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Roman55 [17]
3 years ago
11

what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

the Ka of 2nitrophenol is 5.98 x 10^-8
Chemistry
1 answer:
REY [17]3 years ago
7 0

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

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I can't answer this question without knowing what the specific heat capacity of the calorimeter is. Luckily, I found a similar problem from another website which is shown in the attached picture. 

Q = nCpΔT
Q = (1.14 g)(1 mol/114 g)(6.97 kJ/kmol·°C)(10°C)(1000 mol/1 kmol)
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Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

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4 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms
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To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
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