If you have access to stock solutions of 1.00m h3po4,1.00 m of hcl, and1.00 m naoh solution,(and distilled water of course),what volumesof each would you mix before diluting to a final volume of 2.00 l to prepare 2.00 l of ph 7.40 buffer with a final total concentration of 50 mm of phosphorous contains species (e. g. so that [h3po4] [h2po4-] [hpo42-] [po43-]

Its trait is different and it the organism survive

The answer is a. stable, hope i ya

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Unbalanced equation: fe2o3+ cofe + co21 fe2o3 + 3 co = 2 fe(s) + 3 co212.8g fe2o3 x (1 mol fe2o3 / 159.7g fe2o3) = 0.08015 mol fe2o311.5g co x (1 mol co / 28.01g co) = 0.4106 mol cofrom the balanced equation, 1 fe2o3 reacts with 3 cosince we have 0.08015 mol fe2o3 we therefore need0.08015 mol fe2o3 x (3 mol co / 1 mol fe2o3) = 0.2404 mol cosince we have 0.4106 mol co but only need 0.2404 to completely consume all thefe2o3, we have xs (excess) co and fe2o3 is therefore the limiting reagentfrom the balanced equation, 1 fe2o3 = 2 fe(s)0.08015 mol fe x (2 mol fe(s) / 1 mol fe2o3) = 0.1603 mol fe(s)theoretical mass of fe .1603 mol fe x (55.84g fe / mole fe) = 8.951g fe% yield = (6.10g / 8.951g) x 100% = 68.1%i didn't do this work, so i am not one hundred percent sure if it is correct  seems like a reliable website,