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3 years ago

Which statement best describes a true solution?

2 answers:

3 0

The first statement is true because in a solution, the paticles settle out, and you can like no longer see them. An example of this would be when you make kool-aid becuase the particles settle out after you mix it together, and you can't seperate it either.

Nikitich [7]3 years ago

3 0

The correct answer is C. A true solution will not separate in a centrifuge. I just took the test.

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On hot days, you may have noticed that potato chip bags seem to "inflate", even though they have not been opened. If I have a 25

Slav-nsk [51]

The new volume of the bag will be 789.5 mL.

<u>Explanation:</u>

As per the Charles law, at constant pressure the volume of the gas is directly related to its temperature in Kelvin (K). That is as the temperature increases, the gas expands and vice-versa.

$$\frac{V1}{T1} = \frac{V2}{T2}\\$

V1 = 250 ml

V2 = ?

T1 = 19° C

T2 = 60° C

Now we have to rewrite the equation to get the new volume as,

V2 = $$\frac{V1T2}{T1}$

= $$\frac{250\times60}{19}$

<em> </em>= 789.47 ≈ 789.5 ml.

So the new volume of the bag will be 789.5 mL.

7 0

3 years ago

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha

NARA [144]

There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4

3 0

4 years ago

Read 2 more answers

Which point on the number line represents -3/4

Novosadov [1.4K]

Answer: - 0.75

Explanation:

For this question, you must remember that a fraction can be written of division.

So, -3/4 = -0.75, between -0.7 and -0.8 on the number line.

7 0

3 years ago

Read 2 more answers

How do I do an energy profile diagram

Marina CMI [18]

Answer:

first u have to like draw like a number line uk how dey do in math but instead of num u add wht eva da question askin u

Explanation:

5 0

3 years ago

Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.

natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

$pH=4.76+log(\frac{0.95}{0.92})$

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

$pH=4.76+log(\frac{0.99}{0.88})$

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

$pH=4.76+log(\frac{0.85}{1.02})$

pH = 4.76 - 0.079

pH = 4.681

5 0

4 years ago