Chemistry: Astudent mixes in a test tube 3.00ml of 0.050m cuso4with 7.00ml of 0.20m nh3/nh41 . the solution becomes a deep blue color. assuming all the cu2 is complexed with nh3to form the [cu(nh3)4]2 ion, determine the concentration of the complex in the solution.

Astudent mixes in a test tube 3.00ml of 0.050m cuso4with

Ответ:

gracerich

12.01.2022

Chemistry

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

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Ответ:

NatalieZepeda

13.07.2021

Chemistry

The answer to your question is Q = 1432 Joules

Explanation:

Data

mass of Al = m = 40.0 g

temperature 1 = T1 = 50°C

temperature 2 = T2 = 90°C

Specific heat = C = 0.895 J/g°C

heat = Q = ?

Process

1.- Write the formula to calculate heat

Q = mC(T2 - T1)

2.- Substitution

Q = 40(0.895)(90 - 50)

3.- Simplification

Q = 40(0.895)(40)

4.- Result

Q = 1432 Joules

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Astudent mixes in a test tube 3.00ml of 0.050m cuso4with 7.00ml of 0.20m nh3/nh41 . the solution becomes a deep blue color. assuming all the cu2 is complexed with nh3to form the [cu(nh3)4]2 ion, determine the concentration of the complex in the solution.

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