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Oksi-84 [34.3K]
2 years ago
5

How many joules are needed to heat 20.0 g of Au from 10°C to 50°C? show work

Chemistry
1 answer:
AlexFokin [52]2 years ago
3 0

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of 1 g of a given substance by 1∘C.

In your case, aluminium is said to have a specific heat of 0.90Jg∘C.

So, what does that tell you?

In order to increase the temperature of 1 g of aluminium by 1∘C, you need to provide it with 0.90 J of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by 1∘C. So if you wanted to increase the temperature of 10.0 g of aluminium by 1∘C, you'd have to provide it with

1 gram0.90 J+1 gram0.90 J+ ... +1 gram0.90 J10 times=10×0.90 J

However, you don't want to increase the temperature of the sample by 1∘C, you want to increase it by

ΔT=55∘C−22∘C=33∘C

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

1∘C10×0.90 J+1∘C10×0.90 J+ ... +

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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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