So2+o2 = so3 método de tanteo

A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an

mr_godi [17]

Answer : The value of equilibrium constant (K) is, 424.3

Explanation : Given,

Concentration of $H_2$ at equilibrium = 0.067 mol

Concentration of $CO$ at equilibrium = 0.021 mol

Concentration of $CH_3OH$ at equilibrium = 0.040 mol

The given chemical reaction is:

$CO+2H_2\rightarrow CH_3OH$

The expression for equilibrium constant is:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the given values in this expression, we get:

$K_c=\frac{(0.040)}{(0.021)\times (0.067)^2}$

$K_c=424.3$

Thus, the value of equilibrium constant (K) is, 424.3

4 0

3 years ago

The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?

AlladinOne [14]

Answer:

64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

7 0

3 years ago

Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016Na+ ions arrive at the negative electrode and 3.9

EleoNora [17]

Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

q = charge of sodium ion

n = number of sodium ion = $2.68\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

4 0

3 years ago

What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution

Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

$M=\dfrac{n}{V}$

So , $V=\dfrac{n}{M}$

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

$V=\dfrac{36.25}{1.25}\\\\V=29\ L$

Therefore , volume of water required is 29 L .

5 0

3 years ago

Hell me ASAP please

postnew [5]

Its air and water. if its multiple choice

5 0

2 years ago

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So2+o2 = so3 método de tanteo​

So2+o2 = so3 método de tanteo