Ask question

Ask question

All categories

2 years ago

14

Rank the following aqueous solutions in order of decreasing

1 answer:

True [87]2 years ago

6 0

0.04 m urea > 0.03 m CuSO4 > 0.01 m AgNO3.

<h3>What is colligative property?</h3>

Colligative properties in chemistry are those of a solution that depend on the amount of solute particles to solvent particles in a solution rather than the makeup of the individual particles.
The number ratio can be connected to the several units used to measure a solution's concentration, including molarity, molality, normalcy (chemistry), etc.
For ideal solutions, which are ones that have thermodynamic properties similar to those of an ideal gas, and for diluted actual solutions, the assumption that solution properties are independent of the type of solute particles is exact.

<h3>What factors affect colligative properties?</h3>

A property of a solution is said to be collative if it depends simply on the proportion of solute to solvent particles in the solution and not on the nature of the solute.

Learn more about colligative properties here:

brainly.com/question/10323760

#SPJ4

You might be interested in

1. Antarctica is a frozen land, so cold and icy that no trees can grow there. Yet scientists have discovered fossils(remains pre

algol [13]

Answer:

This means that Antarctica once had a warmer climate.

Explanation:

Trees usually grow in warm climates, and Antarctica has little to no plant life.

8 0

2 years ago

We have air (21% O2 and 79% N2) at 23 bar and 30 C. 4. What is the ideal molar volume (m^3/kmol)? a. b. What is the Z factor? Wh

k0ka [10]

Answer:

The ideal molar volume is $\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The Z factor is $Z = 0.09997$

The real molar volume is $\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

Explanation:

From the question we are told that

The pressure is $P = 23 \ bar = 23 *10^5 Pa$

The temperature is $T = 30 ^ oC = 303 \ K$

According to the ideal gas equation we have that

$PV = nRT$

=> $\frac{V}{n}=V_z= \frac{RT}{P}$

Where $\frac{V}{n }$ is the molar volume and R is the gas constant with value

$R = 8.314 \ m^3 \cdot Pa \cdot K^{-1}\cdot mol^{-1}$

substituting values

$\frac{V}{n} =V_z= \frac{ 8.314 * 303}{23 *10^{5}}$

$\frac{V}{n} =V_z= 0.001095 \ m^3/mol$

The compressibility factor of the gas is mathematically represented as

$Z = \frac{P * V_z}{RT}$

substituting values

$Z = \frac{23 *10^{5} * 0.001095}{8.314 * 303}$

$Z = 0.09997$

Now the real molar volume is evaluated as

$\frac{V_r}{n} = V_k= \frac{Z * RT }{P}$

substituting values

$\frac{V_r}{n} = V_k= \frac{0.09997 * 8.314 * 303}{23 *10^{5}}$

$\frac{V_r}{n} = V_k= 0.0001095\ \frac{m^3}{mol}$

8 0

3 years ago

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What

Fynjy0 [20]

Answer:

$l=3.5*10^{-10}m$

Explanation:

From the question we are told that:

1st Energy $E_1=12eV=4(3eV)$

2nd Energy $E_2=27eV=9(3eV)$

3rd Energy $E_3=48eV=16(3eV)$

Generally the equation for Energy E for electron in one dimensional box at ground state $E_0$ is mathematically given by

$E_0=\frac{h}{8ml^2}$

$E_0=\frac{h}{8ml^2}$

Therefore Length a is mathematically given as

$l=\sqrt{\frac{h^2}{8mE_0} }$

$l=\sqrt{\frac{(6.625*10^{-34})^2}{8(9.1*19^{-31}{(3eV(1.6*10^{-19}))}}$

$l=3.5*10^{-10}m$

8 0

2 years ago

Do you think there could be any disadvantages to wireless charging? If so, what could they be?

Montano1993 [528]

Answer:you could forget to charge the wireless charger but I have one and there’s never a problem so you probably don’t have to worry

Explanation:

4 0

3 years ago

If 41.38 g of sodium combines with

Arlecino [84]

I think it is 3 because Sus wiw word if rn

3 0

2 years ago