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Vikki [24]
3 years ago
13

Calculate the ph of a buffer that is 0.145 m hc2h3o2 and 0.202 m kc2h3o2. The ka for hc2h3o2 is 1.8 × 10-5.

Chemistry
1 answer:
koban [17]3 years ago
6 0

Answer:

pH = 4.8

Explanation:

A buffer is formed by a weak acid (0.145 M HC₂H₃O₂) and its conjugate base (0.202 M C₂H₃O₂⁻ coming from 0.202 M KC₂H₃O₂). The pH of a buffer system can be calculated using Henderson-Hasselbalch's equation.

pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8

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Allisa [31]

The prediction is that B. The electrons will flow to the zinc anode where a negative charge will build up and eventually halt the reaction.

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Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

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4 0
2 years ago
Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min
Olenka [21]
(80+125+45) / 10 = 250/10 =25
25 meters per minute= 0.41 meters/second

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6 0
4 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J
Romashka-Z-Leto [24]

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

0.90

J

g

∘

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

Δ

T

=

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

+

...

+

1

∘

C



10

×

0.90 J



33 times

=

33

×

10

×

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

33

∘

C

will be

q

=

10.0

g

⋅

0.90

J

g

∘

C

⋅

33

∘

C

q

=

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

q

=

m

⋅

c

⋅

Δ

T

, where

q

- the amount of heat added / removed

m

- the mass of the substance

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0
3 years ago
Do nonmetals form anions or cations?
MrRa [10]
Since medals form cations
nonmedals form anions
8 0
3 years ago
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