A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Pressure of the gas mixture = 1 bar
Let us consider 100g of the mixture
So ,mass of hydrogen in the mixture = 20 g
& mass of oxygen in the mixture = 80 g
Using the respective molar masses, we get
nH = 20/ 2 = 10 mol
. nO = 80 / 32 = 2.5 mol
Then, pH = XH x Ptotal
= (nH / nH + nO) x P total
= (10 / 10 + 2.5 ) x 1
= 0.8 bar
Hence, the partial pressure of dihydrogen is 0.8 bar.
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
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(b) MnO4 – (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)
(c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
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(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
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(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l ) for the last subshell that received electrons in building up the electronic configuration.
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(a) P4(s) + OH – (aq) → PH3(g) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO – 2(aq) + O2(g) + H + (aq)
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Also, mention whether change will cause the reaction to go into forward or backward direction.
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(iv) 2H2 (g) + CO (g) ↔ CH3OH (g)
(v) CaCO3 (s) ↔ CaO (s) + CO2 (g)
(vi) 4 NH3 (g) + 5O2 (g) ↔ 4NO (g) + 6H2O(g)
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(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
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2 A(g) + B(g) → 2D(g)
ΔU0 = –10.5 kJ and ΔS0 = –44.1 JK–1.
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CO (g) + H2O (g) ↔ CO2 (g) + H2 (g)
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I cannot understand this concept
Thanks
bro its very simple total pressure of dihydrogen =20% of 1 barr=0.2 barr and pressure of dioxygen =1-0.2=0.8
Very well explained And helped me for my test
Fake ans