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Anna007 [38]
3 years ago
6

Find the pH of a 0.010 M HNO2 solution.

Chemistry
1 answer:
lidiya [134]3 years ago
6 0
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)


Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = 5.0*10^{-4}
\alpha^2 (degree\:of\:ionization) = ?

Ka = M * \alpha^2
5.0*10^{-4} = 0.010* \alpha^2
0.010\alpha^2 = 5.0*10^{-4}
\alpha^2 = \frac{5.0*10^{-4}}{0.010}
\alpha^2\approx500*10^{-4}

\alpha\approx\sqrt{500*10^{-4}}
\alpha \approx 2.23*10^{-3}

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

[ H_{3} O^+] = M* \alpha
[ H_{3} O^+] = 0.010* 2.23*10^{-3}
[ H_{3} O^+] \approx 0.0223*10^{-3}
[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
[ H_{3} O^+] = 2.23*10^{-5}

Formula:
pH = - log[H_{3} O^+]

Solving:
pH = - log[H_{3} O^+]
pH = -log2.23*10^{-5}
pH = 5 - log2.23
pH = 5 - 0.34
\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark

Note:. The pH <7, then we have an acidic solution.
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Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

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Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

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Barium nitrate has the formula Ba( NO3)2. Which statement is true about barium nitrate?
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Then, the number of atoms of O is 2 * 3 = 6.


So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.


From that molecule you can also tell:

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