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EleoNora [17]
3 years ago
10

At what temperature would a 1.50 m nacl solution freeze, given that the van't hoff factor for nacl is 1.9? kf for water is 1.86

∘c/m .
Chemistry
1 answer:
IRINA_888 [86]3 years ago
5 0
Win to ki ne lei ma ki mna mata ki ss kk
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A cylinder with a movable piston originally has a volume of 2805 mL and is filled with nitrogen to a pressure of 4.00
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This problem is providing the initial volume and pressure of nitrogen in a piston-cylinder system and asks for the final pressure it will have when the volume increases. At the end, the answer turns out to be 2.90 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws are used so as to understand the volume-pressure-temperature-moles behavior in ideal gases and relate different pairs of variables.

In this case, we focus on the Boyle's law as an inversely proportional relationship between both pressure and volume at constant both temperature and moles:

P_1V_1=P_2V_2

Thus, we solve for the final pressure by dividing both sides by V2:

P_2=\frac{P_1V_1}{V_2}

Hence, we plug in both the initial pressure and volume and final volume in order to calculate the final pressure:

P_2=\frac{2805mL*4.00atm}{3864mL}\\ \\P_2=2.90atm

Learn more about ideal gases: brainly.com/question/8711877

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What is the percent of O in CO2
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Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)
AURORKA [14]

Explanation:

The net equation will be as follows.

          K(s) + Cl_{2}(g) \rightarrow KCl(s)

So, we are required to find \Delta H_{formation} for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1:  Convert K from solid state to gaseous state

          K(s) \rightarrow K(g),    \Delta H_{1} = 89 kJ

Step 2:  Ionization of gaseous K

           K(g) \rightarrow K^{+}(g) + e^{-},    H_{2} = 418 KJ

Step 3:  Dissociation of Cl_{2} gas into chlorine atom .

            \frac{1}{2} Cl_{2}(g) \rightarrow Cl(g),   \Delta H_{3} = \frac{244}{2} = 122 KJ

Step 4: Iozination of chlorine atom.

              Cl(g) + e^{-} \rightarro Cl^{-}(g),      H_{4} = -349 KJ

Step 5:  Add K^{+} ion and Cl^{-} ion formed above to get KCl .

              K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s),   H_{5} = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      \Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}

                  = 89 + 418 + 122 - 349 - 717

                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

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4 years ago
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The equilibrium constant, Kc, can be determined knowing that the balanced chemical reaction is N2 + 3H2 = 2 NH3. At equilibrium, Kc is equal to (concentration of NH3)^2 / [(concentration of N2) x (concentration of H2)^3]. Using the given values, Kc is computed to be equal to 1.9977 x 10^-4. 
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