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Aleonysh [2.5K]
2 years ago
5

Calculate the volume in L of 12.9 g of NO2 at STP.

Chemistry
1 answer:
d1i1m1o1n [39]2 years ago
3 0

Answer:

6.28 L

Explanation:

Given data:

Volume of NO₂ = ?

Mass of NO₂ = 12.9 g

Temperature and pressure = ?

Solution:

First of all we will calculate the number of moles of gas.

Number of moles = mass/molar mass

Number of moles = 12.9 g/ 46 g/mol

Number of moles = 0.28 mol

At standard temperature and pressure 1 mole of gas occupy 22.4 L volume.

Thus,

0.28 mol × 22.4 L/1 mol

6.27 L ≅ 6.28 L

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According to the Arrhenius concept, an acid is a substance that ________. Group of answer choices can accept a pair of electrons
Schach [20]

Answer:  an increase in the concentration of H^+ in aqueous solutions and  is capable of donating one or more H^+

Explanation:

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions (OH^-) when dissolved in water and an acid is defined as a substance which donates hydrogen ions (H^+) in water.

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

Thus According to the Arrhenius concept, an acid is a substance that causes an increase in the concentration of H^+ in aqueous solutions and  is capable of donating one or more H^+

5 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

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