If the heat of fusion of water is 80 cal/g, the amount of heat energy required to change 15.0 grams of ice at 0°C to

Which example does not involve knowledge of chemistry?

Trava [24]

I would honestly say C. All others have ingredients in them that makes them work, the combination of stuff to put out the fire, film is made from a certain material and drugs are an ovbious mixture. Looking at the stars is the only thing you don't need chemistry for.

6 0

3 years ago

Read 2 more answers

How many grams of KNO3 would recrystallize from 100 g of water, if a saturated solution were cooled from 50 degrees celcius to 2

Lelechka [254]

Answer:

$m_{cryst}=58g$

Explanation:

Hello there!

In this case, according to the attached solubility chart, it is possible for us to realize that about 88 grams of KNO3 are soluble at 50 °C but just 30 grams are soluble at 20 °C in the same 100 g of water.

In such a way, the crystalized mass of this solute can be calculated by subtracting the mass at 50 °C and the mass at 20 °C:

$m_{cryst}=88g-30g\\\\m_{cryst}=58g$

Best regards!

8 0

3 years ago

If the enthalpy value for a reaction is negative, what does that indicate about the reaction?

likoan [24]

This indicates that the reaction is exothermic meaning that it releases heat/energy

7 0

3 years ago

Read 2 more answers

You have 4 moles of a gas in a 50 L container held at 2 atm pressure. Currently the temperature is 27 ºC. R = 0.0821 L*atm/(mol*

Mariulka [41]

The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.

• What is the temperature in Kelvins?
You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.

• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.

 If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
PV=nRT
P= nRT/V
P= 6 moles* 0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm

3 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago