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10 months ago

Chemistry Class 11th

For a chemical reaction A + B $?$ C + D

$(Δ_{r} H^{Θ} = 80 k J m o l^{- 1})$ the entropy change $Δ_{r} S^{Θ}$ depends on the temperature T (in K) as $Δ_{r} S^{Θ} = 2 T (J K^{- 1} m o l^{- 1}) .$

Minimum temperature at which it will become spontaneous is-------------K. (Integer)

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Payal Gupta

Contributor-Level 10

$Δ_{r} H^{o} = 80 k J m o l^{- 1}$

$Δ_{r} S^{o} = 2 T k J m o l^{- 1}$

For a reaction to be spontaneous;

$Δ_{r} S^{o} > \frac{Δ_{r} H^{o}}{T}$

$T > \frac{Δ_{r} H^{o}}{Δ_{r} S^{o}}$

$T > \frac{80 * 1 0^{3}}{2 T}$

$T^{2} > 40000 K$

$T > 200 K$

So, minimum T at which reaction will be spontaneous is 200 K.

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10 months ago

Chemistry Chemistry Some Basic Concepts of Chemistry

The number of significant figures in 50000.020 * 10^-3 is………………

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Payal Gupta

Contributor-Level 10

50000.020 * 10^-3

The significant figure in the given number is 8.

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10 months ago

Chemistry Chemistry Solutions

224mL of SO_2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P_{(H_{2} O)}^{o} = 24 m m o f H g) i s x * 1 0^{- 2} m m$ of Hg, the value of x is-----------. (Integer answer)

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Payal Gupta

Contributor-Level 10

Moles of SO₂= $\frac{224 * 1 0^{- 3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18} = 2$

Using ; relative lowering in V.P

$\frac{P^{o} - P}{P^{o}} = i x_{B}$

Where; $Δ P = P^{0} - P$ is lowering in V.P

$Δ P = P^{0} i x_{B}$

i for NaHSO₃ = 2

here; $x_{B} = \frac{n_{B}}{n_{A}}$ since solution is dilute

$Δ P = 24 * 2 * \frac{0.01}{2} = 0.24$

$Δ P = 24 * 1 0^{- 2} m m H g$

So; x = 24

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New answer posted

10 months ago

Chemistry Class 12th

3.12g of oxygen is adsorbed on 1.2g of platinum metal. The volume oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K in L is-----------.

[R = 0.0821 L atm K^-1 mole^-1]

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Payal Gupta

Contributor-Level 10

PV = nRT

1 * V = $\frac{3.12}{32} * 0.0821 * 300$

V = 2.4 litre

Vol of O₂ adsorbed per gm = 2.4 / 1.2 = 2 litre

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10 months ago

Chemistry Class 12th

Dichromate ion is treated with base, the oxidation number of Cr in the product formed is-----------.

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Payal Gupta

Contributor-Level 10

$C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$

Dichromate ion converted to chromate ion in basic medium and oxidation number of Cr in $C r O_{4}^{2 -}$ is +6.

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10 months ago

Chemistry Chemical Equilibrium

An homogeneous ideal gaseous reaction $A B_{2 (g)} ? A_{(g)} + 2 B_{(g)}$ is carried out in a 25 litre flask at 27°C. The initial amount of AB₂ was 1 mole and the equilibrium pressure was 1.9atm. The value of K_p is x * 10^-2. The value of x is-----------. (Integer answer).

[R = 0.08206 dm³ atm K^-1 mol^-1]

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Payal Gupta

Contributor-Level 10

$A B_{2} (g) ? A (g) + 2 B (g)$

Initial1 mole -

At equilirbium 1-x mole x mole2x mole

V = 25 L and

T = 300 K.

At equilibrium, P = 1.9 atm

Total moles at equilibrium, n = 1 + 2x

V = 25 L

T = 300 K

Using, PV = nRT

1.9 * 25 = (1 + 2x) * 0.08206 * 300

x = 0.465

Now;Partial pressure of AB2 at equilibrium = $\frac{0.535}{1.93} * 1.9 a t m$

Using ; $K_{p} = \frac{P_{A} . P_{B}^{2}}{P_{A B_{2}}}$

$K_{p} = \frac{(\frac{0.465 * 1.9}{1.93}) {(\frac{2 * 0.465}{1.93} * 1.9)}^{2}}{\frac{0.535}{1.93} * 1.9}$ = 0.728

K_P = 0.73

K_p = 73 * 10^-2

So; x = 73

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New answer posted

10 months ago

Chemistry Chemistry Chemical Kinetics

An exothermic reaction X → Y has an activation energy 30 kJ mol^-1. If energy change $Δ E$ during the reaction is -20 kJ, then the activation energy for the reverse reaction in kJ is-----------. (Integer answer)

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Payal Gupta

Contributor-Level 10

X → Y

$\begin{array}{l} E_{a_{f}} - E_{a_{b}} = - 20 \\ 30 - E_{a_{b}} = - 20 \end{array}$

$E_{a_{b}} = 50 k J / m o l e$

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10 months ago

Chemistry Chemistry Electrochemistry

Consider the following reaction

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{+ 2} + 4 H_{2} O, E ° = 1.51 V .$

The quantity of electricity required is Faraday to reduce five moles of $M n O_{4}^{-}$ is--------.

(Integer answer)

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Payal Gupta

Contributor-Level 10

$M n O_{4}^{-} + 8 H + 5 e^{-} \to M n + 4 H_{2} O E^{o} = 1.51 V$

Quantity of electricity required to reduce 1 mole of $M n O_{4}^{-}$ is 5F

So, for 5 mole $M n O_{4}^{-}$ 25F electricity is required.

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New answer posted

10 months ago

Chemistry Class 11th

A certain gas obeys $P (V_{m} - b) = R T .$ The value of ${(\frac{\partial z}{\partial P})}_{T} i s \frac{x b}{R T} .$ The value of x is------------.

(Integer answer) (Z : compressibility factor)

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Payal Gupta

Contributor-Level 10

$P (V_{m} - b) = R T$

$\frac{P V_{m}}{R T} - \frac{P b}{R T} = 1$

$Z = 1 + \frac{P b}{R T}$

${(\frac{\partial z}{\partial P})}_{T} = \frac{b}{R T}$

So, comparing with $\frac{x b}{R T}$

x = 1

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New answer posted

10 months ago

Chemistry Class 12th

Number of bridging CO ligands in $[M n^{2} {(C O)}_{10}]$ is----------.

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Payal Gupta

Contributor-Level 10

[Mn₂ (CO)₁₀]

0 0

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