Consider the ground state of Cr atom (Z= 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are respectively
12 and 4
12 and 5
16 and 4
16 and 5
Sodium bicarbonate on heating decomposes to form sodium carbonate, CO2 and water. If 0.2 moles of sodium bicarbonate is completely decomposed, how many moles of sodium carbonate is formed?
0.1
0.2
0.05
0.025
If standard enthalpies of formation of CaCl (s) (hypothetical) and that of CaCl2 (s) are -188 J mol-1 and -795 kJ mol-1 respectively, calculate the value of standard heat of reaction for the following disproportionation reaction
2CaCl (s) → CaCl2 (s) + Ca (s)
-607 kJ mol-1
+607 kJ mol-1
-419 kJ mol-1
+419 kJ mol-1
Aluminium chloride exists as dimer, Al2Cl6, in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
Al3+ + 3Cl-
[Al(H2O6]3+ + Cl-
[Al(HO)6]3- + 3HCl
Al2O3 + 6HCl
The ions N3-, O2-, F-, Na+, Mg2+ are isoelectronic. Their ionic radii show
an increase from N3- to F- and then decrease from N+ to Mg2+
a decrease from N3- to F- and then increase from Na+ to Mg2+
a significant increase from N3- to Mg2+
a significant decrease from N3- to Mg2+
Which of the following is not true?
In a first order reaction, the half-life is independent to the initial concentration of reactant.
A given piece of charcoal shows increase in its surface area in its powdered form.
In valence bond of H2, each electron spends its time around its own nucleus
In valence bond of H2, both the electrons spend their time around both the nucleus
250 mL of a Na2CO3 solution contains 2.65 g of Na2CO3 10 mL of this solution is added to x mL of water to obtain 0.001 M Na2CO3 solution.The value of x is (molecular weight of Na2CO3 = 106)
1000 mL
990 mL
9990 mL
90 mL
B.
990 mL
Given, volume of Na2CO3 solution = 250 mL
Mass of Na2CO3 solution = 2.65 gm
Molecular mass of Na2CO3 = 106 gm
Eq. wt. of Na2CO3 =
N = 0.2
Sine, 10 mL of this solution is added to x mL of water to obtain 0.001 M Na2CO3 solution, therefore,
Volume of solution taken, V1 = 10 mL
Normality of solution, N1 = 0.2 N
Volume of solution after addition of water, V2 = 10 + x
Normality of solution, N2 = 0.001 M = 0.002 N ( For Na2CO3, N = 2M)
From N1V1 = N2V2
10 × 0.2 = (10 + x) × 0.002
x = 990 mL