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2 years ago

list down some examples of solutions that we need to prepare/make in the form of unsaturated and saturated solution

Chemistry

1 answer:

Vladimir79 [104]2 years ago

5 0

Answer:

1)Carbonated water is saturated with carbon, hence it gives off carbon through bubbles.

2)Adding sugar to water until it no longer dissolves creates a saturated solution.

3)Continuing to dissolve salt in water until it will no longer dissolve creates a saturated solution.

An unsaturated tea and sugar solution would be one into which you could add more sugar and have the sugar still dissolve

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A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container

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Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

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2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

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Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

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2 years ago

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In an equilibrium mixture of the three gases, PCO = PCl2 = 2.22 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is kp=(COCl2)/(CO)(Cl2) which is $1.48*10^=x/(2.24*10-4)^2$ . So, the correct answer is <span>A) 7.34.</span>

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2 years ago

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Answer:

Explanation:

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2 years ago

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<u>Explanation:</u>

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