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Inessa [10]
3 years ago
7

Explain the meaning of the terms “saturated,” “unsaturated,” and “supersaturated” with reference to the amount of solute and sol

vent in solutions.
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

Saturated Solution. A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom. Unsaturated Solution. A solution (with less solute than the saturated solution) that completely dissolves, leaving no remaining substances. Supersaturated Solution.

Explanation: Hope this helps

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Using the equation pH = -log [H+], determine the pH of a solution with a hydrogen ion concentration, or [H+], of 1x10-5. HINT: l
kipiarov [429]

Answer:

the answer is pH of 7 I don't no Hydrochloric hijinx work

8 0
3 years ago
Both diamond and graphite (i.e. pencil lead) consist of carbon atoms. They are only different in their crystalline structures. O
prisoha [69]

Answer:

Moles of carbon atoms  = 3.33  ×  10^{-6} mol

No. of atoms of C in Diamond  = 2.007 ×   10^{28} atom

Atoms of graphite = 6.27 × 10^{23} Atoms

Explanation:

given data

Cost of 0.2g of diamond = $5000

Cost of 25 g of graphite = $ 2

solution

we know cost of 0.2g of diamond is $ 5000 so that for 1$

if buy 1$ = \frac{0.20}{5000}

1$ = 4.0 × 10^{-5} g Carbon

and Moles of carbon atoms  is express as

Moles of carbon atoms = Given mass of Carbon ÷ atomic mass of C      .........1

Moles of carbon atoms  = 4.0  ×  10^{-5}    g/ 2.0g

Moles of carbon atoms  = 3.33  ×  10^{-6} mol

and

No. of atoms of C in Diamond = No. of moles × Avogadro NO    ..............2

No. of atoms of C in Diamond  = 3.33 ×   10^{-6} mol × 6.022 ×   10^{28}

No. of atoms of C in Diamond  = 2.007 ×   10^{28} atom

Graphite

and wew have given Cost of 25 g of graphite is $2 so for but 1$ we get

for buy $1 = 25÷2  = 12.5 g Of graphite

Moles of graphite = 12.5÷12 = 1.04 mol

Atoms of graphite = 1.04 × 6.022 × 1023

Atoms of graphite = 6.27 × 10^{23} Atoms

6 0
3 years ago
A 3.96x10^-24 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000-cm cuvet; a blank solution containi
viktelen [127]

Actual question from source:-

A 3.96x10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.000 cm cuvette.  A blank had an absorbance of 0.029.  The absorbance of an unknown solution of compound A was 0.375.  Find the concentration of A in the unknown.

Answer:

Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}

Explanation:

According to the Lambert's Beer law:-

A=\epsilon l c

Where, A is the absorbance

 l is the path length  

\epsilon is the molar absorptivity

c is the concentration.  

Given that:-

c = 3.96\times 10^{-4}\ M

Path length = 1.000 cm

Absorbance observed = 0.624

Absorbance blank = 0.029

A = 0.624 - 0.029 = 0.595

So, applying the values in the Lambert Beer's law as shown below:-

0.595=\epsilon\times 1.000\ cm\times 3.96\times 10^{-4}\ M

\epsilon=\frac{0.595}{3.96\times 10^{-4}}\ {Ms}^{-1}=1502.53\ {Ms}^{-1}

<u>Molar absorptivity of compound A = 1502.53\ {Ms}^{-1}</u>

4 0
4 years ago
PLEASE HELP! WILL GIVE BRAINLIEST:
Ad libitum [116K]

2.45 °C

From the Ideal gas law (Combined gas law)

PV/T = P'V'/T' .....eq 1

Where:

P - initial pressure

V - initial volume

T - initial temperature

P' - final pressure

V' - final volume

T' - final temperature.

To proceed we have to make T the subject from eq.1

Which is, T = P'V'T/PV.......eq.2

We have been provided with;

Standard temperature and pressure (STP)

P = 760 mm Hg (SP - Standard Pressure)

T = 273.15 K (ST - Standard Temperature)

V = 62.65 L

P' = 612.0 mm Hg

V' = 78.31 L

T' = ? (what we require)

Therefore, we substitute the values into eq.2

T' =

.

T' = 275.60 K

T = (275.60 - 273.15) ......To °C

T = 2.45 °C

>>>>> Answer

Have a nice studies.

5 0
2 years ago
O que são processos físicos de separação das misturas? Dê 3 exemplos
chubhunter [2.5K]

Answer:

Explanation:

The physical methods of separating mixtures are used in sorting a mixture of substances.

It requires no chemical changes occurring between their components and parts in any significant way.

Examples are:

  1. Decantation
  2. Filtration
  3. Sublimation
  4. Magnetism
  5. Centrifugation

The methods simply relies on the physical properties of matter.

8 0
3 years ago
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