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2 years ago

Molar solubility of CuBr Ksp = 6.3 x 10-9

1 answer:

6 0

First, lets write the equation: CuBr--> Cu+2 + Br-
then, write the Ksp expression: Ksp= [Cu+2] [Br-]= 6.3 x 10-9
We let x be the concentration of Ksp that dissolved. Ksp= (x) (x)= 6.3 x 10-9.
Now we solve for x

X^2= 6.3x10-9
x= square root(6.3x10-9)
x= 7.94x10-5

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A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

5 0

2 years ago

At 73.0 ∘c , what is the maximum value of the reaction quotient, q, needed to produce a non-negative e value for the reaction so

damaskus [11]

Here we will use the general formula of Nernst equation:

Ecell = E°Cell - [(RT/nF)] *㏑Q

when E cell is cell potential at non - standard state conditions

E°Cell is standard state cell potential = - 0.87 V

and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2

and Q is the reaction quotient for the reaction
SO42-2(aq) + 4H+(aq) +2Br-(aq) ↔ Br2(aq) + SO2(g) +2H2O(l)

so by substitution :

0 = -0.87 - [(8.314*346K)/(2* 96485)*㏑Q → solve for Q

∴ Q = 4.5 x 10^-26

6 0

2 years ago

A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is kno

frosja888 [35]

Answer:

The 9th percentile is 4.464ml

Explanation:

Hello!

Given the variable X: Amount of dye dispensed into a paint can.

With a normal distribution, mean μ= 5ml and standard deviation δ= 0.4ml

You need to calculate the 9th percentile of the distribution.

The percentile is a measure of the position that indicates the number that separates the distribution or data set in a percentage of interest. In this case, the 9th percentile is the value of the distribution that separates the bottom 9% from the top 91%.

Symbolically:

P(X≤x₀)= 0.09

x₀ represents the value of the percentile.

The best way to calculate this value is by using the standard normal distribution since it is already tabulated, and then "translate" the Z-value to a value of the variable.

Under the standard normal distribution you have to look for:

P(Z≤z₀)= 0.09

The value marks the bottom 9% of the distribution, this means that you'll find it in the left tail of it. Remember the mean of the standard normal distribution is zero, so all values under the mean will be negative. Using the left entry of the Z-table you have to look for 0.09 in the body of the table and reach the margins to find the corresponding value:

z₀= -1.34

Now using the formula of the distribution you can "translate" the Z-value in terms of the variable of interest

Z=(X-μ)/ δ ~N(0;1)

z₀=(x₀-μ)/ δ

z₀*δ=x₀-μ

x₀=(z₀*δ)+μ

x₀=(-1.34*0.4)+5

x₀= 4.464

The 9th percentile is 4.464ml

I hope you have a SUPER day!

4 0

2 years ago