Question

24,30,31 please

1) Gases. 30- Calculate how many grams of glucose (C6H120) can be burned in 750 mL of O2(g) at 769 mm of Hg and 27°C. Use equation 6a. 31-Determine the molar mass of an unknown gas, which has a density of 0.847 g/L at 135°C and 720 mm of Hg.

Answer 1

30) Reaction on glucose with Or is given PepellereePPRACOONOANNA w C6H12O6 + 602 6CO₂ + 6H₂O + Heat Doole olnudus salt Volume (v) of O2 = 750 mL = 0.75 Lt pressure (p) og 02 = 769 mm Hg Temperature (T) = 127°C = 1271 273 = 400k we know the formula pv=nRT where Tomb n= no. of mole, R = gos constant = 62.36 L. mmHg/molk Hence 16 9 x 0. f = 0 x 62 3 6 X 40 0 n= 0.023 mole.og 02 But from the above reaction, 6 mole of 02 burn I mole of C6H12Oo then 0.023 mole of O₂ will burn 3 -

0.023 X 1 EPODOBBBBBB = 0.00385 mole of C6H12O6 mass of CoH12O6 = mole & Molecular wt flere molecular weight of CHIRO6 = 180, So, mass of C6H12O6 = 0.00385 x 180 g A cich o . 693 gram. 31) density of gas = 0.947 911 ore Temperatur (T) = 135°C = 135+273 = 408k al pressure (p) = 720 mm Hg in of the gas. we know the formula. PV = nRT In, no. of mole of gos ] ove is the volume of gos pм - nx м 2T - Me Moleculay weight of pm = NXM RT na u gas P= HRT PADDED hХм

PM = d RT , nxm densitaa) 720 X M20.647 X 62.36 X 408 R = 62.26 L.muty Molek M=29.93 M~ 30 So, the molar mass of unknown gas was is 30 (approx.) 24) the given reaction is - O₂ + NO NO₂ +02 bad lll The different bond dissociation energy involve in the reaction is given by. Enthapy change į (203 - 30) Hodet/2 1 ( 2 20) (AH2) = 1 / 2 NO to NO2 4H₂ - After addition No + O₂ O₂ + NO₂ 4 H = 1 (AH) + (AH) + 4H

OM So, the Enthappy change for the giver reaction is given by terms of bond dissociation Energy AH = 1/2 (441) + 1/ (AH2) HAH3 where AH enthalpy change for the of reaction. CHI is Enthalpy change for 203 3 ₂ 2 AH₂ is bond dissociation energy for 0220 del bud 1, I alla is enthalpy change for NO TO NO AAAAAAA caddARATI dan 0 2014 A