Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
Explanation:
The pH of a weak base falls somewhere between 7 and 10.
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:
We make a simplification because x<<< 0.0225:
x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
