Inorganic Chemistry Time Attack Quiz for NEET

2-Methyl 2-propanol

Given that, formula of first oxide = M₃O₄

Let mass of the metal = x
% of metal in M₃O₄ = (3x/ 3x+64) ×100

but as given % age = (100-27.6) = 72.4 %
so, (3x/ 3x+64)×100 = 72.4

or x = 56.
in 2nd oxide,
oxygen = 30%….so metal = 70%
so, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M₂O₃

Answer:

5.6 liter=11g
22.4 liter=? (x) g
x= 22.4×11/5.6
x=44g
Thus, Molecular mass of gas is 44g

Answer:

Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18
= 142 + 18x

On heating it will lose water and become anhydrous.

55.9% mass is the mass of waater in Na₂SO₄.xH2O.

Mass by Mass% = (mass / molar mass of compound) × 100
mass of water = 18x
55.9 = (18x / 142 + 18x) × 100
55.9(142 + 18x) = 1800x
7937.8 + 1006.2x = 1800x
x = 9.99
x = 10 (approx.)
So, the molecular formula of compound is Na₂SO₄.10H₂O.

Given,

Mass of metal = 0.5 g

Mass of oxide = 0.79 g

First we have to calculate the Mass of Oxygen in metal oxide.

Mass of Oxygen = Mass of oxide – Mass of Metal = 0.79 – 0.5 = 0.29 g

The equivalent weight of oxygen is 8 g/eq.

0.29 g of oxygen combines with the 0.5 g of metal

8 g of oxygen combines with  of metal.

Therefore, the equivalent weight of metal is 13.793 g/eq.

2 C₂H₂ + 5O₂ → 4CO₂ + 2 H₂O

In this reaction, we can see that, 2 moles of C₂H₂reacts with 5 moles of O_2​
Or, 1 mole of ​C₂H₂ reacts with 5/2 moles of O_2​
Under standard conditions, we can say that:
22.44 L of C₂H₂ reacts with (5/2) x 22.4 L moles of O₂
So, 100 cc of C₂H₂ will reacts with (5/2) x 100 cc = 250 cc of O_2​ .

But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C₂H₂ will be = 250 x (100/20) = 1250 cc of air.

The equation of the Mn²⁺ in acidic medium  will be-

MnO₄^– + 8H⁺+ 5 Fe²⁺ → Mn²⁺+ 5Fe³⁺ +4H₂O.

We can balance the reaction by using redox balancing of acid and base.

So, we know from the equation that 5 moles of iron react to form 1 mole of Mn⁺².

Thus, we can say that (10×0.1) moles of iron will react with the (10×0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4.  We will get the same result if we use M₁V₁ = M₂V₂ formulae.

Molecular Weight of CCl₄ = 35.5 + 12 x 4 = 154
1 mole CCl₄ vapour=154 g
According to Avogadro’s hypothesis we know that the molar volume of a gas at STP is 22.4 litres
= 154/22.4 = 6.87

Dalton’s law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust’s law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.

The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.

energy is given by following formulae,

E will be highest when the transition from n₃→n₂

Δx.Δp ≧ h/4π
So by the equation given in the question, the uncertainity value will be the least.

Maximum number of electrons can be accommodated in a shell of principal quantum number n is 2n². Since each orbital can accommodate only 2 electrons, maximum number of orbitals =2n²/2= n²

Total no. of electrons in F- subshell = 14 (when we include both spin +1/2  & – 1/2)
But maximum no. of electrons = 7 if only one spin either – 1/2 or +1/2.

The electronic configuration of Cu is,

1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.

Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons

Two electrons in K-shell will differ in spin quantum number.
They will have same values for the principal quantum number, azimuthal quantum number, and magnetic quantum number.

For first electron, n=1,l=0,m=0,s=+1​/2
For second electron, n=1,l=0,m=0,s=−1/2​

Total no of angular nodes = l (azimuthal quantum no.), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.

<div> <p><strong>Hemocyanin</strong>, a copper-containing protein chemically unlike hemoglobin, is found in some crustaceans.</p> <p><strong>Hemocyanin</strong> is B<strong>lue</strong> in <strong>colour</strong> when oxygenated and colourless when oxygen is removed.</p> </div> <div></div>