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 
.
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>
The rate law for this reaction is [A]².
Balanced chemical reaction used in this experiment: A + B → P
The reaction rate is the speed at which reactants are converted into products.
Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.
Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².
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The correct answer would be 3.49 times 10^ minus 24 molecules
The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
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Answer:- 
atoms.
Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.
For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:
= atoms
So, there are atoms in 12.2 grams of 
.
