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Vedmedyk [2.9K]
3 years ago
7

The first step in coal formation process is the formation of:

Chemistry
1 answer:
Reil [10]3 years ago
4 0
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What are the standard ways to make a solution of known concentration? (select all that apply.)?
MatroZZZ [7]

1- Molar solutions: based on number of moles of chemical in 1 litre of solution

2- Weight % solution:  the weight of chemical divided by the total weight of the solution (chemical + water) and multiplied by 100.

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How many moles of Boron are present in 20g of borax (sodium tetraborate)?​
stira [4]
20 grams of borax contains (20.0g) / (201 g mol -1) =0.10 mol of borax.

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5 0
3 years ago
In this experiment you are directed to add a limited amount of NaOH (aq) and then an excess amount. A similar strategy will be u
Verizon [17]

The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.

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Learn more: brainly.com/question/1527403

7 0
3 years ago
How many grams of aluminum can be heated from 94.2°C to 120.5°C if 593.0 joules are applied? The specific heat of aluminum is 0.
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7 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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