Ask question

Ask question

All categories

Juliette [100K]

3 years ago

8

Can someone pls explain? :'(

1 answer:

telo118 [61]3 years ago

7 0

It is 0.07 g for your answer

You might be interested in

A piece of asphalt has a volume of 5.0 cm3 and a mass of 7.5 g. What is the density of the asphalt?

strojnjashka [21]

Density = mass / volume
Density = 7.5 g / 5.0 cm3
Density = 1.5 g/cm3

6 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago

Which observation indicates that the kinetic-molecular theory has limited use for describing a certain gas? Gas particles are ac

soldier1979 [14.2K]

My answer to the question is "Gas particles are acting like tiny,solid spheres".

7 0

3 years ago

Read 2 more answers

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

3 years ago

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Varvara68 [4.7K]

The dissociation of formic acid is:

$HCOOH \rightleftharpoons HCOO^{-} + H^{+}$

The acid dissociation constant of formic acid, $k_a$ is:

$k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}$

Rearranging the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

pH = 2.75

$pH = -log[H^{+}]$

$[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}$

$pk_a = 3.75$

$k_a = 10^{-3.75} = 1.78\times 10^{-4}$

Substituting the values in the equation:

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}$

$\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}$

Hence, the ratio is $\frac{1}{10}$ .

4 0

3 years ago