Chemistry

Contributor-Level 10

Using $λ = \frac{h}{\sqrt[]{2 q V m}}$

$λ_{L i} = \frac{h}{\sqrt[]{2 * 3 e * V * 8.3 m}}$ m = mass of proton

$λ_{p} = \frac{h}{\sqrt[]{2 * e * V * m}}$

$\frac{λ_{L i}}{λ_{p}} = \sqrt[]{\frac{2 e V m}{2 * 24.9 e V m}}$

= 0.2 = 2 * 10^-1

So; x = 2

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10 months ago

4.5g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is * 10^-1. The value of x is_________ (Rounded off to the nearest integer)

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Contributor-Level 10

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M . W * V} * 1000$

$M = \frac{4.5}{90 * 250} * 1000 = 0.2$

M = 2 * 10^-1 M

So, x = 2

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10 months ago

Chemistry Class 12th

The reaction of sulphur in alkaline medium is given below

$S_{8 (s)} + a O {H^{-}}_{(a q)} \to b {S^{2 -}}_{(a q)} + c S_{2} {O_{3}^{2 -}}_{(a q)} + d H_{2} O_{(I)}$

The values of 'a' is____________. (Integer answer)

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Contributor-Level 10

The balanced equation is :

$S_{8} (s) + 12 O H^{-} (a q) \to 4 S^{2 -} (a q) + 2 S_{2} O_{3}^{2 -} (a q) + 6 H_{2} O (l)$

Here ; a = 12

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10 months ago

The NaNO₃ weighed out to make 50 mL of an aqueous solution containing 70.0mg Na⁺ per mL is…………. g. (Rounded off to the nearest integer)

[Given ; Atomic weight in g mol^-1 -Na : 23; N : 14; O : 16]

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Chemistry Chemistry Solutions

Contributor-Level 10

Mass of Na⁺ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na⁺ is present in 85000 mg of NaNO₃ (1 mole NaNO₃ contains 1 mole Na⁺)

$∴ 3500$ mg Na⁺ will be present in $\frac{85000}{23000} * 3500$

= 12934.78 mg

= 12.93478 gm

$≅ 13$

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10 months ago

When 9.45g of ClCH₂COOH is added to 500ml. of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂ COOH is x * 10^-3. The value of x is__________ (Rounded off to the nearest integer)

$[k_{f (H_{2} O)} = 1.86 K k g m o l^{- 1}]$

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Contributor-Level 10

$Δ T_{f} = 0.5 ° C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$Δ T_{f} = i k_{f} . m$

$0.5 = i * 1.86 * \frac{9.45}{94.5 * 500} * 1000$

i = 1.344

Now, using $α = \frac{i - 1}{n - 1}$

n for ClCH₂COOH = 2

a = $\frac{1.344 - 1}{2 - 1} = 0.344$

Using

$K_{a} = \frac{C α^{2}}{1 - α}$

$K_{a} = \frac{0.2 * {(0.344)}^{3}}{0.66} = 36 * 1 0^{- 3}$

So; x = 36

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10 months ago

Chemistry Class 12th

Emf of the following cell at 298 K in V is x * 10^-2.

$Z n | Z n^{2 +} (0.1 M) | | A g^{+} (0.01 M) | A g$

The value of x is………… (Rounded off to the nearest integer)

$[G i v e n : E_{Z n^{2 +} / Z n}^{0} = - 0.76 V; E_{A g^{+} / A g}^{0} = + 0.80 V; \frac{2.303 R T}{F} = 0.059]$

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Chemistry Coordination Compounds

Contributor-Level 10

${E^{o}}_{C e l l} = {E^{o}}_{A g^{+} / A g} - {E^{o}}_{Z n^{+ 2} / Z n} = 0.8 + 0.76 = 1.56 V$

Anode : $Z n (s) \to Z n^{2 +} (a q) + 2 e^{-}$

Cathode : 2Ag⁺(aq) + 2e^- ® 2Ag(s)

Zn(s) + 2Ag⁺(aq) ® Zn²⁺ (aq) + 2Ag(s)

$E_{c e l l} = {E^{o}}_{C e l l} - \frac{0.0591}{n} l o g \frac{[Z n^{2 +}]}{{[A g^{+}]}^{2}} = 1.56 - \frac{0.0591}{2} l o g (\frac{0.1}{1 0^{- 4}})$

x = 147

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10 months ago

The number of stereoisomers possible for ${[C o {(o x)}_{2} (B r) (N H_{3})]}^{2 -}$ is…………

[ox = oxalate]

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Contributor-Level 10

Kindly consider the following figure

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10 months ago

Chemistry Equilibrium

At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction $?$ Cl_2(g) 2Cl_(g) under the above conditions is x * 10^-1. The value of x is______ (Rounded off to the nearest integer)

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Contributor-Level 10

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

...more

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

Now: $P_{C l_{2}} = \frac{2}{4} * 1 a t m$

$= \frac{1}{2} a t m$

$P_{C l} = \frac{2}{4} * 1 a t m = \frac{1}{2} a t m$

$K_{P} = \frac{{(P_{C l})}^{2}}{P_{C l_{2}}} = \frac{{(\frac{1}{2})}^{2}}{\frac{1}{2}} = \frac{1}{2}$

$K_{P} = 0.5 = 5 * 1 0^{- 1}$

Here : x = 5

less

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10 months ago

The pH of ammonium phosphate solution, if pk_a of phosphoric acid and pK_b of ammonium hydroxide are 5.23 and 4.75 respectively, is…………….

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Contributor-Level 10

pH = $\frac{1}{2} [p K_{w} + p K_{a} - p K_{b}]$ $\frac{}{}_{}_{}_{}$

$= \frac{1}{2} [14 + 4.75 - 5.23]$

$= 6.76 ≅ 7$

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10 months ago

A ball weighing 10g is moving with a velocity of 90 ms^-1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is…………* 10^-33m. (Rounded off to the nearest integer)

[Given: h = 6.63 * 10^-34 Js]

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