The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 4.00 g of butane?

Answers

Answer 1

Answer: Matter is made of very tiny particles. One mole of a substance contains a given amount of particles. This number is constant for all particles- 6.022×10^23 , and is called Avogadro's constant. To get the number of particles in 4g of butane, we need to calculate the number of moles represented by 4 g, then multiply by the Avogadro's constant (Since each mole has particles equivalent to the Avogadro's constant)
Therefore, if 1 mole = 58.1g
how many moles will be equal to 4g?
we cross multiply (4x1) divide by 58.1
equals 0.068847moles
multiply by Avogadro constant = 4.1466 x 10^molecules
But remember the question is not about the number of molecules in butane, its about the number of carbon atoms.
Since there are four carbons in butane, and each has 4.1466 x 10^22 atoms
to get the number of atoms, we multiply by four
equals 1.6586 x 10^atoms

Answer 2

Answer:

$\boxed{1.657 * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}}$ of carbon is present in 4.00 g of butane.

Further Explanation:

The number of atoms or molecules that can be present in one mole of any substance is determined by a number, known as Avogadro’s number. The numerical value of Avogadro’s number is ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ . Such units can either be atoms or molecules.

The formula to calculate the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ is as follows:

${\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \frac{{{\text{Given mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}{{{\text{Molar mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}$ …… (1)

The given mass of butane is 4.00 g.

The molar mass of butane is $58.1 \text{g/mol}.$

Incorporate these values in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{4}}{\text{.00 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\n&= {\text{0}}{\text{.0688 mol}}\n\end{aligned}$

The number of molecules present in one mole of butane is ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ . So the expression to calculate the number of molecules of butane is as follows:

${\text{Molecules of butane}} = \left( {{\text{Moles of butane}}} \right)\left( {{\text{Avogadro's Number}}} \right)$ …… (2)

The number of moles of butane is 0.0688 mol.

The value of Avogadro’s number is ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ .

Incorporate these values in equation (2).

$\begin{aligned}{\text{Molecules of butane}}{\mathbf{ }}&=\left( {0.0688{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\n&= 4.143 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}\n\end{aligned}$

A molecule of butane consists of four carbon atoms in it. So the number of carbon atoms can be calculated as follows:

$\begin{aligned}{\text{Atoms of carbon}}&= \left( {4.143 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of butane}}}}} \right)\n&= 1.657 * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ C atoms}}\n\end{aligned}$

Learn more:

How many grams of potassium were in the fertilizer? brainly.com/question/5105904
Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: $4.00 \text{g}, 58.1 \text{g/mol},$ butane, C4H10, Avogadro’s number, $1.657*10^22$ C atoms, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.

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The molarity of an aqueous sodium phosphate solution is 0.650 M. What is the molality of sodium ions present in this solution? The density of solution is 1.43 g/mL.

Answers

Answer:

molality of sodium ions is 1.473 m

Explanation:

Molarity is moles of solute per litre of solution

Molality is moles of solute per kg of solvent.

The volume of solution = 1 L

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In a 0.01 M solution of HCl, Litmus will be

Answers

In a 0.01 M solution of HCl, Litmus will be red. Litmus paper will turn into red in acidic conditions. Hydrochloric acid is an acid. Litmus is an indicator for acidity and alkalinity made from inchens.

Explanation:

A solution with pH less than 7 will be acidic in nature and it will change color of blue litmus into red.

For example, HCl is a strong acid as it dissociates to give hydrogen ions and chlorine ions.

A solution with pH more than 7 will be basic in nature and it will change color of red litmus into blue.

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A sample of CO2(s) and a sample of CO2(g) differ in their1) chemical compositions
2) empirical formulas
3) molecular structures
4) physical properties

Answers

The correct answer is D.
CO2(s) is a solid.
CO2(g) means it is gas.

So they differ in their physical properties.
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Answers

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Explanation:

A weak acid, HA, has a pKapKa of 4.3574.357 . If a solution of this acid has a pH of 4.0054.005 , what percentage of the acid is not ionized? Assume all H+H+ in the solution came from the ionization of HA.

Answers

Answer:

69.3%

Explanation:

The question should read as follows:

A weak acid, HA, has a pKa of 4.357. If a solution of this acid has a pH of 4.005, what percentage of the acid is not ionized? Assume all H⁺ in the solution came from the ionization of HA.

The Henderson-Hasselbalch equation relates the pKa and pH of a solution to the ratio of ionized (A⁻) and unionized (HA) forms of a weak acid:

pH = pKa + log ([A⁻]/[HA])

Substituting and solving for [A⁻]/[HA]:

4.005 = 4.3574 = log([A⁻]/[HA])

-0.3524 = log([A⁻]/[HA])

[A⁻]/[HA] = 0.444/1

The percentage of acid that is not ionized (i.e. the percentage of acid in the HA form) is calculated:

[HA]/([A⁻] + [HA]) x 100% = 1/(1+0.444) x 100% = 69.3%

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Answers

0K = -273C
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Answer:

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Explanation:

how to convert 37 Celsius to Kevin is to add 273.15 to 37 then you get 310.15 Kevin then you round that number to the nearest whole and you get 310 and that is your answer. please give me brainiest

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