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Oksi-84 [34.3K]
3 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]3 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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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
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0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

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V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

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Carry one insignificant figure (shown in brackets).

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V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

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Here, carrying an insignificant figure doesn't change the value.

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c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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