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2 years ago

What effect does a catalyst have on a reaccion rate an the products formed in a reaccion

1 answer:

8 0

A catalyst will speed up the activation energy and therefore speed up the reaction. The products will form fast because of this.

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Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0

3 years ago

Consider the reaction 2 NO + O2 → 2 NO2

Sonbull [250]

Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

Given that:- $\frac{d[NO]}{dt}=-0.050\ M/s$ (Negative sign shows consumption)

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

5 0

3 years ago

KOH+H 3 PO 4 K3PO4+ H₂O<br> How to solve and answer to this question

sergij07 [2.7K]

Answer:

hmmmm

Explanation: Balance the reaction of KOH + H3PO4 = K3PO4 + H2O using this chemical equation balancer!

4 0

3 years ago

Why is your weight less on the Moon than on Earth, but your mass is the same?

Anon25 [30]

Answer:

Mass is the amount of matter in the object, and is not affected by gravity. Weight, on the other hand, is directly related to gravity and how much you pull to the ground.

8 0

4 years ago

Read 2 more answers

The charge of the stable ion of P

Lady bird [3.3K]

Answer:

its charge could be negative three(-3)

6 0

3 years ago