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NISA [10]
2 years ago
9

Calculate the ph of a solution formed by mixing 150.0 ml of 0.10 m hc7h5o2 with 100.0 ml of 0.30 m nac7h5o2. The ka for hc7h5o2

is 6.5 × 10-5.
Chemistry
1 answer:
m_a_m_a [10]2 years ago
7 0

Answer:

4.49

Explanation:

pKa = - log 6.5 x 10⁻⁵

pKa  =4.19

Given that :

Volume = 150 mL = 0.150 L

For solutions:

number of moles of acid = volume × concentration

number of moles of  acid = 0.150 L × 0.10 M = 0.0150

number of moles  of salt = 0.100 L × 0.30 M = 0.0300

total volume = 150 + 100

= 250 mL

= 0.250 L

Concentration = number of moles/ Volume

∴

For [salt] = 0.0300/ 0.250

= 0.12 M

For [acid] = 0.0150/ 0.250

=0.06 M

pH = 4.19 + log 0.12/0.06

=4.49

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Answer:

Explanation:

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Initial pressure  = P₁  = 0.6atm

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Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

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          P₂   = 0.7atm

         

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In this, the physical state of ice is changing from solid to liquid state.

H_2O(s)\rightleftharpoons H_2O(l)

8 0
2 years ago
Read 2 more answers
Avanti works in a bookstore. She has four books and is going to place them in two stacks. The diagram above shows the books befo
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Answer:

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6 0
2 years ago
NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

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8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

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<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0
2 years ago
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