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4 years ago

Why is the solubility of a neutral organic compound unaltered by exposure to aqueous acid or base?

1 answer:

jeka944 years ago

6 0

Adding an acid or base adds a greater charge in the water, which makes slightly charged compounds dissolve better than they would have on their own. However, if the substance is neutral, it is not affected by the presence of any charge and cannot be made to dissolve by adding acid or base.

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dlinn [17]

Answer:

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3 years ago

The radius of a helium atom is a 31 PM. What is the radius in nanometers?

Paladinen [302]

Answer:

Thus, the radius of the helium atom in nanometers is - 0.031 nm

Explanation:

Given that:-

The radius of the helium atom = 31 pm

Considering the conversion of length in pm to the length in nm as:-

1 pm = 0.001 nm

So,

Applying the above conversion factor in the radius of helium atom as:-

Radius = $31\times 0.001$ nm = 0.031 nm

<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>

7 0

3 years ago

Calculate the number of Calories present in a chocolate chip cookie that contains 19 g carbohydrate, 3 g protein, and 9 g fat.

agasfer [191]

Im unsure :( maybe try looking up a calorie calculator

5 0

3 years ago

What mass of compound must be weighed out to have a 0.0223 mol sample of H2C2O4 (M=90.04 g/mol)?

rodikova [14]

Mass of H2C2O4 :

mm = 90.04 g/mol

number of moles : 0.0223 moles

m = n * mm

m = 0.0223 * 90.04

m = 2.007 g

hope this helps!.

5 0

4 years ago

1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant

vodomira [7]

The masses of the components are obtained as;

Sodium hydrogen carbonate = 3.51 g
Sodium carbonate = 8.708 g

<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

$2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)$

Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g

Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol

= 3.51 g

Learn more about anhydrous sodium carbonate :brainly.com/question/20479996

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6 0

1 year ago