The standard entropy for the substances are as follows:
C6H12O2(s) = -212
<span>O2(g) = -205 </span>
<span>CO2(g) = -214 </span>
<span>H2O(l) = -70
</span>
We calculate the ∆S°r<span>eaction by the expression:
</span>∆S°rxn = ∆S°products - ∆S° reactants
∆S°rxn = (212+6x205)-(6x214+6x70)
∆S°rxn = -262 J/K ------> OPTION 3
Answer:
Continuously recurring, or cycling maybe?
Yeah so you have to start of with converting your first two values into moles (forget the third one)
97.5 g NO * 1 mol/30.01 g NO = 3.25 moles NO
88.0 g O2 * 1 mol/16.00 g O2 = 5.5 moles O2
now we can find the limiting reactant by checking for the amount of product each reactant should give us by using molar ratios
3.25 mol NO * 2 mol NO2/2 mol NO = 3.25 mol NO2
5.5 mol O2 * 2 mol NO2/ 1 mol O2 = 11
so NO is the limiting reactant since it produces less product/gets used up quicker
3.25 mol NO * 2 mol NO2/2molNO = 3.25 mol NO2
so this is our theoretical yield and the question provides us with the actual yield (2.68 moles). since the actual yield is given in moles, we don't have to convert to grams. our percent yield formula goes like: actual yield/theoretical yield * 100
2.68 mol/3.25 mol * 100 = 82.46%
1s2 2s2 2p1
so E & F =2
G =1
Answer:
1) the oxidation state of the metal is +2
2) there are four d electrons
3) metal valence electrons =4 ligand valence electrons= 18
4) sp^3d^2
5) octahedral geometry
Explanation:
[Cr(H2O)6]2+ is an octahedral complex. Octahedral complexes are known to have the metal ion in sp^3d^2 hybridization state. Since there are six ligands each bonding to the central metal ion, there are eighteen ligand valence electrons and four valence electrons from the metal present in the high spin complex.
The crystal field of high spin and low spin octahedral for a d4 ion is shown in the image attached.
