What should you always do at the end of a lab

You are given two aqueous solutions with different ionic solutes (Solution A and Solution B). What if you are told that Solution

klio [65]

Answer:

Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.

It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,

% by mass = mass of solute/mass of solution * 100

Now the formula for molality is,

Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams

Now molality of solution A is,

m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)

m = 2.07

Now the molality of solution B is,

m = 15/58.5 * 1000/85

m = 3.02

Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).

5 0

3 years ago

In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the

notka56 [123]

Answer:

$P_T=112.4torr$

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

$P_{H_2}=x_{H_2}*P_T$

And can be solved for the total pressure as follows:

$P_T=\frac{P_{H_2}}{x_{H_2}}$

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

$x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667$

Then, we can plug in to obtain the total pressure:

$P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr$

Regards!

4 0

3 years ago

Freshwater is distributed in both time and space

Neko [114]

Answer:

Unevenly

Explanation:

Fresh water is distributed unevenly in both time and space.

7 0

2 years ago

The Thompson analogy titled "The Carpet-Seed Children Analogy" attempts to deal with the issue of failed ____________________(A)

vazorg [7]

Answer: (D) contraception

Explanation:

8 0

3 years ago

the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107

oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0

3 years ago