Which element has the largest density at stp

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1 (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P) + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in atm and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol /8.314 J/mol *K))/2.00 atm + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

8 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

Gas chromatography can be useful in determining what about paint?

dexar [7]

Gas chromatography is a type of <span>chromatography that is usually used in analytic chemistry in order to analyze and separate the compounds that can be vaporized without decomposition.

Based on this:
</span><span>Gas chromatography can be useful in determining the chemical composition of paint.</span>

4 0

3 years ago

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When titrating a weak base with a strong acid, the pH at the equivalence point will be:

ahrayia [7]

Answer:

acidic

Explanation:

because the acid is strong whereas the base is weak.

8 0

3 years ago

Read 2 more answers