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LuckyWell [14K]
3 years ago
7

What happens if more solute is added to a saturated solution?

Chemistry
2 answers:
uranmaximum [27]3 years ago
6 0

Answer:

will not change

Explanation:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

Alexxx [7]3 years ago
5 0

Answer:

A saturated solution is a mixture in which the maximum amount of a given solute has been dissolved into the solvent. ... At this point adding more solute will not change the concentration of the solution; adding more solute will simply result in more solid at the bottom of the solution.

You might be interested in
How many liters are in 555 g of O2
victus00 [196]

Answer:

388.5

Explanation:

\frac{555}{32}  = 17.34375 \\ 17.34375  \times 22.4 = 388.5

3 0
4 years ago
H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a
forsale [732]

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = \frac{[SO42-] [H3O+]}{[HSO4-]}

0.011 = \frac{x (0.1 + x)}{0.1 - x}

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = { -b +,-\sqrt{b^2 - 4ac / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = \sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }  / 2(-1)

x = 0.111 +,- \sqrt{0.012321 + 0.0044} / -2

x = 0.111 +,- \sqrt{0.016721} / -2

x = \frac{0.111 +, - 0.1293}{-2}

x = \frac{0.111 + 0.1293}{-2}   , x = \frac{0.111  - 0.1293}{-2}

x = \frac{0.2403}{-2}    , x = \frac{0.0183}{-2}

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

8 0
3 years ago
Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g)
dem82 [27]

<u>Explanation:</u>

Unit is defined as the quantity that is used as a standard for measurement.

S.I. unit of mass of kilograms. Other units in which mass of a substance can be measured are grams, milligrams, etc...

All these units are inter convertible.

  • <u>For a:</u> Mega-

The conversion factor used is:

1g=10^{-6}Mg

  • <u>For b:</u> kilo-

The conversion factor used is:

1g=10^{-3}kg

  • <u>For c:</u> deci-

The conversion factor used is:

1g=10dg

  • <u>For d:</u> centi-

The conversion factor used is:

1g=10^{3}cg

  • <u>For e:</u> milli-

The conversion factor used is:

1g=10^3mg

  • <u>For f:</u> micro-

The conversion factor used is:

1g=10^6\mu g

  • <u>For g:</u> nano-

The conversion factor used is:

1g=10^9ng

  • <u>For h:</u> pico-

The conversion factor used is:

1g=10^{12}pg

8 0
3 years ago
A solute is added to water and a portion of the solute remains undissolved. When equilibrium between the dissolved and undissolv
Zina [86]
When the solute can no longer dissolve in the solution the solvent becomes SATURATED. When no more solute can dissolve and if you look at the bottom of the beaker, test tube, pan, or glass of cold tea you can see the solute permeating out as little particles .
4 0
4 years ago
Read 2 more answers
3. Which part of Thomson's plum pudding model was different from Rutherford's model?
horsena [70]

Answer:

The difference between the Thompson's plum pudding model and the Rutherford atom model is the location of the electrons (option a).

Explanation:

While Thompson compared his atom to a plum pudding, where the electrons floated freely in the pudding, Rutherford established the arrangement of the electrons in orbitals, which were found around the atomic nucleus like the planets around the sun.

Rutherford's findings also established the existence of a small, positively charged nucleus.

<em>Thompson and Rutherford models did not differentiate between the charges of electrons and protons , overall charges or overall size of the atom.</em>

8 0
3 years ago
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