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How many moles of hydrogen atoms are contained in one mole of<img src="https://tex.z-dn.net/?f=C_%7B6%7DH_%7B12%7D" id="TexFormu

Orlov [11]

The number of moles of hydrogen atoms contained in 1 mole of c6h12 is 12.

<h3>What is hydrogen made of?</h3>

Hydrogen can be produced from various domestic resources such as fossil fuels, biomass, and the electrolysis of water by electricity. The environmental sustainability and energy efficiency of hydrogen depends on how hydrogen is produced.

Hydrogen is the simplest element. Each hydrogen atom has only one proton. Hydrogen is also the most abundant element in the universe. Stars like the Sun are mainly made of hydrogen. The sun is essentially a giant sphere of hydrogen and helium gas.

Some facts about hydrogen:

Hydrogen is the most abundant element on earth.

Hydrogen is an alternative fuel with a very high energy content by weight.

The fuel cell can be used to power multiple applications.

Fuel cells are a clean way to generate electricity.

Fuel cell vehicles are very similar to traditional gasoline vehicles

Click here for more information on hydrogen atoms: https: //brainly.com/question/16847428

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4 0

2 years ago

Show the calculation of the mole fraction of each gas in a 1.00 liter container holding a mixture of 8.65 g of CO2 and 4.32 g of

Vlad1618 [11]

Answer:

0.745 and 0.245

Explanation:

Mole fraction (χ) is the number of moles of a component divided by the total number of moles in a mixture.

We must calculate the moles of each component.

Let CO₂ be Gas 1 and SO₂ be Gas 2.

1. Calculate the moles of each gas.

$\text{n} _{1} = \text{8.65 g} \times \dfrac{\text{1 mol}}{\text{44.01 g}} = \text{0.1965 mol}\\\\\text{n} _{2} = \text{4.32 g} \times \dfrac{\text{1 mol}}{\text{64.06 g}} = \text{0.067 44 mol}$

2. Calculate the total moles.

$n_{\text{tot}} = \text{n}_{1} + \text{n}_{2} = \text{0.1965 mol} +\text{0.067 44 mol} = \text{0.2640 mol}$

3. Calculate the mole fraction of each component

$\chi_{1} =\dfrac{0.1965}{0.2640} = \mathbf{0.745}\\\\\chi_{2} =\dfrac{0.06744}{0.2640} = \mathbf{0.255}$

6 0

3 years ago

Does a reaction occur when aqueous solutions of silver(i) nitrate and nickel(ii) iodide are combined?

Serggg [28]

Answer : YES, the reaction will occur.

Explanation : When we mix the aqueous solutions of Silver nitrate with Nickel iodide the reaction that occurs is stated below :-

$2 AgNO_{3} + NiI _{2} ----\ \textgreater \ Ni (NO_{3}) _{2} + 2 AgI$

A double displacement reaction occurs in this case. It means their is exchange of cations (positive ions) and anions (negative ions) between the two reactant compounds. Thus,the two new products which are obtained is the result of switching of ions between the reactants.

7 0

3 years ago

Can someone help please please

Thepotemich [5.8K]

It is your test we can't help you if you have any homework so we can

6 0

2 years ago

Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+

DIA [1.3K]

These are four questons and four answers:

Answers:

1) 7⁺

2) 1⁻

3) 3⁺

4) 5⁺

Explanation:

Question 1) Cl₂O₇:

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

2×X + 7 (-2) = 0

2X - 14 = 0

2X = +14

X = +14 /2 = + 7

Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.

Question 2) AlCl₄⁻

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

1 (+3) + 4X = - 1

+3 + 4X = - 1

4X = - 1 - 3

4X = - 4

X = - 1

Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.

Question 3) Ba(ClO₂)₂

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

+2 + 2X + 4 (-2) = 0

2X +2 - 8 = 0

2X - 6 = 0

2X = +6

X = + 3

Conclusion: the oxidation number of Cl in Ba(ClO₂)₂ is 3⁺.

Question 4) CIF₄⁺

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

X + 4(-1) = +1

X - 4 = +1

X = +1 + 4

X = + 5

Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.

6 0

3 years ago