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ale4655 [162]
2 years ago
10

What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?

Chemistry
1 answer:
Brilliant_brown [7]2 years ago
4 0
Hello!

To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:

pH=pKa+log( \frac{[A^{-}] }{[HA]} )

pH=4,76+log( \frac{0,030M}{0,25M} ) \\  \\ pH=3,84

Now, we use the definition of pH and clear [H₃O⁺] from there:

pH=-log[H_3O^{+}]

[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M

So, the [H₃O⁺] concentration is 0,00014 M

Have a nice day!
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2 years ago
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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
1 year ago
William adds two values, following the rules for using significant figures in computations. He should write the sum of these two
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Therefore 3.5 is the less precise value.
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Ksp of AgCl= 1.6×10⁻¹⁰

AgCl=Ag⁺ +Cl⁻

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Assume [Ag⁺]=[Cl⁻]=x

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7 0
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ASHA 777 [7]
We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

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Hope this answers the question. Have a nice day.
7 0
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