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Nookie1986 [14]
4 years ago
5

Identify the arrows that show removal of thermal energy when matter changes state.

Chemistry
1 answer:
BARSIC [14]4 years ago
6 0
When matter loses heat, one of the following happens:
Liquid changes to solid (freezing), or
Gas changes to liquid (condensation), or
Gas changes to solid (deposition)

1- Freezing: when the temperature of a liquid matter drops below its freezing point, the liquid usually solidifies forming crystals. Example of this is when water changes to ice
2- Condensation: when the temperature of gas is reduced, the molecules slow down and come together changing the matter into liquid. Example of this is when water vapor condenses into water droplets.
3- Deposition: Sometimes when the temperature of a gas drops, it solidifies directly without passing through the liquid phase. An example of this is when water vapor changes directly to ice in sub-freezing areas without passing through the liquid phase. This is the process through which clouds are formed.
The diagram below shows these three changes in matter. 

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A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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3 years ago
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