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Margaret [11]
2 years ago
10

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solut

ion at 18°C? [R = 0.08206 L • atm/K • mol]
Chemistry
1 answer:
kherson [118]2 years ago
3 0

Answer:

P = 17.9618 atm

Explanation:

The osmotic pressure can be calculated and treated as if we are talking about an ideal gas, and it's expression is the same:

pV = nRT

However the difference, is that instead of using moles, it use concentration so:

p = nRT/V ----> but M = n/V so

p = MRT

We have the temperature of 18 °C (K = 18+273.15 = 291.15 K) the value of R = 0.08206 L atm / K mol, so we need to calculate the concentration, and we have the mass of HCl, so we use the molar mass of HCl which is 36.45 g/mol:

n = 13.7/36.45 = 0.3759 moles

M = 0.3759/0.5 = 0.7518 M

Now that we have the concentration, let's solve for the osmotic pressure:

p = 0.7518 * 0.08206 * 291.15

<em><u>p = 17.9618 atm</u></em>

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Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
Acetone major species present when dissolved in water
jek_recluse [69]

Answer: acetone molecule ( CH₃-CO-CH₃)


Explanation:


1) Acetone is CH₃-CO-CH₃


2) That is a molecule (build up of covalent bonds).


3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.


This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.


That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).



3 0
2 years ago
Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?
balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

<h3>C 8.09 SO2 gas </h3>

8 0
2 years ago
The powder-based packet to purify water failed for a number of reasons including Select one: a. Rust b. Expense c. Sales personn
grin007 [14]

Answer:

D. Contamination

8 0
2 years ago
Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many form
ryzh [129]
Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles 3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units

1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04

385 pm = 3.85*10^(-8) cm

The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.

The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.

3 0
2 years ago
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