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2 years ago

Which of the following does not involve colligative properties?

2 answers:

fomenos2 years ago

7 0

Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc.
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.

vodka [1.7K]2 years ago

3 0

Hey there!

The correct answer to your question is:

Option D)
<span>Calcium silicide is added to liquid steel to increase the steel’s ability to freeze.

Hope this helps!
Have a great day (:
</span>

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(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0

2 years ago

Determine the number of moles in 4.21 x 10^23 molecules of CaCl2

Paha777 [63]

<h3>Answer:</h3>

0.699 mole CaCl₂

<h3>Explanation:</h3>

To get the number of moles we use the Avogadro's number.

Avogadro's number is 6.022 x 10^23.

But, 1 mole of a compound contains 6.022 x 10^23 molecules

In this case;

we are given 4.21 × 10^23 molecules of CaCl₂

Therefore, to get the number of moles

Moles = Number of molecules ÷ Avogadro's constant

= 4.21 × 10^23 molecules ÷ 6.022 x 10^23 molecules/mole

= 0.699 mole CaCl₂

Hence, the number of moles is 0.699 mole of CaCl₂

7 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

2 years ago

Green light has a frequency of 6.01*10^14 Hz. What is the wavelength?

NemiM [27]

<span>λν = c

c= speed of light= 3.0x10^8 m/s
</span>λ=wavelength
v= frequency

Plug and Chug.

8 0

2 years ago

On top of one of the peaks in rocky mountain national park the pressure of the atmosphere is 550 torr determine the boiling poin

Dahasolnce [82]

Answer:

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

Explanation:

Step 1: Data given

Pressure = 550 torr

The heat of vaporization of water is 40.7 kJ/mol.

Step 2: Calculate boiling point

⇒ We'll use the Clausius-Clapeyron equation

ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)

ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)

⇒ with P1 = 760 torr = 1 atm

⇒ with P2 = 550 torr

⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin

⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED

ln(550/760) = 4895.4*(1/373.15 - 1/T2)

-0.3234 = 13.119 - 4895.4/T2

-13.4424= -4895.4/T2

T2 = 364.2 Kelvin = 91 °C

The boiling point of water at 550 torr will be 91 °C or 364 Kelvin

4 0

2 years ago