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3 years ago

An object with a mass of 7 g and a volume of 10 mL is most likely what substance?

Chemistry

1 answer:

Archy [21]3 years ago

8 0

Answer:

Ethanol

Explanation:

Given parameters:

Mass of the object = 7g

volume of the object = 10mL

Unknown:

The kind of substance it is = ?

Solution:

Since from the parameters, we can determine the density of the object. Density is an intensive property of substances.

So, we simply find the density and check a density table for the type of substance it is.

Density is the mass per unit volume.

Density = $\frac{mass}{volume}$

Let us take the volume to cm³;

10mL = 10cm³

Density = $\frac{7}{10}$ = 0.7g/cm³

This is most likely ethanol

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The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

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

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

3 years ago

Rearrange the equation for the area of a rectangle (A=l x w) to solve for length, L

Mariana [72]

Not sure if this is what you mean but l= A/w
(length equals area divided by width)

4 0

3 years ago

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Darya [45]

Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

k = rate constant = $0.610M^{-1}s^{-1}$

$a_0$ = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

$\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}$

$t=3.120s$

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

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3 years ago

Carbon hexafluroide formula

ikadub [295]

The formula for Hexafluoride is F6S.

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3 years ago

Propanoic acid has a boiling point of 141 °C, propanamide has a boiling point of 213 °C, and propanal has a boiling point of 48

deff fn [24]

Explanation:

As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.

As in $-NH_{2}$ , the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.

Hence, these bonds are so strong that high heat needs to given to break them.

A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.

Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.

In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.

As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.

4 0

3 years ago