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yan [13]
3 years ago
9

solution of 0.160 MNaOH is used to titrate 21.0 mL of a solution of H2SO4:H2SO4(aq)+2NaOH(aq)→2H2O(l)+Na2SO4(aq)If 38.8 mL of th

e NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution?
Chemistry
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

0.0432 M H2SO4

Explanation:

First, we want to find the moles of MNaOH used. We know that Molarity x Liters = moles. 0.160M x 0.0210L = 0.00336 moles MNaOH

to find the moles of H2SO4, we can use a mol ratio.

0.00336mol MNaOH x (1Mol H2SO4 /2mol MNaOH)

= 0. 00168 mol H2SO4

I found the mol ratio by looking at the coefficients in front of the molecules I knew(MNaOH) and the molecule I needed to find(H2SO4)

then, to find Molarity, we do mol/Liters

0.00168 mol/ 0.0388L =. 0.0432 M H2SO4

You can convert mL to L by dividing by 1000

the significant figures of this problem is 3, so my final answer will also have 3 sig figs.

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A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

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<span>523 grams First, calculate the molar mass of CH3OH Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass of CH3OH = 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now determine how many moles you have by dividing the number of atoms by avogadro's number moles = 9.83 x 10^24 / 6.0221409 x 10^23 = 16.3231 Now multiply the number of moles by the molar mass 16.3231 * 32.04146 = 523.0159557 Since we only have 3 significant figures, round the result to 3 significant figures. 523.0159557 grams = 523 grams</span>
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The chemical formula of the ionic compound is XY₂.

<h3>Ionic compounds</h3>
  • Ionic compounds are compounds which are formed from the combination of negatively charged and positively charged ion.
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<h3>Formation of ionic compounds</h3>

During the formation of the ionic compound between an ion X that has a charge of 2+ and ion Y has a charge of 1-, two negatively charged ions of Y are required to form a neutral ionic ionic compound when they react with X.

The chemical the formula of the ionic compound they form is as follows;

X²⁺  + 2Y⁻ ---> XY₂

Therefore, the chemical formula of the ionic compound is XY₂.

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