Calculate the mass of 4.60 x 10^25 atoms of neon?

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

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

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

4 0

3 years ago

If 190dm of hydrogen gas collected at 20°c and 760mmHg .Calculate it's volume at stp (standard pressure=760mmHg

Andrei [34K]

Answer:

177.1 L

Explanation:

The excersise can be solved, by the Ideal Gases Law.

P . V = n . R . T

In first step we need to determine the moles of gas:

We convert T° from, C° to K → 20°C + 273 = 293K

We convert P from mmHg to atm → 760 mmHg = 1atm

1Dm³ = 1L → 190L

We replace: 190 L . 1 atm = n . 0.082 . 293K

(190L.atm) / 0.082 . 293K = 7.91 moles.

We replace equation at STP conditions (1 atm and 273K)

V = (n . R .T) / P

V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L

We can also make a rule of three:

At STP conditions 1 mol of gas occupies 22.4L

Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L

3 0

3 years ago

This image shows the development of what climate phenomenon?

Tpy6a [65]

Answer:

Rain shadow

Explanation:

Climate is the average weather condition of a place over a long period of time.

The image shows a rain shadow effect around a mountainous region. The type of rainfall controlled this way is known as an orographic rainfall.

On the leeward side which is the opposite side of where the rain cloud forms, there is dryness and lack of rainfall.
As moist air rises along the edge of the windward side, cloud forms.
This windward side receives a significant amount of rainfall
The windward side is hit with cold and dry wind and it is barren.

8 0

3 years ago

The following equation is balanced according to the Law of Conservation of Mass:

igor_vitrenko [27]

The answer would be:
True

6 0

2 years ago

Water is placed into a sealed jar and has a mass of 22.0 grams. The jar is placed in the freezer and the water turns to ice. The

Serhud [2]

exactly 22.0 grams

because freezing is a physical change not a chemicl change..so nothing is taken away or added

6 0

2 years ago