Answer:
After three hours, concentration of C₂F₄ is 0.00208M
Explanation:
The rate constant of the reaction:
2 C2F4 → C4F8 is 0.0410M⁻¹s⁻¹
As the units are M⁻¹s⁻¹, this reaction is of second order. The integrated law of a second-order reaction is:
![\frac{1}{[A]} =\frac{1}{[A]_0} +Kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2BKt)
<em>Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time of the reaction (In seconds).
</em>
3.00 hours are in seconds:
3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds
Initial concentration of C2F4 is:
0.105mol / 4.00L = 0.02625M
Replacing in the integrated law:
![\frac{1}{[A]_0}= \frac{1}{0.02625} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]_0}=480.9M^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%3D%20%5Cfrac%7B1%7D%7B0.02625%7D%20%20%2B0.0410M%5E%7B-1%7Ds%5E%7B-1%7D%2A10800s%5C%5C%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%3D480.9M%5E%7B-1%7D)
[A] = 0.00208M
<h3>After three hours, concentration of C₂F₄ is 0.00208M</h3>
Answer:
1 and 2
Explanation:
Message me for explanation.
Answer:
about 19 or 20 g
Explanation:
To do this, is neccesary to watch a solubility curve of this compound. This is the only way that you can know how many grams are neccesary to dissolve this compound in 50 mL of water to a given temperature.
Now, if you watched the attached graph, you can see the solubility curve of many compounds in 100 g of water (or 100 mL of water). So, to know how many do you need in 50 mL, it's just the half.
So watching the curve, you can see that at 20 °C, we simply need between 35 g and 40 g. Let's just say we need 38 grams of NH4Cl to be dissolved in 100 mL of water.
So, in 50 mL, it's just the half. So, we only need 19 g or 20 g of NH4Cl at 20 °C, to dissolve this compound in water.
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
