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Annette [7]
3 years ago
11

A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are adde

d to 225 mL of this buffer, what is the pH of the resulting solution
Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

                                                   = 82.8

Millimoles of potassium cyanide = 225 x 0.360

                                                   = 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

        = 4.74 + log[82.8 - 51.3 / 81 + 51.3]

         = 4.102

Therefore, pH = 9.05

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KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

x + y = 0.042moles

Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

x=0.032moles\\y=0.010 moles

0.010 moles of KBr multiplied it molar mass is

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Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

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3 years ago
If you are given an ideal gas with pressure (P) = 259,392.00 Pa and temperature (T) = 2.00 oC of 1 mole Argon gas in a volume of
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Answer:

 R = 0.064 dm³ atm K⁻¹ mol⁻¹

Explanation:

Answer:

Explanation:

Data Given:

volume of gas V = 8.8 dm³

no. of mole of gas (n) = 1 mole

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Convert Pascal (Pa) to atm (atmospheric pressure)

As

101,325 Pascals = 1 atm

So,

259,392.00 Pa  = 2 atm

Then Pressure (P) = 2 atm

Temperature T = 2.00 °C  

change the temperature from °C to K

As  to convert °C to K the below formula used

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So, for 2 °C

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formula used for Ideal gases

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as we have to find R of the gas:

we will rearrange the ideal gas equation as below:

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Put value in equation (1)

                 R = 2atm x 8.8 dm³ / 1 mole x 275.15 K

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So the value of R is 0.064 dm³ atm K⁻¹ mol⁻¹

and the unit of R (ideal gas constant) is dm³ atm K⁻¹ mol⁻¹

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