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raketka [301]
3 years ago
12

Anyone who could help me would be great!!! Plss!!!

Chemistry
1 answer:
garri49 [273]3 years ago
7 0

Answer:

I think its a.The hockey puck loses heat molecule's.

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Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.
mars1129 [50]

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

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3 years ago
Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m
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3 0
3 years ago
if 5430 J of energy is used to heat 1.25 L of room temp. water (23.0 °C) whats the final temp of the water?
skelet666 [1.2K]

<span>We can use the heat equation,
Q = mcΔT </span>

 

<span>Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature difference (°C).</span>


Density = mass / volume


The density of water = 0.997 g/mL

<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>

<span>                                                                   = 1246.25 g</span>


Specific heat capacity of water = 4.186 J<span>/ g °C.</span>


Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

      5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
               T = 1.04 °C + 23 °C
               T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.
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Which of the following could be considered a scientific statement?
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