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mestny [16]
2 years ago
14

The molar mass of KCl is 74. 55 g/mol. If 8. 45 g KCl are dissolved in 0. 750 L of solution, what is the molarity of the solutio

n? Use Molarity equals StartFraction moles of solute over liters of solution EndFraction. 0. 113 M 0. 151 M 6. 62 M 11. 3 M.
Chemistry
1 answer:
rosijanka [135]2 years ago
4 0

Molarity is the strength of the solution or the ratio of the moles and the volume of the solution. The molarity of the solution is 0.1516 mol per Liter.

<h3>What is molarity?</h3>

Molarity is the molar concentration of the solution that tells about the amount of the solute or the substance dissolved in the given amount of the solution.

Given,

  • Mass of potassium chloride = 8.45 g
  • Molar mass of potassium chloride = 74. 55 g/mol
  • Volume of the solution = 0.750 L

Moles can be calculated by:

\begin{aligned}\rm Moles &= \rm \dfrac{Mass }{\rm Molar \;mass}\\\\&= \dfrac{8.45}{74.55}\\\\&= 0.1133\;\rm  mol\end{aligned}

Calculate molarity by:

\rm Molarity = \dfrac{moles}{\text{Volume in L}}

Substituting values in the equation:

\begin{aligned}\rm  Molarity &= \dfrac{0.1133}{0.750}\\\\&= 0.1516 \;\rm mol\;L^{-1}\end{aligned}

Therefore, the molarity of the solution is 0.1516 mol per Liter.

Learn more about molarity here:

brainly.com/question/2560307

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I always thought it was a mixture but can also be a compound 

5 0
3 years ago
The melting point of the metal potassium is 63.5c, while that of titanium is 1660c. what explanation can be given for this great
kiruha [24]

The melting point of potassium = 63.5^{0}C

Melting point of titanium = 1660^{0}C

Titanium has a stronger metallic bonding compared to potassium. Titanium being a transition metal has greater number of valence electrons (4 valence electrons) contributing to the valence electron sea compared to potassium which has only one valence electron. The atomic size of Titanium much lower than that of potassium, so the bonding between Titanium atoms is stronger than that of potassium. Hence, the melting point of Titanium is much higher than that of potassium.

3 0
3 years ago
Which organisms in the food web shown compete for the same food source in this environment?
11111nata11111 [884]

Answer:

there is no photo

Explanation:

3 0
3 years ago
Data was collected by students in an acid base titration lab. They used 1.63 M Ca(OH)2(AQ)
jenyasd209 [6]

Answer:

3.8 M

Explanation:

Volume of acid used VA= 57.0 - 37.5 = 19.5 ml

Volume of base used VB= 67.8 - 45.0 = 22.8 ml

Equation of the reaction

2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)

Number of moles of acid NA= 2

Number of moles of base NB= 1

Concentration of acid CA= ???

Concentration of base CB= 1.63 M

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA= CBVBNA/VANB

CA= 1.63 × 22.8 × 2/ 19.5 × 1

CA= 3.8 M

HENCE THE MOLARITY OF THE ACID IS 3.8 M.

6 0
3 years ago
How many molecules of H2O and O2 is present in 8.5g of H2O?​
Igoryamba

Answer:

2.8 x 10²³ molecules H₂O

1.4 x 10²³ molecules O₂

Explanation:

First, you will need the balanced chemical equation for the formation of water:

2H₂ + O₂ -> 2H₂O

This will help in determining the mole ratios between water and oxygen, which we will need later.

Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O

Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O

Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂

Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂

3 0
3 years ago
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