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harina [27]
2 years ago
6

A 2.0 molal sugar solution has approximately the same freezing point as 1.0 molal solution of 1) CaCl2 2) CH3COOH 3) NaCl 4) C2H

5OH
Chemistry
1 answer:
sertanlavr [38]2 years ago
8 0

Answer:

3) NaCl.

Explanation:

<em>∵ ΔTf = iKf.m</em>

where, <em>i</em> is the van 't Hoff factor.

<em>Kf </em>is the molal depression freezing constant.

<em>m</em> is the molality of the solute.

<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>

<em></em>

  • For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<em>So, for sugar: i = 1.</em>

<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>

<em></em>

  • For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.

<em>So, i for NaCl = 2.</em>

<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>

<em></em>

<em>So, the right choice is: 3) NaCl.</em>

<em></em>

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dimaraw [331]

Answer : The percent difference between the ideal and real gas is, 4.06 %.

Explanation : Given,

Ideal pressure (true value) = 49.3 atm

Real pressure (measured value) = 47.3 atm

The formula used to calculate percent difference is :

Percent difference = \frac{\text{True value - Measured value}}{\text{True value}} \times 100

Percent difference = \frac{(49.3- 47.3)atm}{49.3atm}\times 100

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Therefore, the percent difference between the ideal and real gas is, 4.06 %.

4 0
2 years ago
2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
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Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag
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<span>NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
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This is chemical change (chemical reaction), because new substance (magnesium oxide MgO) is formed, the atoms are rearranged and the reaction is followed by an energy change (exothermic reaction because energy is released).

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2 years ago
Read 2 more answers
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

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\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

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Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

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

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
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