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TiliK225 [7]
3 years ago
12

One's motivation can lead people to interpret the same situation differently.

Chemistry
2 answers:
lina2011 [118]3 years ago
7 0
The answer to the question is true.
Len [333]3 years ago
3 0

Answer: the answer is true

Explanation:

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What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (Kf = 1.86°C/m) –1.86°C –7.44°C –5.58°C –3.72°C
tresset_1 [31]

Answer:

  • Last choice: <em><u>- 3.72°C</u></em>

Explanation:

The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.

The equation to predict the freezing point depression in a solvent is:

  • ΔTf = Kf × m × i

Where,

  • ΔTf is the freezing point depression of the solvent,
  • m is the molality,
  • Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.

The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.

Download pdf
6 0
3 years ago
How many Ni atoms are in 3.6 mol of Ni
Sliva [168]
I believe that the answer is 1.8^24 of Ni atoms in 3.6 mol of Ni.

Hope this helps. :)

6 0
3 years ago
Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be
amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0
3 years ago
The various atomic masses of the same element are called neutrino positron critical mass isotopes
Mariana [72]
The various atomic masses of the same element are called<u> isotopes</u>. They have the same number of protons but different number of neutrons.
3 0
3 years ago
Read 2 more answers
Which of the following methods can be used to identify the suspected use of heat lamps in a house without entering the house?
LuckyWell [14K]

Answer:

b I see some of ask you to do it

7 0
3 years ago
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