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Tju [1.3M]
3 years ago
11

How does the mass number of an element differ from its average atomic mass?

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
4 0
The mass number is the total number of protons and neutrons in a nucleus of an element. The average atomic mass is the sum of the masses of an element's isotopes, each multiplied by their abundance.
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classify the following compounds as ionic (metal + nonmetal) , covalent (nonmetal + nonmetal) or both (compound containing a pol
musickatia [10]

Answer:

1. ionic

2. covalent

3. covalent

4. both

Explanation:

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Define intensive properties​
natka813 [3]

Answer:

An intensive property is a property of matter that does not change as the amount of matter changes. It is a bulk property, which means it is a physical property that is not dependent on the size or mass of a sample.

In contrast, an extensive property is one that does depend on sample size. Examples of extensive properties include mass and volume.

5 0
3 years ago
Identify a reason that chemical reactions release energy during the reaction process.
Bess [88]

Answer:

option b

Explanation:

When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.

5 0
3 years ago
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How many atoms are in 1.5 moles of iron?
alina1380 [7]

Answer:

<em>One mole of anything is 6.022×1023 everythings, including atoms. Multiply the moles Fe by 6.022×1023 atoms/mol. 3.77mol Fe ×6.022×1023atoms Fe1mol Fe =2.27×1024 atoms Fe rounded to three significant figures.</em>

3 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
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