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garik1379 [7]
4 years ago
6

N the molecule below, how many atoms could make hydrogen bonds with water? the compound has a ch2, double bond, chc, double bond

, chch2ch2cch2c, double bound, chch2sh backbone, with a cooh group attached to the ninth carbon. in this group one oxygen is attached to carbon by a double bond and another oxygen is attached by a single bond, and the seventh carbon of this backbone is double-bonded to oxygen. the group -ch2op is attached to the 3rd carbon. in this group p is double-bonded to oxygen and single-bonded to two –oh groups.

Chemistry
1 answer:
Anna35 [415]4 years ago
3 0
<h3>Answer:</h3>

                10 Atoms

<h3>Explanation:</h3>

The structure of said compound is sketched according to guide lines provided in statement and is attached below.

Hydrogen Bond Interactions:

                     Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atom (i.e. F, O and N) of one molecule and the partial negative most electronegative atom of another molecule.

                      In given structure we are having seven most electronegative oxygen atoms (labelled red) and three partial positive hydrogen atoms (labelled blue) directly attached to most electronegative atom (i.e. oxygen atoms).

                      Therefore, the oxygen atoms will make hydrogen bonds with water's hydrogen atoms and the partial positive hydrogen atoms will make hydrogen bonds with water's oxygen atoms respectively.

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5 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
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m_{H_2O}=4.86gH_2O

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Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
4 years ago
Please help, much appreciated.
Artyom0805 [142]
This is a synthesis reaction, the general pattern is: A+B=AB. It fits, you see :)?
5 0
4 years ago
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