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LenKa [72]
2 years ago
13

Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to

one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 6.380 mL of the resulting mixture to a second compartment of the microwell plate.
Sally knows n (the charge on the metal ion) = +2
She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M

Required:
How many moles of [Cl-] must be dissolved in that compartment?
Chemistry
1 answer:
hram777 [196]2 years ago
3 0

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$

In that compartment $[M^{n+}]=[m^{2+}]=8.279 \ M$

The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

                                                      = 52.82 m mol

                                                      = 0.05280 mol

$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

                                       = 2 x 0.05282

                                       = 0.1056 mole

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Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

                     P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}

                         P =15 \ atm

The next thing is to obtain the new pressure of the gas , using boyle's law

              P_1V_1 = P_2V_2

                  P_2 = \frac{P_1 V_1}{V_2}

                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

                  P_2 = 2.5 \ atm    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

                \Delta H = n C_p \Delta T

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

7 0
2 years ago
Carbon tetrachloride contains one carbon and four chlorine atoms. for this compound 11.818 g of chlorine combine with 1.000 g of
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5 0
2 years ago
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The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
ziro4ka [17]

Answer:

The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group

The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists

Explanation:

5 0
2 years ago
A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside
Inessa05 [86]
<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

  • Initial volume, V1 = 100.0 mL
  • Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

  • Initial pressure, P1 = 1.80 atm
  • Final temperature , T2 = -25°C

                                     = 248.15 K

  • Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

  • According to the combined gas law equation;

\frac{P1V1}{T1}=\frac{P2V2}{T2}

Rearranging the formula;

V2=\frac{P1V1T2}{T1P2}

Therefore;

V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}

V2=112.08mL

Therefore, the new volume of the gas is 112.08 mL

8 0
2 years ago
Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0
ExtremeBDS [4]

Answer:

The freezing point will be 2.9^{0}C

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

Depressioninfreezing point=K_{f}Xmolality

Where

Kf= 20\frac{^{0}C}{m}

the molality is calculated as:

molality=\frac{moles_{solute}}{mass_{solvent}}

moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol

massofcyclohexane=25g=0.025Kg

molarity=\frac{0.0045}{0.025}=0.18m

Depression in freezing point = 20X0.18=3.6^{0}C

The new freezing point = 6.5^{0}C-3.6^{0}C=2.9^{0}C

5 0
2 years ago
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