Is it true or false I need help.

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution

Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0

3 years ago

The electron dot diagram for a neutral atom of chlorine (atomic number 17) is shown below.

Tom [10]

The answer is B ! Hoped I helped!

8 0

3 years ago

Read 2 more answers

The atoms of Group 7A elements gain electrons when they form ions.<br><br> True<br> False

artcher [175]

Answer: True.

Explanation.

The group 7A is actually named group 17.

That group is the halogens: F, Cl, Br, I, At, and Ts (Ts is one of the last elements discovered).

Those elements have 7 valence electrons (notice that it is the same number as the second digit in 17).

The atoms with 7 valence electrons will "easily" gain one electron to get the configuration of the next noble gas (8 valence electrons). That is why these elements gain electrons to form ions.

When atoms gain electrons form anions (negative ions). For example: F(-), Cl(-), Br(-), I(-).

6 0

3 years ago

Read 2 more answers

The equation for photon energy, E, is E=hcλ where h = 6.626×10−34 J⋅s (Planck's constant) and c = 2.99×108 m/s (the speed of lig

masya89 [10]

Answer:

9.133×10^-6 m

Explanation:

A photon is the smallest discrete amount or quantum of electromagnetic radiation. It is the basic unit of all light. Photons are always in motion and, in a vacuum, travel at a constant speed to all observers of 3 x 10^8 m/s.

From the question, the following details were supplied;

Energy of the photon (E) = 4.61×10^−19 J

Speed of light (c) = 2.99×10^8 m/s

Plank's constant (h) = 6.626×10^−34 J⋅s

Wavelength of the photon (λ)= ??? The unknown

From

E=hc/λ

λ= hc/E

Substituting values

λ= 6.626×10^−34 × 2.99×10^8/ 4.61×10^−19

λ= 91.33×10^-7 m

λ= 9.133×10^-6 m

3 0

3 years ago