Question 5 OT 5

Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =

julia-pushkina [17]

The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
Density of water = 1 g/mL
Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

Mass of water = 400 g
Molar mass of water = 18.02 g/mol
Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>

Mass of KBr = 0.3 g
Molar mass of KBr = 119 g/mol
Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>

Mole of KBr = 0.0025 mole
Mole of water = 22.2 moles
Total mole = 0.0025 + 22.2 = 22.2025 moles
Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

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6 0

1 year ago

HURRY FAST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

miss Akunina [59]

B. The sand increases friction by increasing roughness.

8 0

3 years ago

Read 2 more answers

Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N

Rina8888 [55]

Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

7 0

3 years ago

Simply parmanent tissue real life application please help it's for my project

WITCHER [35]

The simple tissues are parenchyma, sclerenchyma and collenchyma. Chlorenchyma is a parenchyma, having chloroplast. It is a simple permanent tissue, having chloroplast.

8 0

3 years ago

Interstellar space has an average temperature of about 10 K , and an average density of hydrogen atoms of about one hydrogen ato

Bas_tet [7]

Answer:

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.Explanation:

The mean free path equation is given as:

$\lambda =\frac{1}{\sqrt{2}\pi d^2 n}$

Where"

d = diameter of hydrogen atom in meters

n = number of molecules per unit volume

We are given: d = 100 pm = $100\times 10^{-12}= 10^{-10} m$

$n = 1 m^{-3}$

$\lambda =\frac{1}{\sqrt{2}\pi (10^{-10} m)^2\times 1 m^{-3}}$

$\lambda =2.2508\times 10^{19} m$

$2.2508\times 10^{19} m$ meter the mean free path of hydrogen atoms in interstellar space.

7 0

3 years ago