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denis-greek [22]
3 years ago
5

A geochemist in the field takes a small sample of the crystals of mineral compound X from a rock pool lined with more crystals o

f X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist dissolves the crystals in 3.00 L of distilled water. He then filters this solution and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.36 kg
1) Using only the information above can you calculate the solubility of X in water at 26 degrees Celsius? yes or no

2) If yes calculate the solubility. Round answer to 2 signifacnt digit
Chemistry
1 answer:
dedylja [7]3 years ago
3 0

Answer:

1. <u>No, you cannot calculate the solubility of X in water at 26ºC.</u>

Explanation:

You cannot calculate the solubility of X in <em>water at 26 degrees Celsius </em>because you do not know whether the solution formed by dissolving the crystals in 3.00 liters of water is saturaed or not.

The only way to determine the solubility of the compound X is by dissolving the crystals in certain (measured) amount of water and making sure that some crystals remain undissolved, as a solid on the bottom of the beaker.

Next, you should filter the solution to remove the undissolved crystals. Then, weigh the solution, evaporate, wash, dry, and weigh the crystals.

Then you have the mass of the crystals dissolved and the mass of the solution which will let you calculate the mass of pure water, and then the solubility.

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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
In principle, the equilibrium in the dehydration of an alcohol could be shifted to the right be removal of water. Why is this ta
Trava [24]

Answer:

See explanation

Explanation:

In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.

See figure 1

I hope it helps!

3 0
3 years ago
Do you know any of the answers
zzz [600]

Answer:

soorry   i ddsmfcj know

Explanation:

4 0
3 years ago
How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
2 years ago
A 1.0 M H2S solution has a pH = 3.75 at equilibrium. What is the value of Ka?
MrRa [10]
The answer is 3.16*10-8
7 0
3 years ago
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