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3 years ago

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PLEASE HELP! THIS IS DUE VERY SOON! Whats a few examples of comparing and contrasting solutions with different concentrations of

a solute? It's for my science class. I really need help.

1 answer:

Assoli18 [71]3 years ago

4 0

In osmosis, water moves from areas of low concentration of solute to areas of high concentration of solute.

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In what area of the titration curve for phosphoric acid (h3po4) is the molecule fully deprotonated?

Arada [10]

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

Initial pH

This is determined by the most acidic of the Ka values and the initial

concentration of the acid. (Same as a monoprotic acid)

Half-way points

At each halfway point, the pH = pKa of the group you are titrating. At this point in

the titration curve, we are in a buffering region, and the curve will be relatively

flat.

Equivalence points

At each equivalence point, the pH is the average of the pKa values above and

below. At the last equivalence point (the endpoint), the pH is determined by the

Kb of the conjugate base of the weakest acid.

Plotting the titration of 100 mL of 0.10 M phosphoric acid with 1.0 M NaOH.

H3PO4 + H2O  H2PO4

-

+ H3O+

Ka1 = 7.5 x 10-3

H2PO4

-

+ H2O  HPO4

2- + H3O+

Ka2 = 6.2 x 10-8

HPO4

2- + H2O  PO4

3- + H3O+

Ka3 = 4.2 x 10-13

Plot these points and connect them to determine the titration curve of phosphoric acid. The curve should be relatively flat around each of the halfway points when we are in a buffering region.

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

For more information on the titration curve click on the link below:

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4 0

2 years ago

Si unit of heat and energy

lara31 [8.8K]

The answer is Joule.

7 0

3 years ago

Read 2 more answers

Sugar is easily soluble in water and has a molar mass of 342.30 g/mol. what is the molar concentration of a 252.6 ml aqueous sol

lions [1.4K]

0.811 M is the molar concentration of a 252.6 ml aqueous solution prepared with 70.3 g of sugar.

<h3>Define molarity of a solution.</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Given data:

252.6 mL = 0.2526 L

Next, we shall determine the number of moles of the sugar. This can be obtained as follow:

Molar mass of sugar = 342.30 g/mol.

Mass of sugar = 70.3 g

Mole of sugar =?

Mole = $\frac{mass}{molar \;mass}$

Mole = $\frac{70.3 g}{342.30}$

Mole of sugar = 0.205 mole

Finally, we shall determine the molar concentration of the sugar. This can be obtained as follow:

Mole of sugar = 0.205 mole

Volume = 0.2526 L

Molarity =?

Molarity = $\frac{Moles \;solute}{Volume of solution in litre}$

Molarity = $\frac{0.205 mole}{ 0.2526 L}$

Molarity = 0.811 M

Therefore, the molar concentration of the solution is 0.811 M

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8 0

2 years ago

A balloon that contains 0.750 moles of gas has a volume of 16.8 l. if the balloon expands to a volume of 25.4 l at a constant pr

fomenos

The answer is 1.13 mol

6 0

3 years ago

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2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g) H = -827.4 kJ What is the volume of sulfur dioxide produced at STP if 975 kJ of heat are l

Akimi4 [234]

Answer:

52.79 dm3 of SO2 will be produced when 975kJ/mol of heat are liberated.

Explanation:

In the reaction given;

2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g) H = -827.4 kJ/mol

-827.4 kJ/mol of heat liberates 2 moles of SO2 in the reaction involving lead and oxygen

At STP, -827.4 kJ/mol of heat liberate 22.4 * 2 dm3 of SO2

So therefore, 975 kJ/mol of heat will liberate

= 975 * 22.4 * 2 / -827.4

= 43 680 / -827.4

= 52.79 dm3 of SO2.

52.79 dm3 of SO2 will be produced at STP if 975 kJ/mol of heat are liberated.

6 0

3 years ago

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