How would we tackle an alien invasion?
We'd lose. Horribly horribly badly. There's ZERO chance of us winning or even posting a threat. We'd likely not even know we were invaded. Stop for a moment to consider the technology needed to traverse the stars. They would be orders of magnitude more advanced than us. They'd be more advanced than us than we are of the Sentinelese people. If we sought to invade those islands with our full military force, all at once, every nation on planet earth . . . that MIGHT compare to the degrees of technological disparity I'm talking about.We wouldn't even know we were invaded. Stop thinking "Independence Day" and/or "V" and/or "Skyline" or any other fictional account. Just like the Sentinelese people have no ability to imagine our technology and prepare for or predict or counter such an invasion, we too would have no ability to imagine, predict or counter an alien invasion from such an advanced species.I've already stated this, by any point in time that an alien race would traverse the stars and invade us, they'd have long since learned with 100% accuracy how to create micro (or nano) contagions. They'd simply be able to release a probe with a few hundred different strains of whatever, that target us, or all animals or all biology and begin wiping us out. We'd simply think some horrible plague had happened and would be gone within short order. We'd be extinct before any detectible ship came within 100 million kilometers of Earth. A small, capsule-sized probe, dropped over any part of the planet and winds to carry the contagion to the biosphere.Aliens capable of traversing the stars would be patient and not in any rush. They'd not need be. They'd simply annihilate us with biology. And then wait for us to die.
Oxidation Reduction?
A. H2SO3 = 2H+ and (SO3)2- lookin at just (SO3)2-, the oxidation number of S is +4 in (SO4)2-, the oxidation number of S is +6 since the oxidation number of S increases from +4 to +6, then u can conclude that S undergoes oxidation (or SO3). since SO3 undergoes oxidation, u can assume that it loses 2 electrons. H2SO3 + ______ -------> SO4(2-) + 2e- +_____ balance out the right hand side of the equation (which has 4 negative charges in total) by adding 4H+ H2SO3 + ______ --------> SO4(2-) + 2e- + 4H+ now, when u look at the equation, u'll notice that on the right side there is an additional of O in SO4, and 2 extra H. so u realise, hey! that's a water molecule, which makes sense since u never added anything to H2SO3 in the first place, but its an aq solution, so logically water is part of the reactants! thus, H2SO3(aq) + H2O(l) --------> SO4(2-)(aq) + 2e- + 4H+(aq) B. in Cr(OH)3, since its a neutral compound, and OH carries a charge of -1 each, then the oxidation number of Cr is +3. for chromate ion, Cr has an oxidation number of +6. so, since the oxidation number increase, Cr, like the previous question, undergoes oxidation. similarly, we work out the balanced equation in the same manner. since oxidation increases by 3, u can assume that Cr(OH)3 loses 3 electrons Cr(OH)3 + ______ ------> CrO4(2-) + 3e- + _____ now, the right side has 5 additional negative charges. but since tis is a basic solution, instead of using H+ to balance the charges, OH- is used instead Cr(OH)3 + 5OH- ------> CrO4(2-) + 3e- + ______ so, now u can see that u haf an additional 8H and 4O at the left side. tat's 4 water molecules that should be added to the right side! Cr(OH)3(aq) + 5OH-(aq) ------> CrO4(2-)(aq) + 3e- + 4H2O(l) tadaa!