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2 years ago

B) Wax is a solid at room temperature. Air is a gas. Air's density is 1.30kg/m³

Chemistry

2 answers:

never [62]2 years ago

7 0

The density of wax is <u>much higher</u> than the density of air because they are rigid and denser.

<h3>What is Density? </h3>

<em>Density </em>is defined as the mass of a unit volume of a material substance.

It is given by the formula,

Density = Mass / Volume

The units of density are g/cm³, kg/m³ etc.

Gases are not as dense compared to solids. This is because the particles in <u>solids</u> are very close to each other and the intermolecular gaps are very less. This makes them rigid and dense.

Thus, solids are denser than gases.

Wax is a solid and air is gas.

Hence, the density of wax is <u>much higher</u> than the density of air

Learn more on Density here: brainly.com/question/1897586

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Gwar [14]2 years ago

3 0

The density of wax much higher than the density of air because of its more mass.

<h3>Why is the density of wax much higher than the density of air?</h3>

The density of wax is much higher than the density of air because of its state of matter. We know that wax is a solid substance whereas air is a gaseous substance and we also know that solid substance is heavier than gas so wax has higher density as compared to air.

So we can conclude that the density of wax much higher than the density of air because of its more mass.

Learn more about density here: brainly.com/question/1354972

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How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?

podryga [215]

Moles of Oxygen= 2.8075 moles

<h3>Further explanation</h3>

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

= 5/2 x mol C₂H₂

= 5/2 x 1.123

= 2.8075 moles

7 0

2 years ago

Citric acid is one component of some soft drinks. Suppose that 8 L of solution are made from 0.24 g of citric

olya-2409 [2.1K]

Answer:

0.0156

Explanation:

7 0

3 years ago

Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal

german

Answer:

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

Explanation:

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $(62.5-61.7)^oC=0.8^oC$

Mass of acenapthalene = 0.515 g

Mass of $CHCl_3$ = 15.0 g = 0.015 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

$0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}$

$\text{Molar mass of acenapthalene}=155.7875 g/mol$

Let the molecule formula of the Acenapthalene be $C_{6n]H_{5n}$

$6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol$

n = 2.0

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

4 0

3 years ago

Which of the following is not a form of heat transfer?

aev [14]

D.) "Vaporization" <span>is not a form of heat transfer.

It is just conversion of the state of matter from Liquid to Vapor

Hope this helps!</span>

4 0

3 years ago

How much heat per mole is generated in the combustion of glucose?

Yuri [45]

<span>When 1 mol of glucose is burned, 2802.5 KJ of energy is released.

</span>

6 0

3 years ago