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borishaifa [10]
3 years ago
15

What is benzene and alcohol

Chemistry
2 answers:
jek_recluse [69]3 years ago
7 0

Benzene is an organic chemical compound with the molecular formula C₆H₆. The benzene molecule is composed of six carbon atoms joined in a planar ring with one hydrogen atom attached to each. As it contains only carbon and hydrogen atoms, benzene is classed as a hydrocarbon

An alcoholic drink is a drink that contains the recreational drug ethanol, a type of alcohol produced by fermentation of grains, fruits, or other sources of sugar. The consumption of alcohol plays an important social role in many cultures.

hope mine can help you

JulijaS [17]3 years ago
6 0

Answer:

No benzene is not an alcohol

Explanation:

Any alcohol by organic chemistry standards contains a hydroxyl group.

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A certain skin lotion is a fine mixture of water and various oils. This lotion is cloudy and cannot be separated into oil and water by any filtration or centrifuge, it is a colloid mixture. The colloid consists of large molecules & ultra microscopic particles of a substance that has been dispersed by another substance. 
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3 years ago
Explain how the structure and bonding in bromine account for it's relatively low melting point.
svetlana [45]

Melting point is dependent on the intermolecular forces which means the bonds between the molecules of bromine as it is a simple molecular structure

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3 years ago
When people smoke, carbon monoxide is released into the air. In a room of volume 70 m3, air containing 5% carbon monoxide is int
KonstantinChe [14]

Answer:

\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002

Explanation:

Given that the volume is 70\ m^3

The air flows (F) at the rate of 0.002\ m^3/min. The concentration of carbon monoxide in the air is 5%.

Let Q be the amount of the carbon monoxide present in the room at time t. then,

\frac {dQ}{dt}=Rate\ of\ Q_{in}-Rate\ of\ Q_{out}

Also,

Rate\ of\ Q_{in}=5\%\ of\ 0.002=0.0001

Rate\ of\ Q_{out}=\frac {Q}{70}\times 0.002

Thus,

\frac {dQ}{dt}=0.0001-\frac {Q}{70}\times 0.002

6 0
4 years ago
Convert 0.109 mol of lithium to grams
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3 0
3 years ago
The standard enthalpy of combustion of naphthalene is –5157 kj mol-1. calculate its standard enthalpy of formation. (use data in
Artemon [7]

The combustion of naphthalene is given as:

C10H8 (s) + 12 O2 (g) --> 10CO2 (g) + 4 H2O (l)

Enthalpy of combustion ΔH = -5157 kJ/mol

Now, the enthalpy change of a reaction is given

ΔH = ∑nΔHf (products) - ∑nΔHf (reactants)

where n = number of moles

            ΔHf = enthalpy of formation

Therefore,

ΔH = [10*ΔHf(CO2) + 4*ΔHf(H2O)] - [1*ΔHf(C10H8) + 12*O2]

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ΔHf(C10H8) = 78.68 kJ/mol

4 0
4 years ago
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