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Chemistry Organic Chemistry - Some Basic Principles and Tech

A 0.166 g sample of an organic compound was digested with conc. H₂SO₄ and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H₂SO₄. The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is ___________.

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Vishal Baghel

Contributor-Level 10

Meq of NH₃ = Meq of used H₂SO₄ = Meq of NaOH = 0.25 * 30 = 7.5

Millmoles of N = millimoles of NH₃ = 7.5 (As n factor = 1)

Mass of nitrogen = 7.5 * 14 * 10^-3 = 0.105 gm

% of Nitrogen = $\frac{0.105}{0.166} * 100 % = 63.25 % ~ 63 %$

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Chemistry Coordination Compounds

In the cobalt-carbonyl complex : [Co₂(CO)₈], number of Co-Co bonds is “X” and terminal CO lignads is “Y”. X + Y = __________.

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Vishal Baghel

Contributor-Level 10

Number of Co-Co Bond = X = 1

Number of terminal ligand = Y = 6

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Chemistry Coordination Compounds

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Vishal Baghel

Contributor-Level 10

Number of Co-Co Bond = X = 1

Number of terminal ligand = Y = 6

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Chemistry Class 12th

The difference in oxidation state of chrominum in chromate and dichromate salts is _________.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Chemistry Class 12th

The difference in oxidation state of chrominum in chromate and dichromate salts is _________.

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Chemistry Chemistry Chemical Kinetics

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 – 1000 K. The data has been analysed by plotting ln k vs $\frac{1 0^{3}}{T}$ graph. The value of activation energy for the reaction is ___________ kJ mol^-¹.

(Nearest integer)

(Given : R = 8.31 J K^-¹ mol^-¹)

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Vishal Baghel

Contributor-Level 10

$k = A . e^{- (\frac{E a}{R T})}$

$l n k = l n A - \frac{E a}{R T}$

$\Rightarrow l n k = l n A + (- \frac{E a}{1000 R}) . \frac{1000}{T}$

Slope = $\frac{- E a}{1000 R} = - 18.5$

=> Ea = 18.5 * 1000 * 8.31 = 153.735 * 10³ J = 154 KJ

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Chemistry Class 11th

Some gases are responsible for heating of atmosphere (green house effect). Identify from following the gaseous species which does not cause it.

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alok kumar singh

Contributor-Level 10

CH₄, O₃ & H₂O are green-house gases.

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Chemistry Electrochemistry

The cell potential for the following cell

$P t | H_{2} (g) | H^{+} (a q) | | C u^{2 +} (0.01 M) | C u (s)$

Is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)

(Given : $E_{C u^{2 +} / C u}^{0} = 0.34 V a n d \frac{2.303 R T}{F} = 0.06 V$ )

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Vishal Baghel

Contributor-Level 10

Anode, H₂ (g) -> 2H⁺ + 2e^-

Cathode, Cu²⁺ + 2e -> Cu (s)

Net cell reaction H₂ + Cu²⁺ -> 2H⁺ Cu (s)

$E_{c e l l}^{0} = E_{C u^{2 +} / C u}^{0} - E_{H^{+} / H_{2}}^{0}$

= 0.34 V – 0

= 0.34 V

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.06}{2} l o g \frac{{[H^{+}]}^{2}}{[C u^{2 +}]}$

$\begin{array}{l} \Rightarrow 0.576 = 0.34 - \frac{0.06}{2} l o g \frac{{[H^{+}]}^{2}}{0.01} \\ \Rightarrow - l o g [H^{+}] = 4.93 \\ \Rightarrow p H o f s o l u t i o n = 4.93 ~ 5 \end{array}$

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Chemistry Solutions

The osmotic pressure of blood is 7.47 bar at 300K. To inject glucose to a patient intravenously, it has to isotonic with blood. The concentration of glucose solution in gL^-¹ is __________.

(Molar mass of glucose = 180 g mol^-¹ R = 0.083 L bar K^-¹ mol^-¹) Nearest integer)

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Vishal Baghel

Contributor-Level 10

For isotonic solution

$?_{i n j e c t o r} = ?_{B l o o d}$

$? C * 0.082 * 300 = 7.47$

$? C = 0.3 ? ? m o l e / l$

$= 0.3 * 180 g m / l = 54 g m / l$

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Chemistry Thermodynamics

2O₃(g) $?$ 3O₂(g)

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-)________. J mol^?1. (Nearest integer)

[Given: In 1.35 = 0.3 and R=8.3 J K^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

2O₃(g) -> 3O₂

t = 0 a moles 0

-0.5a mole +0.75a mole

At Eq. 0.5a mole 0.75a mole

Total moles at eq = 0.5a + 0.75 a = 1.25a

$P_{O_{3}} = x_{O_{3}} * P_{T} = \frac{0.5 a}{1.25 a} * 1 = \frac{2}{5} a t m$

$P_{O_{2}} = x_{O_{2}} * p_{T} = \frac{0.75 a}{1.25 a} * 1 = \frac{3}{5} a t m$

$K_{p} = \frac{{(P_{O_{2}})}_{e q}^{3}}{{(P_{O_{3}})}_{e q}^{3}} = \frac{{(\frac{3}{5})}^{3}}{{(2 / 5)}^{2}} = 1.35 a t m$

$Δ G^{°} = - R T l n K P$

$\begin{array}{l} = - 8.3 * 300 l n 1.35 \\ = - 747 J / m o l e \end{array}$

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