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vodomira [7]
3 years ago
8

How many joules in total would be required to boil 11.8 g of water if I begin heating it at a temperature of 65.5 C ? ( two step

s required)
Chemistry
1 answer:
solong [7]3 years ago
3 0
In this solution we will need two steps, 1) calculate the heat required to get water to the temp of 100 C and 2) find the latent heat of vapourization required. 

First, change the 11.8g to kg: = 0.0118kg
Equations: Q = mcT, Q=mL

c for water = 4186 J/(kg C)
L for water = 22.6 x 10^5 J/kg

1) Q = mcT
Q = 0.0118 kg x 4186 J/(kg C) x (100 C - 65.5 C)
Q = 1704.12 J

2) Q = mL
Q = 0.0118 kg x 22.6 x 10^5 
Q = 26668 J

Qf = 1704.12J + 26668J
Qf = 28372.12J

With sig figs: <u>2.84 x 10^4 J</u>
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What does Gibbs free energy predict?
7nadin3 [17]

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2 years ago
Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
dexar [7]

8.8 × 10-5 M is the  [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = \frac{[ClO-][H3O+]}{[HClO]}

2.9 × 10^-8 = \frac{[x] [x]}{[0.265-x]}

x^{2} = 7.698 x10^{-9}

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M  in 0.265 M solution of HClO.

8 0
3 years ago
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