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Softa [21]
3 years ago
12

A generic gas, x, is placed in a sealed glass jar and decomposes to form gaseous y and solid z. 2x(g)â½ââây(g)+z(s) how are thes

e equilibrium quantities affected by the initial amount of x(g) placed in the container? assume constant temperature.
Chemistry
1 answer:
Elan Coil [88]3 years ago
3 0
The chemical equation given is:

<span>2x(g) ⇄ y(g)+z(s)</span>

Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of  products y(g) and z(s) will be obtained at equilibrium.

Justification:

As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.

The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
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The rate constant for this second‑order reaction is
o-na [289]

Answer:

daddadaddddddddddddddddddddddddddddda

Explanation:

dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

3 0
3 years ago
If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f
I am Lyosha [343]

Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L

3 0
3 years ago
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0
3 years ago
Fe2O3+2Al=Al2O3+2Fe
True [87]
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
6 0
3 years ago
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Nat2105 [25]
The answer is Solute.  The solution is the whole thing.  The solvent is the major component (the liquid doing the dissolving).
6 0
3 years ago
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