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Chemistry Class 12th

At 310 K, the solubility of CaF₂ in water is 2.34 * 10^-3 g/ 100 mL. The solubility product of CaF₂ is ____________* 10^-8 (mol/L)³. (Give molar mass : CaF₂ = 78 gmol^-1)

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Payal Gupta

Contributor-Level 10

Solubility of CaF₂ = S mol/L

$S = \frac{2.34 * 1 0^{- 3}}{0.1 * 78} = \frac{2.34}{78} * 1 0^{- 2} = 3 * 1 0^{- 4} m o l / L$

$K_{s p} (C a F_{2}) = 4 S^{3} = 4 {(3 * 1 0^{- 4})}^{3} = 108 * 1 0^{- 12}$

= 0.0108 * 10^-8 (mol/L)³

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Chemistry Class 11th

According to MO theory, number of species / ions from the following having identical bond order is___________.

CN^-, NO⁺, O₂, $O_{2}^{+}, O_{2}^{2 +}$

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Payal Gupta

Contributor-Level 10

$C N^{-}, N O^{+} a n d O_{2}^{2 +}$ have bond order = 3

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a year ago

Chemistry Class 11th

The molar heat capacity for an ideal gas at constant pressure is 20.785JK^-1 mol^-1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K. The number of moles of the gas at constant volume is __________. (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$C_{P, m} = C_{v, m} + R$

$C_{v, m} = 20.785 - 8.314 = 12.471 J K^{- 1} m o l^{- 1}$

$n = \frac{5000}{12.471 * 200} = \frac{25}{12.471} \approx 2$

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a year ago

Chemistry Class 11th

Amongst the following, the number of oxide(s) which are paramagnetic in nature is Na₂O, KO₂, NO₂, N₂O, ClO₂, NO, SO₂, Cl₂O (Nearest Integer)

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Payal Gupta

Contributor-Level 10

$K O_{2}, N O_{2}, C l O_{2}, N O$ are paramagnetic.

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a year ago

Chemistry Chemistry Chemical Kinetics

2NO + 2H₂ → N₂ + 2H₂O

The above reaction has been studied at 800°C. The related data are given in the table below.

Reaction serial number	Initial Pressure of H₂ / kPa	Initial Pressure of NO / kPa	Initial rate (-dP/dT)/(kPa/s)
1	65.6	40.0	0.135
2	65.6	20.1	0.033
3.	38.6	65.6	0.214
4.	19.2	65.6	0.106

The order to the reaction with respect to NO is_______________.

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Payal Gupta

Contributor-Level 10

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$∴$ Order of reaction w.r.t 'NO' = 2

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a year ago

Chemistry Class 12th

20 mL of 0.02 M K₂Cr₂O₇ solution is used for the titraction of 10 mL of Fe²⁺ solution in the acidic medium. The molarity of Fe²⁺ solution is ___________* 10^-2 M. (Nearest Integer)

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Payal Gupta

Contributor-Level 10

Meq of K₂Cr₂O₇ = Meq of Fe²⁺

(Molarity * Volume * nf) of K₂Cr₂O₇ = (molarity * volume * nf) of Fe²⁺

0.02 * 20 * 6 = M * 10 * 1

M = 0.24 M

Molarity = 24 * 10^-² M

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a year ago

Chemistry Physics

Is Math calculation important for BiPC students?

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Baskaran M

Contributor-Level 9

Math calculation is important for BiPC students as it is used in:

1. Physics and chemistry problem-solving

2. Biology for statistical analysis and modeling

3. Medical and health sciences

4. Research and data analysis

Basic math skills can benefit BiPC students in their academic and professional pursuits.

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a year ago

Chemistry Class 12th

The total number of monobromo derivatives formed by the alkanes with molecular formula C₅H₁₂ is (excluding stereo isomers)___________.

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Payal Gupta

Contributor-Level 10

Consider the image below

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a year ago

Chemistry Class 12th

The separation of two coloured substances was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 3.25cm, 2.08cmand 1.05cm, respectively. The ratio of R_f values of A to B is ________.

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Payal Gupta

Contributor-Level 10

$R_{f}^{0} = distance travelled by the solute$ $/$ $/ distance travelled by the solvent$

${(R_{f})}_{A} = \frac{2.08}{3.25} {(R_{f})}_{B} = \frac{1.05}{3.25}$

$\frac{{(R_{f})}_{A}}{{(R_{f})}_{B}} = \frac{2}{1}$

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a year ago

Chemistry Class 12th

A sample of 4.5mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ____________ g/mol.

[Nearest integer]

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Payal Gupta

Contributor-Level 10

(Calculation is done considering STP condition)

$R O H + C H_{3} M g l \to C H_{4} + R O M g l$

No. of moles ROH = no. of moles of CH4

$\frac{4.5 * 1 0^{- 3}}{M} = \frac{31}{22400} \Rightarrow 32.52 \approx 33 g m / m o l e$

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