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3 years ago

The lower the ignition temperature, better the fuel. True or false?Why?

Chemistry

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

True

Explanation:

This means the fuel is highly flammable hence it has a low ignition temperature. This, therefore, means it burns fast without putting in a lot of energy to ignite it. Even in cars, therefore, the compression ratios, in engines using these of these fuels, is lesser.

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Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 o

Elenna [48]

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q=?
m= 45 g
c= 4.184 $\frac{J}{g*C}$
ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J

8 0

3 years ago

PLEASE HURRY

Aloiza [94]

Answer:

C. The conditions in which a substance exists in a certain phase.

Explanation:

7 0

3 years ago

Draw the structure of the compound C9H10O2 that might exhibit the 13C-NMR spectrum below. Impurity peaks are omitted from the pe

zhenek [66]

Complete question

Draw the structure of the compound $C_{9}H_{10}O_{2}$ that exhibits the $^{13}C-NMR$ spectrum shown on the first uploaded image(on the second and third uploaded image is closer look at the $^{13}C-NMR$ spectrum ) . Impurity peaks are omitted from the peak list. The triplet at 77 ppm is $CDC_{l3}$ .

Answer:

The structure that might exhibit the $^{13}C-NMR$ spectrum is shown on the fifth uploaded image

Explanation:

In order to get a good understanding of the answer above we need to know that

• Proton NMR spectrum: proton NMR spectroscopy is one of the techniques, which is useful to predict the structure of the compound.

• In $^{\rm{1}}{\rm{H NMR}}$ spectroscopy, peaks are observed at the point where the wavelength of proton nuclei matched to substance nuclei wavelength.

• In same manner there are other spectroscopies are present like $^{{\rm{13}}}{\rm{C NMR}}$

, IR and mass spectroscopy.

• Infrared spectroscopy is used to determine the functional groups present in a compound.

• Infrared bands observed when there is change in dipole moment occurs between the atoms. Infrared bands describe about the bond stretches, which causes due to the dipole moment present in the molecule.

Fundamentals

Double bond equivalence: number of double bonds or number of rings in the structure can be calculated by using double bond equivalence formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ number of carbon atoms

$N_{H}=$ number of hydrogen atoms

$N_{Cl} =$ number of chlorine atoms

$N_{N}=$ number of nitrogen atoms

The table for the $^{{\rm{13}}}{\rm{C NMR}}$ is shown on the fourth uploaded image

Molecular formula of the compound is ${{\rm{C}}_9}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{2}}}$

Double bond equivalence of the compound is calculated below.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ 9

$N_{H}=$ 10

$N_{Cl} =$ 0

$N_{N}=$ 0

$DBE = N_{c} + 1 - (\frac{(10+0) -0}{2}})$

$DBE =5$

Therefore, the compound has five double bonds, which indicating that there is chance of getting aromatic rings too.

Note:

Double bond equivalence is calculated as 5 which indicates that there are 5 double bond (may rings) in the structure of the compound.

Double bond equivalence is calculated by using this formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

13C NMR data of the compound is explained below.

1.A peak at 166.5 ppm, which indicates the presence of ester group

2.Peaks at 132.7, 130.5, 129.5, 128.2 ppm (aromatic carbons) are indicating a mono substituted aromatic ring

3.A peak at 60.9 ppm means methylene group attached to oxygen atom

4.A peak at 14.3 ppm, which indicates the presence of methyl group

According to this data and the using the double bond equivalence, structure of the compound shown on the fifth uploaded image .

Note:

According to given spectral data, structure of the compound has been predicted. It is clear that; -ester functional group is present in the structure because there is a peak at 166.5ppm. According to given proton $^{13}C NMR$ data, above structure has been drawn. Therefore, the compound is ethyl benzoate.

7 0

4 years ago

Scientists observe patterns in how molecules interact with each other. At the same temperature, for example, liquids have more a

Alla [95]

Answer:

Hello

I am Parth

Can we became friends

and ur intro plz

7 0

3 years ago

Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read

lapo4ka [179]

Answer:

1) 18.91 mL

2) 2.35 %

Explanation:

1. The volume of base that was required is equal to the difference between the reading at the end of the titration and the reading at the beginning:

V = 20.95 mL - 2.04 mL = 18.91 mL

2. The mass percent can be written as:

Mass Percent = Mass solute / Total mass * 100%

For this problem:

Mass solute = 2.28 g

Total mass = 96.92 g

We input the data and calculate the mass percent:

% mass = 2.28/96.92 * 100% = 2.35 %

7 0

3 years ago