CHEMISTRY HIGH SCHOOL

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 1.50 g of butane?

Answers

Answer 1

Answer:

6.21 × 10²² Carbon Atoms

Solution:

Data Given:

Mass of Butane (C₄H₁₀) = 1.50 g

M.Mass of Butane = 58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 1.50 g ÷ 58.1 g.mol⁻¹

Moles = 0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

Moles = Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

Number of C₄H₁₀ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of C₄H₁₀ Molecules = 0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

Number of C₄H₁₀ Molecules = 1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

1 Molecule of C₄H₁₀ contains = 4 Atoms of Carbon

So,

1.55 × 10²² C₄H₁₀ Molecules will contain = X Atoms of Carbon

Solving for X,

X = (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X = 6.21 × 10²² Atoms of Carbon

Answer 2

Answer:

$\boxed{6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ atoms}}}$ of carbon is present in 1.50 g of butane.

Further Explanation:

Avogadro’s number indicates how many atoms or molecules a mole can have in it. In other words, it provides information about the number of units that are present in one mole of the substance. It is numerically equal to ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ . These units can 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)

Substitute 1.50 g for the given mass and 58.1 g/mol for the molar mass of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{1}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\n&= {\text{0}}{\text{.0258 mol}}\n\end{aligned}$

Since one mole of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ has ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . Therefore the formula to calculate the molecules of butane is as follows:

${\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \left( {{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}} \right)\left( {{\text{Avogadro's Number}}} \right)$ …… (2)

Substitute 0.0258 mol for the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ and ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ for Avogadro’s number in equation (2).

$\begin{aligned}{\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\mathbf{}}&=\left( {0.0258{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\n&= 1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}} \n\end{aligned}$

The chemical formula of butane is ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . This indicates one molecule of butane has four atoms of carbon. Therefore the number of carbon atoms can be calculated as follows:

$\begin{aligned}{\text{Atoms of carbon}} &= \left( {1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}} \right)\n&= 6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ C atoms}} \n\end{aligned}$

Learn more:

Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: 1.50 g, 58.1 g/mol, butane, C4H10, Avogadro’s number, $6.216*10^22$ C atoms, $1.554*10^22$ molecules, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.

Related Questions

Which of the following elements is not a metalloid?A. boron B. sulfur C. silicon D. arsenic
Why might calcium be important in the diet of many livingthings?
Balance this chemical equation: Fe(s) + O2(g) ----> FeO(s) *The 2 is in subscript
3 During a flame test, a lithium salt produces a characteristic red flame. This red color is produced when electrons in excited lithium atoms(1) are lost by the atoms (2) are gained by the atoms (3) return to lower energy states within the atoms (4) move to higher energy states within the atoms
The Steller sea lion population has been decreasing in parts of Alaska. New evidence suggests that Steller sea lions need herring as a part of their diet in order to survive. Pollock, another type of fish, also feed on herring. Which would most likely help the Steller sea lion population to increase?

Most ionic substances are soluble in water because water molecules are?

Answers

There exists the same question that has the following choices.

(1)nonpolar
(2)inorganic
(3)ionic
(4)polar

The correct answer is (4) polar. Water itself is polar. Ionic compounds are very polar, that is, one part is positively charged, the other negatively charged.

Identify the bond types between the carbon and the two oxygen Atoms in the carbon dioxide molecule.

Answers

The type of bond between the atoms in the carbon dioxide molecule is COVALENT BOND.
Covalent bond involves sharing of electrons between the atoms that are involved in the chemical reaction. The one carbon atom and the two oxygen atom that formed the carbon dioxide molecule share electrons in such a way that the outermost shell of each atom is made up of eight electrons.

a. How much energy is needed to raise the temperature of a 75 g sample of aluminum from 22.4°C to 94.6°C?

Answers

To calculate the energy required to raise the temperature of a substance, you can use the formula:

Q = m * c * ΔT

Where:
Q is the energy (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in joules per gram-degree Celsius)
ΔT is the change in temperature (in degrees Celsius)

For aluminum, the specific heat capacity is approximately 0.897 J/g°C.

Given:
Mass of aluminum (m) = 75 g
Specific heat capacity of aluminum (c) = 0.897 J/g°C
Change in temperature (ΔT) = 94.6°C - 22.4°C = 72.2°C

Substituting the values into the formula:

Q = 75 g * 0.897 J/g°C * 72.2°C

Calculating the result:

Q = 4846.35 J

Therefore, approximately 4846.35 joules of energy are needed to raise the temperature of a 75 g sample of aluminum from 22.4°C to 94.6°C.

