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3 years ago

You need to determine the

Chemistry

1 answer:

velikii [3]3 years ago

5 0

Answer:

3.7 mL

Explanation:

Step 1: Given data

Volume of water before the chalk is added: 8.7 mL

Volume of water after the chalk is added: 12.4 mL

Volume of the chalk: ?

Step 2: Calculate the volume of the chalk

The volume of the chalk is equal to the volume of water after the chalk is added minus the volume of water before the chalk is added.

V = 12.4 mL - 8.7 mL = 3.7 mL

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Explain the term lattice energy as applied to ionic solids

Elena-2011 [213]

Answer:

The lattice energy is the energy required for the atoms of an element to attract to each other therefore making it a solid

4 0

4 years ago

When 229.0 J of energy is supplied as heat to 3.00 mol of Ar(g) at constant pressure the temperature of the sample increases by

bazaltina [42]

Answer:

The molar heat capacity at constant volume is 21.62 JK⁻¹mol⁻¹

The molar heat capacity at constant pressure is 29.93 JK⁻¹mol⁻¹

Explanation:

We can calculate the molar heat capacity at constant pressure from

$C_{p,m} = \frac{C_{p} }{n}$

Where $C_{p,m}$ is the molar heat capacity at constant pressure

${C_{p} }$ is the heat capacity at constant pressure

and $n$ is the number of moles

Also ${C_{p} }$ is given by

${C_{p} } = \frac{\Delta H}{\Delta T}$

Hence,

$C_{p,m} = \frac{C_{p} }{n}$ becomes

$C_{p,m} = \frac{\Delta H }{n \Delta T}$

From the question,

$\Delta H$ = 229.0 J

$n$ = 3.00 mol

$\Delta T$ = 2.55 K

Hence,

$C_{p,m} = \frac{\Delta H }{n \Delta T}$ becomes

$C_{p,m} = \frac{229.0}{(3.00) (2.55)}$

$C_{p,m} =$ 29.93 JK⁻¹mol⁻¹

This is the molar heat capacity at constant pressure

For, the molar heat capacity at constant volume,

From the formula

$C_{p,m} = C_{v,m} + R$

Where $C_{v,m}$ is the molar heat capacity at constant volume

and $R$ is the gas constant ( $R$ = 8.314 JK⁻¹mol⁻¹)

Then,

$C_{v,m} = C_{p,m} - R$

$C_{v,m} = 29.93 - 8.314$

$C_{v,m} =$ 21.62 JK⁻¹mol⁻¹

This is the molar heat capacity at constant volume

5 0

4 years ago

At what minimum temperature do rocks melt into lava?

navik [9.2K]

The answer is D 800 degrees Celsius

7 0

3 years ago

What is chemistry and what types of questions do chemists try to answer

Reil [10]

Answer:

Chemistry is the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.

They answer questions like, like "What makes up this material?," "How does this material change when heated or cooled,?" What happens when I mix this material with another material,?" and "What rules determine how materials change in different situations?" What makes up Chemical X?

6 0

3 years ago

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc

White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says "burn", it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

55 L of gasoline (assuming gasoline to be pure octane).
The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the conversión factor 1 L = 1000 cm³.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since density = mass/volume, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

For octane:

molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

12 g/mol . 8 + 1g/mol . 18 = 114 g/mol

According to the balanced equation reacts 1 mol of octane, which means 114 grams of it.

For oxygen:

According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is 114 g octane : 12.5 moles of oxygen

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$

4 0

4 years ago