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oksano4ka [1.4K]
3 years ago
8

A solution that is 0.17m in HCHO2 and 0.11 Min in NaCHO2

Chemistry
1 answer:
Elanso [62]3 years ago
7 0
Hello!

(Missing Previous Enunciate) <span>Use the Henderson-Hasselbalch equation to find the pH of: 

To solve this question we need the pKa of HCOOH, which is 3,75, now, we can apply the Henderson-Hasselbach equation:

</span>pH=pKa+ log( \frac{[HCOONa]}{[HCOOH]} )=3,75 + log( \frac{0,11M}{0,17M} )=3,56
<span>
So, the pH of a solution that is 0,17 M in HCHO</span>₂ and 0,11 M in NaCHO₂ is 3,56

Have a nice day!
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Answer:

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Explanation:  

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The ph scale for acidity is defined by ph=−log10[h+] where [h+]is the concentration of hydrogen ions measured in moles per liter
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The complete question probably is

The pH scale for acidity is defined by pH = − log₁₀[H⁺] where [H⁺] is the concentration of hydrogen ions measured in moles per liter (M). A solution has a pH of 2.55. Find the hydrogen ion concentration.

Answer: -

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Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0
grandymaker [24]

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

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Explanation:

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\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

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\text{Number of moles}=\frac{88.0g}{44g/mol}=2.0moles

b) moles of H_2O

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According to stoichiometry :

6 moles of CO_2 require = 6 moles of H_2O

Thus 2.0 moles of CO_2 require=\frac{6}{6}\times 2.0=2.0moles  of H_2O

Thus CO_2 is the limiting reagent as it limits the formation of product.

As 6 moles of CO_2 give =  1 moles of glucose

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Please help, its urgent.
san4es73 [151]

Answer:

Mass of excess reactant left = 179.6 g

Limiting reactant = nitrogen

Mass of ammonia formed = 200.6 g

Explanation:

Given data:

Mass of nitrogen = 165.0 g

Mass of hydrogen = 215.0 g

Limiting reactant = ?

Mass of ammonia formed = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 165.0 g/  28 g/mol

Number of moles = 5.9 mol

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 215.0 g/  2 g/mol

Number of moles = 107.5 mol

Now we will compare the moles of ammonia with both reactant.

                  H₂      :      NH₃

                   3        :       2

                  107.5  :      2/3×107.5 = 71.7 mol

                   N₂      :      NH₃

                    1        :       2

                  5.9      :      2/1×5.9 = 11.8 mol

Less number of moles of ammonia are formed by the nitrogen it will act as limiting reactant.

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 11.8 mol × 17 g/mol

Mass = 200.6 g

Mass of hydrogen left:

We will compare the moles of hydrogen and nitrogen.

               N₂         :        H₂

                1           :          3

               5.9        :         3/1×5.9 = 17.7 mol

Out of 107.5 moles 17.7 moles of hydrogen react with nitrogen.

Number of moles left unreacted = 107.5 - 17.7 mol = 89.8 mol

Mass of hydrogen left:

Mass = number of moles × molar mass

Mass = 89.8 mol × 2 g/mol

Mass = 179.6 g

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3 years ago
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