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3 years ago

The molecule ch3-ch2-ch2-ch2-ch2-ch2-ch3 is best described as a(n) ________.

Chemistry

1 answer:

Gwar [14]3 years ago

7 0

Given molecule:

CH3-CH2-CH2-CH2-CH2-CH2-CH3

- The molecule contain only C and H atoms, hence it is a hydrocarbon

- The C atoms in the linear chain are linked together by carbon-carbon single bonds, hence it is a saturated compound

- It has 7 C atoms and 16 H atoms, hence the molecular formula is C7H16

-The molecular formula can be generally represented as CnH2n+2 where n = 7. This corresponds to the general molecular formula for alkanes

The given molecule is best described as an <u>alkane</u>

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Please write down the definitions of elements, compounds, mixtures, homogenous mixtures, and heterogenous mixtures.

MA_775_DIABLO [31]

A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture.

A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture.

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3 years ago

Read 2 more answers

What is the lengthening of a muscle without damage? And what is the shortening of a muscle?

siniylev [52]

Answer:

Stretching is the lengthening and contraction is the shortening

Explanation:

Unless it's asking for eccentric as the stretching and concentric as the shortening

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3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

$= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

$= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

= $\frac{11659.06*1000}{62.07*10577.95}$

= 17.757 m

≅ 17.76 m

∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ = 100°C + 17.56 × 0.512

= 109.09 °C

6 0

3 years ago

A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2

mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: $2NO \Leftrightarrow N_{2}O_{2}$ (fast)

Rate expression for step 1 is as follows.

Rate = k $[NO]^{2}$

Step 2: $N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O$

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. $N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O$ (fast)

Here, $N_{2}O_{2}$ is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0

3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

3 years ago