HELP FAST PLZ!!!!

N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f

kow [346]

Answer : The final concentration of $N_2O_5$ is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $5.89\times 10^{-3}\text{ min}^{-1}$

t = time passed by the sample = 3.5 min

a = initial concentration of the reactant = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

$3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}$

$a-x=2.9M$

Thus, the final concentration of $N_2O_5$ is, 2.9 M

3 0

2 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago

Determine the number of molecules in a 100. gram sample of CCl4

enyata [817]

100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)

Hope this helps~

8 0

3 years ago

The most important ore of aluminum is _____ show the chemical formula .

3241004551 [841]

Answer:

The most important ore of aluminum is<u> Bauxite.</u>

And its chemical formula is $Al_2O_3.2H_2O$ .

Explanation:

Bauxite is the most important ore of Aluminium from which Aluminium is extracted. Bauxite is a rock and composed of Aluminium bearing mineral.

And its chemical formula is $Al_2O_3.2H_2O$ .

It contains Gibbsite, Bohmite and Diaspore along with iron.

It is a soft material with somewhat white to grey to reddish brown in colour.

It has an earthy lustre and low specific gravity.

<u />

5 0

3 years ago

A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be

Ann [662]

The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.

r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.

d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.

The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

4 0

3 years ago