NaOH
Na=23
O=16
H=1
SO,NaOH=23+16+1
NaOH=40g/mol
42% of 40=42/100x40
oxygen is the compound
density=mass/volume
1.30=mass/6
mass=7.8g
<span>(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
19.55 x 196.97 =
3850.76 g Au
I hope this helps you and have a great day!! :)
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Answer:
a. New alpha- 1,6 linkages can only form if the branch has a free reducing end
b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages
c. At least four glucose residues separate alpha-1,6 linkages
e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.
Explanation:
Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.
The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.
The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.
Answer:
14.04 L.
Explanation:
- The balanced reaction to form Al₂O₃ is:
<em>4Al + 3O₂ → 2Al₂O₃,</em>
4.0 moles of Al react with 3.0 moles O₂ to produce 2.0 moles Al₂O₃.
- Firstly, we need to calculate the no. of moles in (46.54 grams) of Al₂O₃:
<em>n = mass/molar mass </em>= (46.54 g) / (101.96 g/mol) = <em>0.4565 mol.</em>
<em></em>
<u><em>Using cross multiplication:</em></u>
3.0 moles of O₂ produce → 2.0 mole Al₂O₃, from the stichiometry.
??? moles of O₂ produce → 0.4565 mole Al₂O₃.
<em>∴ The no. of moles of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃</em> = (3.0 mol)(0.4565 mol)/(2.0 mol) = <em>0.6847 mol.</em>
- To calculate the volume of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃, we can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 0.6847 mol).
R is the general gas constant (R = 0.082 L/atm/mol.K),
T is the temperature of the gas in K (T = 300.0 K).
<em>∴ V = nRT/P</em> = (0.6847 mol)(0.082 L/atm/mol.K)(300.0 K)/(1.2 atm) =<em> 14.04 L.</em>
Answer:
The incomplete and varying inversion of configuration takes place at the chirality center.
Explanation:
When optically active alcohols react with HBr an SN1 reaction occurs.
In SN1 reactions an intermediate carbocation is formed in which the nucleophile can attack it on either side of the molecule. Therefore, there is a partial inversion of the center of chilarity of the molecule.
