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2 years ago

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Examples of molecular compound

1 answer:

Marta_Voda [28]2 years ago

7 0

Answer:

Molecular compounds are inorganic compounds that take the form of discrete molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2). These compounds are very different from ionic compounds like sodium chloride (NaCl).

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Mechanical waves need mediums. or else it isn't a mechanical wave

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3 years ago

The statement 2 2 = 4 would be considered a _____ in the science world

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The statement would be considered a law.

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Hydrogen is 99% 1H, 0.8% 2H, and 0.2% 3H. Calculate its average atomic mass.

AlekseyPX

The average atomic mass of hydrogen (99% H-1, 0.8% H-2, and 0.2% H-3) is 1.012 amu.

The average atomic mass of hydrogen can be calculated with the following equation:

$A = m_{H-1}\%_{H-1} + m_{H-2}\%_{H-2} + m_{H-3}\%_{H-3}$

Where:

$m_{H-1}$ : is the mass of protium (H-1) = 1

$m_{H-2}$ : is the <em>mass</em> of deuterium (H-2) = 2

$m_{H-3}$ : is the <em>mass </em>of tritium (H-3) = 3

$\%_{H-1}$ : is the abundance percent of H-1 = 99%

$\%_{H-2}$ : is the <em>abundance percent</em> of H-2 = 0.8%

$\%_{H-3}$ : is the <em>abundance percent</em> of H-3 = 0.2%

The average atomic mass is:

$A = 1*99\% + 2*0.8\% + 3*0.2\%$

Changing all the <u>percent values</u> into <u>decimal ones</u>, we have:

$A = 0.99 + 2*0.008 + 3*0.002 = 1.012$

Therefore, the average atomic mass of hydrogen is 1.012 amu.

Learn more about average atomic mass here brainly.com/question/8575403?referrer=searchResults

I hope it helps you!

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2 years ago

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vodomira [7]

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3 years ago

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n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph

olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

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