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Usimov [2.4K]
2 years ago
12

Which is larger? C atom or F atom? Explain why.

Chemistry
1 answer:
tekilochka [14]2 years ago
3 0

Answer:

Carbon or Fleuronie?

Explanation:

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Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce
adoni [48]

Answer:

The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }

In the given acid nitric:

Owing to the nitric acid mass = 75.9 g

Nitric acid molar weight= 63\  \frac{g}{mol}

If they put values above the formula, they receive:

\text{moles in nitric acid} = \frac{75.9}{63}

                               =1.204 \ mol

In the given chemical equation:

3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

= \frac{1}{2} \times 1.204

=0.602\ \ \text{mol of water}.

We are now using Equation 1 in determining the quantity of water:

Water moles = 0.602\  mol

Water weight molar = 18.02 \ \frac{g}{mol}

\to 0.602 = \frac{\text{mass of mols}}{ 18.02}\\\\\to \text{mass of mols} = 0.602 \times 18.02\\\\

                         =10.84 \ g

8 0
3 years ago
Find the pH of a 0.010 M HNO2 solution.
lidiya [134]
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.
6 0
3 years ago
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