Chemistry, 22.08.2019 07:00 buddyben422

How would you prepare 1.00 l of a 0.400m solution of copper(ii)sulfate, cuso4?

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Answer from: Thelearner100

We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of CuSO_4

we have to dilute??

It's simple.

$\eta=\frac{m}{MM}$

Using a periodic table we can find the molar mass of Cu,~~S~~and~~O

$MM_{CuSO_4}=153.9~g/mol$

Then

$m=\eta*MM$

now we can replace it

m=0.400*159.6

$\boxed{\boxed{m=63.84~g}}$

Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M

Answer from: malissa0325

To prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of copper(II)sulfate solution = 0.400 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.400M=\frac{\text{Moles of copper(II)sulfate}}{1.00L}\\\\\text{Moles of copper(II)sulfate}=(0.400mol/L\times 1.00L)=0.400mol$

To calculate the mass of copper(II)sulfate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of copper(II)sulfate = 159.61 g/mol

Moles of copper(II)sulfate = 0.400 moles

Putting values in above equation, we get:

$0.400mol=\frac{\text{Mass of copper (II) sulfate}}{159.61g/mol}\\\\\text{Mass of copper (II) sulfate}=(0.400mol\times 159.61g/mol)=63.8g$

Hence, to prepare the solution of given molarity, we add 63.8 g of copper (II) sulfate to the solution.

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

answer:

technetium is the most reactive. this is followed by silver. the least reactive out of these three elements is strontium.

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