Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
Answer:
A = 0.75 ×10² KJ.
B = 3.9 ×10³ dg
C = 0.22 × 10² μl.
Explanation:
A = 7.5 ×10⁴ j to kilo joules
7.5 ×10⁴ / 1000 = 0.75 ×10² KJ.
Joule is the smaller unit while kilo joule is the larger unit. One kilo joule equals to the thousand joule that's why we will divide the given value by 1000 in order to convert into KJ.
B = 3.9 ×10⁵ mg to decigrams.
3.9 ×10⁵ / 100 = 3.9 ×10³ dg
Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into decigram we have to divide the value by 100.
C = 2.21 ×10⁻⁴ dL to micorliters
2.21 ×10⁻⁴ ×10⁵ = 0.22 × 10² μl.
Deciliter is bigger unit then micro liter . One deciliter equals to the 100000 micro liters. In order to convert the dL into micro liter we have to multiply the given value with 100000.
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Answer:
mass PbI₂ formed = 1383 grams
Explanation:
Pb(NO₃)₂ + 2NaI => 2NaNO₃ + PbI₂(s)
6 mol NaI => 1/2(6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)
