Chemistry: Topic-wise Test- 1 - NEET

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

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

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 water 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.

Equation
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 formula
E = hc / λ
1 / λ = RZ² [ 1 / n²₁ - 1 / n₂² ]
E = RZ² hc [ 1 / n²₁ - 1 / n²₂]
This energy will be highest when the transition is 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.

Answer : 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.

Boiling Point is directly proportional to VanderWaals forces of attraction. Down the group, this forces increases. so force of attraction in CHF₃ is weaker than CHL₃, that's why boiling point of CHF₃ is lowest.

SiO₂ is a covalent 'network' solid. So the energy required to break bonds will be higher. Therefore the melting point will be higher.

Due to intramolecular H-bonding  in o-nitrophenol it exists as a monomer while due to intermolecular H-bonding in p-nitrophenol it exists as an associate molecule. As a result, vapour pressure of a o-nitrophenol is higher than that of a p-nitrophenol.

When the number of hybrid orbitals, H  is 4, the hybridization is sp³ hybridization
H = 1 / 2[V + M - C + A]
where, V = number of monovalent atoms
C = total positive charge
A = negative charge
a) For CH₄
H = 1 / 2[4 + 4 - 0 + 0] = 4
sp³ 
b) For SF₄,
H = 1 / 2 [6 + 4 - 0 + 0] = 5
sp³d
c) For BF^-₄,
H = 1 / 2 [3 + 4 - 0 + 1] = 4, thus sp³
d) For NH⁺₄
H = 1 / 2[5 + 4 - 1 + 0] = 4 thus sp³
Thus, only in SF₄ the central atom does not have sp³ hybridisation.

According to VESPR theory repulsion order

  lp - lp > lp - bp > bp - bp

As the number of lone pairs of electrons increases, bond angle decreases due to repulsion between lp-lp, Moreover, as the electronegativity of central atom decreases, bond angle decreases.
The correct order of increasing bond angle is Cl₂O < ClO₂^- < ClO₂

In ClO₂^- there are 2 Ione pairs of electrons present on the central chlorine atom. Therefore the bond angle in ClO₂^- is less than 118° which is the bond angle in ClO₂ which has less number of electrons on chlorine. 

Option b is correct  because as the charge increases covalent character also increases.

p_x-p_x and p_y-p_y will form pi bond when z-axis is molecular axis as the options have only one of these so option (d) is the correct answer.

Li⁺ have larger size in aq.solution due to the presence of the water molecules present inside the solution surronds easily and become the solution to be more hydrated comparing to the corresponding element present in the aqeous solution.

Ionisation Potential is directly proportional to positive oxidation state and inversely proportional to negative oxidation state.So, in O^2- removal of electron is easiest.

On moving down the group, there is an increase in the size of atoms. As a result, the tendency to accept an additional electron decreases and hence, the electron gain enthalpy decreases from Ne to Rn. However, He has small size and much higher tendency to accept an additional electron. Hence, its electron gain enthalpy is least positive of all the noble gases. 

Thus, it is concluded that neon has the highest positive electrons gain enthalpy.

Actually electron gain enthalphy and ionisation energy both are dependent on electronic configuration.
Electronegativity means tendency to attract electron it doesnot matter where it will keep the electron
Electron gain entalphy basically refer to attract e it must have space to keep electron in sub shell like spdf should be present
order of electronegativity is
F > O > N > Cl > Br  > I > S > C > P

With respect to Chlorine hydrogen will be electropositive. Chlorine being a strong electronegative element tends to force hydrogen to loss electron and hydrogen acquire electropositive character so that chlorine can complete its octet .

Francium has the greatest tendency to lose electrons.

37 = Rb
55 = Cs
87 = Fr
All belong to alkali metal group.
So answer is (d)

Lithium nitrate when heated decomposes to give lithium oxide and nitrogen dioxide. The reaction is as follows :

4LiNO₃ → 2Li₂O+ 4NO₂+ O₂

LiNO₃ on heating, gives reddish brown fumes of NO₂

Carnallite is an ore of potassium i.e., KCl.MgCl₂.6H₂O.

BaCO₃ decomposes at highest temp.

All the carbonates decompose on heating to give CO₂ and metal oxide.

The stability of carbonate towards heat depends upon the stability of the resulting metal oxide. More is the stability of the resulting metal oxide lesser is the stability of the carbonate towards heat and vice versa.

Calcium cyanamide (CaCN₂) mixed with carbon (C) is called nitrolim.

 

Gypsum is CaSO₄.2H₂O and Plaster of Paris is CaSO₄.1 / 2H₂O. Difference in number of water molecules = 3 / 2 or 1 x 1 / 2 

In Diphenyl the two Benzene rings share a carbon -carbon bond but not the two carbon atoms.

As the double bond has to be taken in main chain and also the carbon to with functional group has to be numbered first thus option A is the correct one.

The IUPAC name of the given compound  is Ethyl-2-methyl-2-(3-nitrophenyl) propanoate,because the side chain contains the functional group , then the phenyl ring is considered as a substituent.

1 linear and 1 cyclic benzene ring.

For an aldehyde or a ketone to exhibit tautomerism it is essential that it must have at least one α−Hα−H atom.

So, C₆H₅COC(CH₃)₃

Hence D is the correct answer.

Two pairs of cis and trans forms.

The most stable free radical is C₆H₅CHCH₃ because of resonance.

The electronegativity of F, O, and N follows the order: N < O < F

Therefore the negative inductive effect of - NR₂, -OR and -F follows the order:

-NR₂ < -OR < - F

Bond Length ∝ 1 / Bond Order

So, the correct order of bond length is

I > IV > II > III

Methyl grps are electron rich, donate electrons and create a +I effect, and
NUCLEOPHILE means a electron rich atom