I'd be an outlet so that i'd never run out of energy and power like batteries and generators do cx
Answer:
a) 8.33 ml of the original stomach acid is neutralized
b) 191.67 ml of the stomach acid was neutralized
c) 249.68 ml acid would be neutralized by the original tablet
Explanation:
a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?
25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid
8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid
b) how much stomach acid was neutralized y the 4.3628 g tablet?
It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)
This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml
c) how much stomach acid would have been neutralized by the original 5.6832 g tablet
4.3628 g antacid is equivalent to 191.67 ml acid ( this is the answer for question 2)
5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid
Your answer should be heat energy!
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 gram0.90 J+1 gram0.90 J+ ... +1 gram0.90 J10 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∘C10×0.90 J+1∘C10×0.90 J+ ... +