Answer: Amylose is a form of starch which has only α-1,4-links bonds glucose units.
Explanation:
Amylose is a polysaccharide made up of α(1-4) bound glucose molecules. The carbon atoms on glucose are numbered, starting at the aldehyde (C=O) carbon, so, in amylose, the 1-carbon on one glucose molecule is linked to the 4-carbon on the next glucose molecule.
To determine the standard heat of reaction, ΔHrxn°, let's apply the Hess' Law.
ΔHrxn° = ∑(ν×ΔHf° of products) - ∑(ν×ΔHf° of reactants)
where
ν si the stoichiometric coefficient of the substances in the reaction
ΔHf° is the standard heat of formation
The ΔHf° for the substances are the following:
CH₃OH(l) = -238.4 kJ/mol
CH₄(g) = -74.7 kJ/mol
O₂(g) = 0 kJ/mol
ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
ΔHrxn° = +163.7 kJ
Answer:
Explanation:
If an atom has 13 electrons then it belongs to p-block of periodic table.
s level can accommodate 2 electrons.
p level can accommodate 6 electrons.
13 means 1s2 2s2 2p6 3s2 3p1.
As you can see there totally 5 sub-shells.
Total number of shells are 3(1,2,3).
Answer:
The Earth, Stars, and Mars I guess...
Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
