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2 years ago

At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?

1 answer:

4 0

The moles of fluorine present are 71/19 = 3.74
Now, we know that one mole of gas at 273 K and 101.3 kPa (S.T.P.) occupies 22.4 liters
Volume of 3.74 moles at S.T.P = 3.74 x 22.4
Volume = 83.776 L = 83,776 mL

Now, we use Boyle's law, that for a given amount of gas,
PV = constant

P x 6843 = 101.3 x 83776
P = 1,240 kPa

You might be interested in

Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules. True False

Kitty [74]

Answer:

Explanation:

General reaction of acid in water is as follows:

HCl + H2O = H3O+ + Cl-

Thus Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules is true

3 0

1 year ago

Read 2 more answers

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

1 year ago

How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0

1 year ago

The common constituent in all acid solutions is

Oksi-84 [34.3K]

Answer:

H+/H3O , H2O

Explanation:

The ability to be a proton donor is the Bronsted-Lowry definition of acids. The Lewis definition of an acid is an electron pair acceptor, which covers molecules liKE BF3

The ability to accept a pair of electrons is what is common to all acids, not the ability to be a proton donor.

All acid solutions contain hydronium ions (H3O+), hydroxide ions (OH-) and water molecules. Each different acid solution will then have an anion that is exclusive to that acid. For example, hydrochloric acid solution will contain all of the above and chloride ions (Cl-).

All acids contain the acidic substance dissolved in water. Water naturally dissociates to a small amount, creating hydronium and hydroxide ions. But most of the water remains as water molecules.

Then when we add an acid, like HCl, the oxygen on the water attracts the hydrogen from the HCl. The electrons in the covalent bond remain with the chlorine, giving it a negative charge and thus it becomes the chloride ion (Cl-). The hydrogen now has a positive charge and as said before, is attracted to the water (specifically the lone pair of electrons on the oxygen) to create hydronium ions.

This creates extra hydronium ions, making the solution acidic. But remember, there are still water molecules, hydroxide ions and the negative ion all in solution for all acids.

8 0

1 year ago

2.A solid block with a length of 6.0 cm, a width of 3.0 cm, and a height of 3.0 cm has a mass of 146 g. What is the block’s dens

posledela

2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3

</span>

3 0

1 year ago

Read 2 more answers