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mojhsa [17]
3 years ago
7

In the naturally occurring D series of ketoses, the carbonyl group is found on carbon number ___________, whereas in aldoses, th

e carbonyl group is found on carbon number _________.
Chemistry
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

second carbon atom from the  end

end carbon atom

Explanation:

Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen. The general molecular formula of Carbohydrates is C_x(H_2O)_y.

Carbohydrates can be classified based on structures,

Carbohydrates with the structure of alkanals (-CHO) are known as aldose while those of the structure of alkanones (C=O) are known as ketose.

In stereochemistry , D series is a kind of configurational arrangement where the hydroxyl group attaches itself to the right hand side.

Thus; in naturally occurring D series of ketoses, the carbonyl group is found on carbon number <u>second carbon atom from the  end </u>whereas in aldoses, the carbonyl group is found on carbon number <u> end carbon atom.</u>

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Hunter-Best [27]

Answer:

The answer is 0.36 kg/s NO

Explanation:

the chemical reaction of NH3 to NO is as follows:

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

We have the following data:

O2 Volume rate = 645 L/s

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we calculate the moles found in 645 L of O2:

P*V = n*R*T

n = P*V/R*T

n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2

With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:

14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO

8 0
3 years ago
For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration
Illusion [34]

Answer:

the original concentration of A = 0.0817092  M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

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where;

k = the specific rate coefficient  = 3.4 × 10⁻⁴ s⁻¹

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3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}

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\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}=   log \dfrac{a}{0.00018}

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a =453.94 \times 0.00018

a = 0.0817092  M

Thus , the original concentration of A = 0.0817092  M

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Answer:

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