Only animals and not plants can adapt to their environment true or false

A solution contains 1.817 mg of CoSO4 (155.0 grams/mole) per mL. Calculate the volume (in mL) of 0.009795 M Zn2 needed to titrat

Nana76 [90]

Answer:

85.952 ml $Zn^2^+$ needed to titrate the excess complexing reagent .

Explanation:

Lets calculate

After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml

As the number of moles of $CoSO_4$ = $\frac{Given mass }{molar mass}$

= $\frac{1.817}{155}$

=0.01172

Molarity = $\frac{no. of moles}{volume of solution}$

= $\frac{0.01172}{20}$

=0.000586 moles

Excess of EDTA = concentration of EDTA - concentration of CoSO4

= 0.009005 - 0.000586

= 0.008419 M

As M1V1 ( Excess of EDTA ) = M2V2 $(Zn^2^+)$

$0.008419\times100ml=0.009795\times V2$

$V2=\frac{0.008419\times100}{0.009795}$

V2 =85.952 ml

Therefore , 85.952 ml $Zn^2^+$ needed to titrate the excess complexing reagent .

3 0

3 years ago

How will the vapor pressure of an aqueous solution of sodium chloride compare to that of pure water?

solmaris [256]

The vapor pressure of a compound refers to the pressure generated by the vapour of a compound in equilibrium with its condensed stage. The major determinant of a compound vapour pressure is temperature. The vapour pressure of an aqueous solution of sodium chloride will be lower than that of pure water. This is because, dissolution of solute in a solvent always reduces the vapour pressure of the solution as a result of interactions between the molecule of the solute and the solvent. Thus, a pure compound will always have a higher vapour pressure than a solution.

7 0

4 years ago

Read 2 more answers

If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration

Ray Of Light [21]

Answer:

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

Data Given:

Ksp of CaSO4 = 2.4 × 10∧-5

CaSO4 ⇔ Ca+2 + SO4∧-2

Solution:

Ksp = [Ca+2].[ SO4∧-2]

2.4 × 10∧-5 = [x].[x]= x²

x = 4.9×10∧-3 M

Result:

The conc. of Ca+2 ion is 4.9×10∧-3 M

3 0

3 years ago

an ideal gas is at a pressure 1.00 x 10^5 N/m^2 and occupies a volume 11.00 m^3. If the gass is compressed to a volume of 1.00 m

bogdanovich [222]

Answer:

$P_2=1.1x10^6Pa$

Explanation:

Hello.

In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:

$P_1V_1=P_2V_2$

In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:

$P_2=\frac{P_1V_1}{V_2}$

Consider that the given initial pressure is also equal to Pa:

$P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa$

Which stands for a pressure increase when volume decreases.

Regards.

4 0

3 years ago

Himalayan pink salt has a density of 1.03 g/mL.

morpeh [17]

The volume, in liters, per kg of the Himalayan pink salt would be 0.97 L

Recall that: density = mass/volume

Hence: volume = mass/density

In this case, the density of the Himalayan pink salt is 1.03 while the mass we are working with is 1 kg.

1 kg is equivalent to 1000 g

Thus: volume = 1000/1.03

= 970.87 mL

Divide by 1000 to convert to L

970.87/1000 = 0.97 L

More on density can be found here: brainly.com/question/2040396

8 0

3 years ago