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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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When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m

Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

$500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2$

$42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl$

Have a nice day!

#LearnwithBrainly

7 0

2 years ago

A semipermeable membrane separates two solutions of different concentrations. Identify which of the following statements are cor

GenaCL600 [577]

Answer:

The correct options are A, and C.

Explanation:

Osmosis: It is defined as the movement of solvent with the help of selectively semipermeable membrane into a region of where high solute concentration is present to equalize the concentration of solute on the both compartments.

Reverse osmosis: It is defined as the movement of the high concentration solvent is forced onto the lighter concentration side with the help of mechanical pressure.

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2 years ago

Read 2 more answers

The hybrid orbital set used by the central atom in h2co is ________. the hybrid orbital set used by the central atom in h2co is

Feliz [49]

The Lewis structure for H₂CO is shown in the attached picture. The central atom is the carbon. However, I'm not sure which bond you're referring to. There can be two answers. The two C-H bonds are sp³ hybridized because it is a single bond. The C=O bond is sp² hybridized because it is a double bond.

5 0

2 years ago

2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0

2 years ago

To prepare a 2 M solution of potassium nitrate (KNO3), which quantities must be measured? The mass of the and the volume of the

Olin [163]

Answer:

The mass of the solute and the volume of the solution.

Explanation:

Hello,

In this case, given the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.

Best regards.

4 0

2 years ago

Read 2 more answers

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

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