Write out the full electron configuration for an atom that has four valence electrons in the second energy level. Use ^ to denote a superscripted value.

Cis the answer ur welcome if i get it wrong u don't have to me

The answer is c) pyrite

Reaction involved in present system is :   c12h22o11 + h2o → 2 c6h12o6the reaction system is obeying following rate law:   rate = k[c12h22o11]thus, it is a first order reaction.now, for 1st order reactionrate constant (k) =  given: initial conc. = 3.0 g, final conc. = 2.7 g, t = 2.5 h∴rate constant (k) =  ∴ k = 0.04215 h-1now, for 1st order reaction, half life = 0.693/k = 0.693/0.04215 = 16.4 hthus correct answer is option b

2poclâ‚(g) ⇜ 2 pclâ‚(g) + oâ‚‚(g) â†gâ° = (2 mol)î”gfâ°(pclâ‚) – (2 mol)î”gfâ°(poclâ‚) = (2 mol)(–270 –(–502) kj/mol) â†gâ° = + 464 kj = â†hⰠ– tî”sâ° î”sâ° = 179 j/k ⇒ â†hâ° = â†gâ° + (298 k)(179 j/k) = –411 kj assuming that î”hâ° and î”sâ° do not depend on temperature, the temperature at which â†gâ° = 0 is where t = â†hâ°/î”sâ° = (–411 kj)/(0.179 j/k) = 2294 k. at that temp. keq = 1 for the reaction.