Which of the following lines from the story indicates a flashback?

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

5 0

3 years ago

Read 2 more answers

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (

Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^{x}[B]^{y}[C]^{z}$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}$

$\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}$

$3 = 3^{x} \\\\x =1$

Order w.r.t B : Use trials 2 and 5

$\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}$

$\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}$

$4 = 2^{y} \\\\y =2$

Order w.r.t C : Use trials 1 and 2

$\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}$

$1 = (0.5)^{z}$

z = 1

Therefore, order w.r.t C = 1

8 0

3 years ago

A student needs to prepare 50.0 mL of a 1.30 M aqueous H2O2 solution. Calculate the volume of 5.0 M H2O2 stock solution that sho

lisov135 [29]

Answer:

13.0mL

Explanation:

So this is a classic M1V1=M2V2 problem, where M is molarity and V is volume and the subsequent numbers represent the two sets of condition (1 being before dilution and 2 being after dilution)

So M1 is going to be 5.0M because it is our initial molarity and V1 is what we are trying to find since we are trying to find how much of initial volume should be diluted.

M2 is 1.30M since it is what molarity is after dilution and M2 is what volume is after dilution which is 50mL.

So M1V1=M2V2 (becomes an algebra problem)

5*V1=1.30*50

V1= 13.0mL

Now this answer should make sense since to dilute something with large molarity to small you only need very few mL than the final volume because you add water to dilute it.

7 0

3 years ago

Sophia and her friend Nora have the same unknown solid. Sophia has progressed a little faster with the purification process, but

Katarina [22]

Answer:

Nora used a minimal volume of hot solvent, this hot mixture was filtered to remove insoluble impurities. She had to perform a rapid filtration with a minimum of evaporation in a short-stemmed funnel previously heated in an oven, and provided with a pleated filter to increase the filtration rate, so in this way she prevented the product from crystallizing prematurely on the funnel stem.

If with this technique does not reach a pure substance, the process can be repeated by heating the solvent, changing the funnel or looking for a solvent with high dissolution power of the substance.

Explanation:

The solid products obtained in a reaction are usually accompanied by impurities that must be removed in order to dispose of the desired product in the highest possible purity. The most suitable method for removing impurities that contaminate a solid is by crystallization either in a pure solvent or in a mixture of solvents. This procedure is given the generic name of recrystallization.

3 0

3 years ago

What is the number of moles in beryllium in 6.0 g of Be

STALIN [3.7K]

Answer:

see picture

Explanation:

see picture above. character filling moment

3 0

2 years ago