grams of NaHCO(s) wanted = 118.8 grams NaHCO Clarification: HCHO NaHCO => NaCHO(s) HO(l) CO(g) (extra) ( grams) (112.0 grams = 112.0g/82.03gmol = 1.37mol) For the reason that response ratios are all 1:1 the moles of NaCHO(s) produced equals the variety of moles of NaHCO(s) used. Subsequently, for 1.37 moles of NaCHO(s) produced one will want 1.37 moles of NaHCO(s). Changing moles to grams, multiply by method weight of NaHCO (87.003 g/mole) grams of NaHCO(s) wanted = 1.37 mole x 87.003 g/mole = 118.8 grams NaHCO
Reply 6
Reply 6
Reply 6
B Clarification: the thickness of the lava
grams of NaHCO(s) wanted = 118.8 grams NaHCO Clarification: HCHO NaHCO => NaCHO(s) HO(l) CO(g) (extra) ( grams) (112.0 grams = 112.0g/82.03gmol = 1.37mol) For the reason that response ratios are all 1:1 the moles of NaCHO(s) produced equals the variety of moles of NaHCO(s) used. Subsequently, for 1.37 moles of NaCHO(s) produced one will want 1.37 moles of NaHCO(s). Changing moles to grams, multiply by method weight of NaHCO (87.003 g/mole) grams of NaHCO(s) wanted = 1.37 mole x 87.003 g/mole = 118.8 grams NaHCO
A dome volcano would kind.
Dome volcano will kind Im fairly positive
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