Answer:
6H2 + P4→ 4PH3
Explanation:
Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.
the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).
therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)
Answer:
Colourless
Explanation:
We know that Y^3+ has the electronic configuration of;
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (the 5s and 4d levels are empty).
According to the crystal field theory, the colour of complexes result from transitions between incompletely filled d orbitals.
As a result of this, complexes with empty or completely filled d orbitals are colourless. Thus, [Y(H2O)6]3 is colourless according to the Crystal Field Theory.
Tt is the genotype that will appear in boxes two and three.
If you look at the column and row that intersect to form boxes two and three, you will see that they are T and t. That is the best way I can describe it, sorry if it’s confusing.
Answer:
Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.
Explanation:
Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.
First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)
we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters
we can now isolate mass and get
M * Liters * molar mass = mass
now we need to find the molar mass of CaCl2 which is 110.98 g/mol
plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass
mass = 252.4795g however the 6.5L has only 2 sig figs so i would say
mass CaCl2 = 2.5 * 10 ^2 g
