The first molecule is a sensible molecule having complete octet of each atom such as C, H and O whereas the second molecule having hydrogen present between the aldehyde and methyl group and thus showing hydrogen is making bond with aldehyde and methyl as well which is not possible because hydrogen only having one electron in its octet due to which it can only form a single bond by sharing its valence electron.
Claim: Earth's atmospheric CO2 levels were measured over the past 1000 years and correlated with average world temperatures. The data demostrate that Earth's temperature did not change during the period CO2 levels were steady. But the past 100 years show a dramatic rise in CO2 levels, with a corresponding rise in world temperatures. The CO2 rise can be traced back to the start of the industrial revolution, when machines began doing much of the work. These machine burn fossil fuels and the increased CO2 levels have led to a dramatic rise in world temperatures. The data clearly show the importance of reducing fossil fuel useage, as well as other CO2 emitters. A world of incresing temperatures will lead to greater natural disasters, such as storms, flooding, hurricanes, and drought - all of which upset the ecological balance of the planet.
2.1648 kg of CH4 will generate 119341 KJ of energy.
Explanation:
Write down the values given in the question
CH4(g) +2 O2 → CO2(g) +2 H20 (g)
ΔH1 = - 802 kJ
2 H2O(g)→2 H2O(I)
ΔH2= -88 kJ
The overall chemical reaction is
CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ
CH4 +2 O2 → CO2 +2 H20
(1mol)+(2mol)→(1mol+2mol)
Methane (CH4) = 16 gm/mol
oxygen (O2) =32 gm/mol
Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O
which generate 882 KJ /mol
Therefore to produce 119341 KJ of energy
119341/882 = 135.3 mol
to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require
=135.3 *16
=2164.8 gm
=2.1648 kg of CH4
2.1648 kg of CH4 will generate 119341 KJ of energy
Answer:
whah the other dude said :) also stay safe
Explanation:
B. false is the answer of this question
