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New answer posted
a year ago
Contributor-Level 10
Let x be the oxidation number to the underlined elements in the given species:
(a) NaH2PO4
(+1) + 2 (+1) + x + 4 (-2) = 0
x + 3 – 8 = 0
x = +5
(b) NaHSO4
(+1) + (+1) + x + 4 (-2) = 0
x – 6 = 0
x = +6
(c) H4P2O7
4 (+1) + 2x + 7 (-2) = 0
2x -10 =0
x = +5
(d) K2MnO4
2 (+1) + x + 4 (-2) = 0
x – 6 = 0
x = +6
(e) CaO2
2 + 2x = 0
x = -1
(f) NaBH4
1 + x + 4 (-1) = 0 (Since H is present as hydride ion.)
x = +3
(g) H2S2O7
2 (+1) + 2x + 7 (-2) = 0
x = +6
(h) KAl (SO4)2.12H2O
+1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) = 0
x = +6
New answer posted
a year agoContributor-Level 10
5.18. Given, P1 = P2 and V1 = V2
We know that P1V1 = P2V2
Or, n1RT1 = n2RT2
i.e., n1T1 = n2T2
Substituting n = w/M, we get
(W1/M1) x T1 = (W2/M2) x T2
(2.9/M1) x (95 + 273) = (0.184/2) x (17 + 273)
M1 = (2.9 x 368 x 2) / (0.184 x 290) = 40 g mol-1
New answer posted
a year agoContributor-Level 10
5.17. No. of moles of CO2 = Given mass of CO2 / Molar mass
= 8.8g / 44g mol-1 = 0.2 mol
Pressure of CO2 = 1 bar
RR = 0.083 bar dm3 K–1 mol–1
T = 273 + 31.1 K = 304.1 K
According to ideal gas equation,
PV = nRT
Therefore, V = nRT/P
= (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L
New answer posted
a year agoContributor-Level 10
5.16. Radius of the balloon = 10 m
Therefore, volume of the balloon = (4/3)? r3 = (4/3) x (22/7) x (10 m)3
= 4190.5 m3
Volume of He filled at 1.66 bar and 27 °C = 4290.5 m3
To calculate the mass of He,
PV = nRT = (w/M) RT, where M is molar mass of He i.e. 4 g per mole or 4 x 10-3 kg mol-1
=> w = [ (4 x 10-3 kg mol-1) (1.66 bar) (4190.5 m3)] / [ (0.083 bar dm3 K-1 mol-1) (300K)]
= 1117.5 kg
Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg
Maximum mass of the air that can be displaced by balloon to go up = volume x density
= 4190.5 m3 x 1.2 kg m-3 = 5028.6 kg
New answer posted
a year agoContributor-Level 10
5.15. Molar mass of O2 = 32 g/mol
It means, 8 g of O2 has 8/32 mol = 0.25 mol
Molar mass of H2 = 2 g/mol
It means, 4 g of H2 has 4/2 mol = 2 mol
Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol
Given, V = 1dm3, T = 27°C = 300 K, R = 0.083 bar dm3 K-1 mol-1
Applying PV = nRT,
P = nRT / V
= (2.25) (0.083 bar dm3 K-1 mol-1) (300 K) / (1dm3)
= 56.025 bar
New answer posted
a year agoContributor-Level 10
5.14. Time taken to distribute 1010 wheat grains = 1s
Time taken to distribute Avogadro number of wheat grains = (1s x 6.022 x 1023) / 1010
= 6.022 x 1013 s
= (6.022 x 1013 / 60 x 60 x 24 x 365) year
= 1.9 x 106 years
New answer posted
a year agoContributor-Level 10
5.13. Molecular mass of N2 = 28g
28 g of N2 has No. of molecules = 6.022 x 1023
1.4 g of N2 has No. of molecules = (6.022 x 1023 x 1.4 g)/28 g= 3.011 x 1022 molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N2 has electrons = 7 x 2 = 14
3.011 x 1022 molecules of N2 have electrons= 14 x 3.011 x 1022= 4.215 x1023 electrons.
New answer posted
a year agoContributor-Level 10
5.12. Given,
P= 3.32 bar
V= 5 dm3
n= 4 mol
R= 0.083 bar dm3 K-1 mol-1
PV = nRT
Or T = PV / nR = 3.32 x 5 dm3 / (4.0 mol x 0.083 bar dm3 K-1 mol-1)
= 50 K
New answer posted
a year agoContributor-Level 10
9.56. Dihydrogen is prepared from water by the action of alkali metals like Na and K which is a strong reducing agent.
2Na + 2H2O → 2NaOH + H2
2K + 2H2O → 2KOH + H2
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