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andrew11 [14]
3 years ago
6

A diprotic acid (H2X) titrated with a solution of NaOH for 25 ml of the acid 87.42 ml of a 1.95 M solution of NaOH was required

to reach the equivalent point, the concentration of the diprotic acid equals
0.14
6.8
3.4
0.29
Chemistry
1 answer:
larisa [96]3 years ago
4 0

The concentration of the diprotic acid equals 3.4 M

We'll begin by writing the balanced equation for the reaction.

H₂X + 2NaOH —> Na₂X + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₂X (nA) = 1
  • The mole ratio of the base, NaOH (nB) = 2

From the question given above, the following data were obtained:

  • Volume of acid, H₂X  (Va) = 25 mL
  • Volume of base, NaOH (Vb) = 87.42 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 1.95 M
  • Molarity of acid, H₂X (Ma) =?

MaVa / MbVb = nA/nB

(Ma × 25) / (1.95 × 87.42) = 1/2

(Ma × 25) / 170.469 = 1/2

Cross multiply

Ma × 25 × 2 = 170.469

Ma × 50 = 170.469

Divide both side by 50

Ma = 170.469 / 50

Ma = 3.4 M

Therefore, the concentration of the acid, H₂X is 3.4 M

Learn more about titration:

brainly.com/question/25624144

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Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C_{rms}= √\frac{3RT}{M}=√\frac{3PV}{M}=√\frac{3P}{d}, where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.

Given,  T=273 K, P=1.00 x 10⁻² atm, d=1.24 x 10⁻⁵ g/cm³.

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(b) Molar mass can be determined by using the formula C_{rms}=√{3RT}{M}

49.18=√\frac{3X8.314X273}{M}

49.18²=√(3X8.314X273)/M

M=\frac{3X8.314X273}{49.18^{2} }

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Molecular mass is 2.

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Need help with this.
Jobisdone [24]

Answer:

18.2 g.

Explanation:

You need to first figure out how many moles of nitrogen gas and hydrogen (gas) you have. To do this, use the molar masses of nitrogen gas and hydrogen (gas) on the periodic table. You get the following:

0.535 g. N2 and 1.984 g. H2

Then find out which reactant is the limiting one. In this case, it's N2. The amount of ammonia, then, that would be produced is 2 times the amount of moles of N2. This gives you 1.07 mol, approximately. Then multiply this by the molar mass of ammonia to find your answer of 18.2 g.

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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

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Hence, molarity = \frac{\text{no. of moles}}{volume}

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As reaction equation for the given reaction is as follows.

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So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

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                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

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