What kind of frog is this!?

Why is a solution that is 0.10M HCN and 0.10M LiCN considered a good buffer system?

const2013 [10]

Because of the common ion effect. Adding a common ion decreases the solubility of the salt which is the intention of a buffer.

8 0

3 years ago

What is the boiling point of an aqueous solution that has a vapor pressure of 23.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘

UNO [17]

Answer:

Boiling point of the solution is 100.964°C

Explanation:

In this problem, first, you must use Raoult's law to calculate molality of the solution. When you find the molality you can obtain the boiling point elevation because of the effect of the solute in the solution (Colligative properties).

Using Raoult's law:

Psol = Xwater × P°water.

As vapour pressure of the solution is 23.0torr and for the pure water is 23.78torr:

23.0torr= Xwater × 23.78torr.

0.9672 = Xwater.

The mole fraction of water is:

$0.9672 = \frac{X_{H_2O}}{X_{H_2O}+X_{solute}}$

Also,

$1 = X_{H_2O}+X_{solute}$

You can assume moles of water are 0.9672 and moles of solute are 1- 0.9672 = 0.0328 moles

Molality is defined as the ratio between moles of solute (0.0328moles) and kg of solvent. kg of solvent are:

$09672mol *\frac{18.01g}{1mol}* \frac{1kg}{1000g} = 0.01742kg$

Molality of the solution is:

0.0328mol Solute / 0.01742kg = 1.883m

Boiling point elevation formula is:

ΔT = Kb×m×i

<em>Where ΔT is how many °C increase the boiling point regard to pure solvent, Kb is a constant (0.512°C/m for water), m molality (1.883m) and i is Van't Hoff factor (Assuming a i=1).</em>

Replacing:

ΔT = 0.512°C/m×1.882m×1

ΔT = 0.964°C

As the boiling point of water is 100°C,

<h3>Boiling point of the solution is 100.964°C</h3>

<em />

5 0

3 years ago

Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o

konstantin123 [22]

Answer:

- From octane: $m_{CO_2}=1.54x10^5gCO_2$

- From ethanol: $m_{CO_2}=9.57x10^4gCO_2$

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

$C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O$

Thus, the produced mass of carbon dioxide is:

$m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2$

Now, for ethanol:

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

$m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2$

Best regards.

3 0

3 years ago

I’m driving down the highway at 75.0 miles/hour, how fast is that in m/seconds?

Aliun [14]

Answer:

1.25 miles/ 1 minute

60x1.25=7.50

Explanation:

4 0

3 years ago

at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were

Nataly [62]

<u>Answer:</u> The percent change in volume will be 25 %

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$