Introduction:

Understanding the amount of energy required to change the temperature of a substance is fundamental in many fields, from chemistry and physics to engineering and everyday applications. In this case, we're looking at how much energy it takes to heat a 75 g sample of aluminum.

Specific Heat Capacity of Aluminum:

To determine the energy required, we first need to consider the specific heat capacity of aluminum. The specific heat capacity (c) is a unique property of each material and represents the amount of heat energy needed to raise the temperature of 1 gram of that substance by 1 degree Celsius (or 1 Kelvin). For aluminum, the specific heat capacity (c) is approximately 0.897 J/g°C (joules per gram per degree Celsius).

Mass of the Sample:

The next piece of the puzzle is the mass of the aluminum sample. You mentioned that it's 75 grams, so we'll use that value in our calculations.

Change in Temperature:

We're looking to raise the temperature of the aluminum from 22.4°C to 94.6°C. To find the change in temperature (ΔT), we subtract the initial temperature from the final temperature:

ΔT = 94.6°C - 22.4°C = 72.2°C

Calculating the Energy:

Now, we can use the specific heat capacity formula to calculate the energy (Q) needed to raise the temperature of the aluminum sample:

Q = m * c * ΔT

Where:

Q is the energy in joules (J).

m is the mass of the sample (75 g).

c is the specific heat capacity of aluminum (0.897 J/g°C).

ΔT is the change in temperature (72.2°C).

Plugging in these values:

Q = 75 g * 0.897 J/g°C * 72.2°C

Q ≈ 4863.15 J

Conclusion:

Therefore, approximately 4863.15 joules of energy are needed to raise the temperature of a 75 g sample of aluminum from 22.4°C to 94.6°C. This calculation is essential in various scientific and practical applications, from cooking to materials engineering, and helps us understand the energy requirements for temperature changes in different substances.

What is 16.00 kPa in atm?

Answers

There are 101.325 kPa in 1 atm. You are given 16 kPa and convert it to atm. All you need to do is divide 16 kPa to 101.325 kPa and you will get 0.16 atm

Carbon disulfide (CS2) undergoes a single displacement reaction with O2 to form CO2. If 100 grams of CS2 reacts with 38 grams of O2, what will the limiting reagent be?CS2 + O2 CO2 + 2S

Answers

The balanced chemical reaction is:

CS2 + O2 = CO2 + 2S

We are given the amounts of the reactants. These amounts will be the starting point for the calculations. First, we convert these amounts into moles.

100 g CS2 (1 mol / 76.15 g ) = 1.3132 mol CS2
38 g O2 (1 mol / 32 g) = 1.1875 mol O2

From the reaction, the mole ratio of the reactants is 1:1. Therefore, the limiting reactant is CS2.

Answer: $O_2$

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $CS_2=(100g)/(76g/mol)=1.3moles$

Moles of $O_2=(38g)/(32g/mol)=1.2moles$

For the given chemical reaction, the equation follows:

$CS_2+O_2\rightarrow CO_2+2S$

By Stoichiometry:

1 mole of oxygen reacts with 1 mole of carbon disulphide

So, 1.2 moles of oxygen reacts with = $(1)/(1)* 1.2=1.2moles$ of carbon disulphide

As, the required amount of carbon disulphide is more than the required amount. Hence, it is considered as the excess reagent. (1.3-1.2)= 0.1 mole will be left unused.

Oxygen is considered as a limiting reagent because it limits the formation of product.

Choose the correct one plz!

Answers

Answer:

Explanation:

Answer:

13 N to the right

Explanation:

Other Questions

Which is NOT an indicator of a chemical change? Heat or light is produced. State of matter changes. A new substance is formed. A gas is formed.

As the pressure on the surface of a liquid decreases, the temperature at which the liquid will boil(1) decreases(2) increases(3) remains the same

Water, H2O, frequently forms Hydrogen Bonds with other non metal atoms. Why does water do this? A: the Oxygen has a partial negative charge which attracts atoms and molecules that have a position or partial positive charge B: The Hydrogen has a partial negative charge which attracts atoms and molecules that have a positive or partial positive charge C: Because Oxygen is extremely electronegative it attracts atoms and molecules with a negative or partial negative charge D: Because Hydrogen has a very low electronegative rating it attracts atoms and molecules with a positive or partial positive charge

. Find the mass of 0.159 mol silicon dioxide.