Ask question

Ask question

All categories

2 months ago

6

3.0 mL of 0.02 M Fe(NO3)3 solution is mixed with 3.0 mL of 0.002 M NaNCS and diluted to the mark with HNO3 in 10 mL volumetric f

lask. The blood-red [Fe(NCS)]2+ ion that forms has an equilibrium molar concentration of 2.5*10-4 mol/L as determined from the calibration plot. Calculate the Kc for [Fe(NCS)]2+ formation. Assume the volumes are additive.

1 answer:

lorasvet [2.7K]2 months ago

5 0

Answer:

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$ .

Explanation:

$Moles=Concentration\times Volume (L)$

$Fe(NO_3)_3(aq)\rightarrow Fe^{3+}(aq)+3NO_3^{-}(aq)$

$[Fe(NO_3)_3]=0.02 M=[Fe^{3+}]$

Concentration of ferric ions = $[Fe^{3+}]=0.02 M$

Volume of the ferric solution is 3.0 mL or 0.003 L.

Calculating moles of ferric ions gives $=0.02 M\times 0.003 L$

1 mL = 0.001 L

$NaNCS(aq)\rightarrow Na^+(aq)+NCS^-(aq)$

$[NaNCS]=0.002 M=[NCS^-]$

Concentration of $NCS^-$ ions = $[NCS^{-}]=0.002 M$

Volume of the $NCS^-$ ion solution is also 3.0 mL or 0.003 L.

The moles of $NCS^-$ ions yields $0.002 M\times 0.003 L$

Volume of the nitric acid solution is 10 mL or 0.010 L.

Upon mixing, the concentrations of both ferric ions and $NCS^-$ ions will adjust.

Total volume of the solution = 0.003 L + 0.003 L + 0.010 L = 0.016 L.

Initial concentration of ferric ions before reaching equilibrium:

= $\frac{0.02 M\times 0.003 L}{0.016 L}=0.00375 M$

Initial concentration of $NCS^-$ ions before equilibrium is reached:

= $\frac{0.002 M\times 0.003 L}{0.016 L}=0.000375 M$

$Fe^{3+}+NCS^-\rightleftharpoons [Fe(NCS)]^{2+}$

In the beginning:

0.00375 M 0.000375 M 0

At equilibrium:

(0.00375-x) (0.000375-x) x

Concentration of $[Fe(NCS)]^{2+}=x=2.5\times 10^{-4} M$ at equilibrium.

The expression for the equilibrium constant regarding the formation of $[Fe(NCS)]^{2+}$ is formulated as:

$K_c=\frac{[[Fe(NCS)]^{2+}]}{[Fe^{3+}][NCS^-]}$

$K_c=\frac{x}{(0.00375-x)\times (0.000375-x)}$

$K_c=\frac{2.5\times 10^{-4} }{(0.00375-2.5\times 10^{-4})\times (0.000375-2.5\times 10^{-4})}$

$K_c=5.7\times 10^2$

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$ .

You might be interested in

Which list of radioisotopes contains an alpha emitter, a beta emitter, and a positron emitter?

alisha [2963]

Answer: The correct option is 3.

Explanation: Radioisotopes that emit alpha-particles are termed alpha-emitters. These isotopes undergo alpha-decay.

Those radioisotopes that emit beta-particles $(_{-1}^0\beta )$ are called beta-emitters. They undergo beta-minus decay, in which a neutron converts to a proton and an electron.

Isotopes that emit positrons $(_{+1}^0\beta )$ are known as positron-emitters, undergoing beta-plus decay where a proton becomes a neutron.

From the options given,

Option 1: All three isotopes undergo beta-minus decay.

Option 2: Cs-137 and Tc-99 undergo beta-minus decay.

Fr-220 undergoes alpha-decay.

Option 3: Kr-85 undergoes beta-minus decay.

$_{36}^{85}\textrm{Kr}\rightarrow _{37}^{85}\textrm{Rb}+_{-1}^0\beta$

Ne-19 undergoes positron decay.

$_{10}^{19}\textrm{Ne}\rightarrow _{9}^{19}\textrm{F}+_{+1}^0\beta$

Rn-222 undergoes alpha decay.

$_{86}^{222}\textrm{Rn}\rightarrow _{84}^{218}\textrm{Po}+_{2}^4\alpha$

Option 4: All three isotopes undergo beta-minus decay processes.

Therefore, the correct choice is 3.

6 0

1 month ago

Read 2 more answers

A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota

castortr0y [3046]

The partial pressure of nitrogen gas is calculated to be 21.16 MPa. The partial pressure of oxygen equates to 5.62 MPa, and the overall gas pressure is stated as 26.78 MPa. This adheres to the principle that the total pressure in a gas system equals the sum of all individual gas partial pressures. Thus, the total pressure in the system reflects the sum of the partial pressures of nitrogen and oxygen. Accordingly, the partial pressure for nitrogen can be derived as follows: Total pressure minus the partial pressure of oxygen. Thus resulting in: 26.78 - 5.62, which gives a partial pressure of nitrogen at 21.16 MPa.

7 0

1 month ago

Read 2 more answers

On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.

KiRa [2933]

Answer:

Please refer to the attached image for the graph with labeled axes and points.

Explanation:

This is a guide to fulfilling the instructions along with essential notes for understanding how to create such graphs:

1) The horizontal axis should indicate pressure ranging from 0 mb to 760 mb, while the vertical axis corresponds to volume ranging from 0 to 1 mL.

The x-axis captures the independent variable, and the y-axis records the dependent variable. Both axes must be accurately labeled, showing the variable names and their respective units.

In this context, the origin, (0,0), signifies the intersection of the axes at a pressure of 0 mb and a volume of 0.0 milliliters.

2) Allocate values for the divisions on the axes to maximize the usage of space on both.

An effective graph aims to utilize the entire space on both axes; for this, identify the maximum values for pressure and volume, and determine the corresponding marks.

The pressure range along the x-axis is [90, 760 mb], suggesting large divisions of 100 mb, with the farthest right mark at 800 mb. You can then subdivide each 100 mb interval into 10 smaller sections, using small divisions of 10 mb (my example employs 4 sections of 25 mb, but 10 mb is preferable).

The volume's range for the vertical axis is [0.1, 0.8], so it’s best to use divisions set at 0.1 ml.

3) Next, identify and label the points as follows:

(90, 0.9) ⇒ 90 mb, 0.9 ml

(100, 0.8) ⇒ 100 mb, 0.8 ml

(400, 0.2) ⇒ 400 mb, 0.2 ml

(600, 0.15) ⇒ 600 mb, 0.15 ml

(760, 0.1) ⇒ 760 mb, 0.1 ml

The points represented as (x, y) are referred to as ordered pairs, indicating that the sequence is significant: the first number denotes the independent variable whereas the second denotes the dependent variable.

Thus, for the point (90, 0.9), 90 indicates a pressure of 90 mb and 0.9 indicates a volume of 0.9 ml.

To find (600, 0.15), since the horizontal increments are valued at 0.1, you should place the second coordinate of the point between the marks corresponding to 0.1 and 0.2 ml.

This allows you to accurately plot each point on the graph.

5 0

1 month ago

Read 2 more answers

7.744 Liters of nitrogen are contained in a container. Convert this amount to grams.

lorasvet [2795]

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2

7 0

3 months ago