fe2O3 + 15 Moles 3CO = 10 Moles Fe + 15 Moles 3CO2

What is the ratio of carbon monoxide to solid iron

Ratio is 3:2 or 1.5 CO = 1 Fe

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3 years ago

How many of sodium (Na) are needed to make 4.5 liters of a 1.5mol/L of Na solution?

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Answer:

Explanation:

First you will find the mole from the molarity and then the desired mass from the mole.

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2 years ago

The ionization energies for removing successive electrons from sodium are 496 kJ/mol, 4562 kJ/mol,

ddd [48]

Answer:

Explanation:

The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.

The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.

Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.

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3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago