Answer:
The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)
Explanation:
Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5
Answer:
From highest to lowest:
butanol: 117.7 degree Celsius
butanone: 79.64 degree Celsius
diethyl ether: 34.6 degree Celsius
n-butane: -0.4 degree Celsius
Sr(s)+Mg²+(aq)→Sr²+(aq)+Mg(s)
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
=-(-70922.3)/2×96500)
=0.367v.
The product of a reaction between these two elements is 
.
Explanation:
The oxidation state of an ion in a compound is equal to its charge.
The aluminum having a charge of +3 because oxidation state is +3
The oxide is having charge of -2
The product of these reactants will produce a chemical compound.
The compound formed is i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.
The <em>+</em><em>3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. </em>
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Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:
Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
