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rewona [7]
2 years ago
10

Which equation demonstrates that nuclear fusion forms elements that are heavier than helium?

Chemistry
1 answer:
zysi [14]2 years ago
4 0

Answer:

16 over 8 O+4/2 He------->20 over 10 Ne

Explanation:

when nuclei is heavier than helium.

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Geologists use radioactive dating to...
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determine the absolute age of rocks.

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
The organic compound di‑n‑butyl phthalate, C 16 H 22 O 4 ( l ) , C16H22O4(l), is sometimes used as a low‑density ( 1.046 g ⋅ mL
damaskus [11]

Answer:

36.63 Torr

Explanation:

You need to use two expressions, one for pressure and the other with the relation of density and height of the column.

For the pressure:

P = h * d * g  (1)

h is height.

d density

g gravity

The second expression put a relation between the densities and height of the column so:

d1/d2 = h1/h2 (2)

let 1 be the phthalate, and 2 the mercury.

Let's calculate first the relation of density:

d1/d2 = 13.53 / 1.046 = 12.93

Now with the first expression, we can calculate the pressure so:

P = hdg

We have two compounds so,

h1d1g = h2d2g ---> gravity cancels out

From here, we can solve for h2:

h2 = h1*(d1/d2)

replacing:

h2 = 459 / 12.53

h2 = 36.63 mm

1 mmHg is 1 torr, therefore the pressure of the gas in Torr would be 36.63 Torr

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