Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 3(NH4)2SO4(aq) + Cr2(CO3)3(s)
<span>Ionic: 2Cr+3 + 3SO4^-2 + 6NH4+ + 3CO3^-2 ----> 6NH4+ + 3SO4^-2 + Cr2(CO3)3 (spectator ions are NH4+, SO4^-2) </span>
<span>Net Ionic: 2Cr^+3(aq) + 3CO3^-2(aq) -------> Cr2(CO3)3(s) </span>
Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.
Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.
Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.
Let the molar solubility be S upon dissociation.
PbBr₂ or Lead Bromide dissociates in pure water as follows:
PbBr₂ ----------> Pb⁺² + Br⁻
S 2S
Ksp = [Pb⁺²] [ Br⁻]
Ksp = (S) (2S)²
Ksp = 4S³
6.60 × 10⁻⁶ = 4S³
S = 0.0118M
Hence, the Molar solubility S is 0.0118M.
Learn more about Molar solubility here, brainly.com/question/16243859
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The answer is 7. Valence electrons are the electrons in the very last shell, so we need to look at the outer “circle” and count the electrons, or the little black dots. There are 7 in the last shell.
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
I believe 212.5m, but I may be wrong, I’m a little rusty with moles
