Answer:
D) 0 k.cal per 4 grams
Explanation:
Genarally, glucose can be classified into two enantiomers such as d-glucose and l-glucose. The d-glucose is the most common sugar that bodies of living organisms use as source of energy. However, l-glucose is an organic compound and it is one of the aldohexose monosaccharides. It is the l-isomer of glucose and commonly refer to as a low-calorie sweetener. The l-glucose is relatively indistinguishable in taste from d-glucose but cannot be used as a source of energy. Therefore, in the given problem:
if d-glucose has an estimated caloric value of 1 k.cal per 4 grams of carbohydrate, then the caloric value of l-glucose will be 0 k.cal per 4 grams.
for one you could do a bell, ringing noise, then say someone rang it.
For two you could do tunnel, echoing, then say someone yelled and it echoed.
for three you could do a bowl, ringing, spoon hitting it
hope i could help
Answer:
3.3535 g
Explanation:
Considering:
Or,
Given :
For iron(III) nitrate :
Molarity = 0.404 M
Volume = 52.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 52.0×10⁻³ L
Thus, moles of iron(III) nitrate :
Moles of iron(III) nitrate = 0.021 moles
1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.
Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles
Also, molar mass of iron(III) oxide = 159.69 g/mol
So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g
Covalent bonds are the bonds formed from the sharing of electrons between atoms.
