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asambeis [7]
1 year ago
12

Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are

mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 ∘C to 27.7 ∘C.
a) Before mixing, how many grams of Cu are present in the solution of CuSO4?
b) Predict the identity of the precipitate in the reaction.
c) Write complete equation for the reaction that occurs when the two solutions are mixed.
d) Write net ionic equation for the reaction that occurs when the two solutions are mixed.
e) From the calorimetric data, calculate ΔH for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water.
Chemistry
1 answer:
kolbaska11 [484]1 year ago
7 0

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

1.00M=\frac{n}{0.050 L}

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

1\times 0.050 mol=0.050 mol of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)

d) CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)..[1]

KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)..[2]

Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)

Common ion both sides are removed. The net ionic equation is given as:

Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

m=d\time V=1 g/ml\times 100 mL = 100 g

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

Q=mc\Delta T

Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ

Enthalpy of the reaction = ΔH = \frac{Q}{\text{Moles of copper}}

ΔH = \frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

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Which compound would be expected to show intense IR absorption at 2710 and 1705 cm-1? (Ph = phenyl group)
Serhud [2]

Answer:

B. PhCHO

Explanation:

Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .

The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .

The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .

7 0
2 years ago
A new student is planning to use thin layer chromatography (TLC) for his research project. After setting up the apparatus the st
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Answer:

The open system evaporates the solvent in the solution

Explanation:

An open system is a system in which exchange of materials and energy can occur. If a TLC set up is left open, then the set up constitutes an open system.

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Most of these solvents used used TLC are volatile organic compounds. Therefore, if the TLC set up is left open, the solvent will evaporate leading to poor results after running the TLC experiment.

7 0
1 year ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

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6 0
2 years ago
How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?
Snezhnost [94]

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852

So, there are 0.852 moles in 19.6 g of Sodium.

6 0
2 years ago
Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio
kkurt [141]
should be 7.57*10^3 years
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