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Vsevolod [243]
4 years ago
12

Which formulas represent two polar molecules?

Chemistry
1 answer:
Tresset [83]4 years ago
7 0
The correct answer is shown in option 3. Water and hydrochloric acid are polar molecules. These molecules are polar because of the presence of bonds that are partially ionic or polar covalent bonds. Other examples are hydrogen fluoride and ammonia. 
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Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF
algol [13]

The molar concentration of the original HF  solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342

7 0
3 years ago
Worksheet Ch.3: % Composition, Molarity, and Other Units for
Ann [662]

The percentage composition by mass of the elements in the compound are:

Ai. The percentage composition by mass of Na in 3.65 g of NaF is 54.8%

Aii. The percentage composition by mass of F in 3.65 g of NaF is 45.2%

Bi. The percentage composition by mass of Zinc in the compound is 67.1%

Bii. The percentage composition by mass of Sulfur in the compound is 32.9%

Percentage composition by mass of an element in a compound can be obtained by using the following formula:Percentage = \frac{mass of Element}{mass of compound}  * 100

Ai. Determination of the percentage of Na in 3.65 g of NaF

Mass of Na = 2 g

Mass of NaF = 3.65 g

<h3>Percentage of Na =? </h3>

Percentage = \frac{mass of element }{mass of compound} * 100\\\\Percentage of Na = \frac{2}{3.65} * 100

<h3>Percentage of Na = 54.8%</h3>

Aii. Determination of the percentage of F in 3.65 g of NaF

Mass of F = 1.65 g

Mass of NaF = 3.65 g

<h3>Percentage of F =? </h3>

Percentage = \frac{mass of element }{mass of compound } * 100\\\\Percentage of F = \frac{1.65}{3.65} * 100\\\\

<h3>Percentage of F = 45.2%</h3>

Bi. Determination of the percentage composition of Zn in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

<h3>Percentage of Zn =? </h3>

Percentage = \frac{mass of element}{mass of compound} * 100\\\\Percentage of Zn = \frac{32.69}{48.72} *100\\\\

<h3>Percentage of Zn = 67.1%</h3>

Bii. Determination of the percentage composition of Sulphur in 48.72 g of the compound

Mass of Zn = 32.69 g

Mass of compound = 48.72 g

Mass of S = 48.72 – 32.69 g

Mass of S = 16.03 g

<h3>Percentage of S =? </h3>

Percentage = \frac{mass of element}{mass of compound}  * 100\\\\Percentage of S = \frac{16.03}{48.72}  * 100\\\\

<h3>Percentage of S = 32.9%</h3>

Learn more: brainly.com/question/1350382

3 0
3 years ago
Which is an example of radiation?
SpyIntel [72]

Answer:

Number 3.

Explanation: The sun produces radiation in the car, which then becomes hot because much radiation is coming in, but little goes out.

4 0
3 years ago
What are the steps to evaporation but without heat just evaporation steps to follow
densk [106]
First, the sun shines liquid (ocean) Next, the water evaporates 
8 0
4 years ago
Read 2 more answers
How many grams of copper would have the same number of atoms as 139.82 grams of argon
lianna [129]

Answer:

Lesson 3 Quiz ReturntoAssessmentList Part 1 of 4 - 4.0/ 8.0 Points Question 1 of ... in the balloon containing regular air D.Each balloon has an equal number of atoms. ... The purity of those pre-1982 pennies are really about 95% copper (5% zinc ... that had a mass of 3.11 grams, how many atoms of copper would you have?

Explanation:

7 0
3 years ago
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