Answer:
Explanation:
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
Initial (M) 2.00 2.00 0 0
Change (M) −x −x +x +x
Equil (M) 2.00 − x 2.00 − x x x
2 2 c 3
2
c 2
[SO ][NO ]
[SO ][NO]
(2.00 )
=
= −
K
x K
x
Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of
NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.
2.00 − x = 1.30
x = 0.70
Substituting back into the equilibrium constant expression:
2c 2 2c 2
(2.00 )
(0.70)
(2.00 0.70)
= − = −
x KxK
Kc = 0.290
Answer:
idk but this is what i know only
Explanation:
he half-life is the amount of time it takes for one-half of an isotope sample to decay into a different element. The half-life of 238U is 4.5 billion years. Other radioactive isotopes decay in a much shorter time. By comparison, the half-life of 14C (carbon 14) is approximately 5,700 years. Radiometric dating has calculated the age of the earth at 215 approximately 4.6 billion years. 1. Which circle in Figure 13-2 represents the amount of the original isotope before decay began?A 2. Assume that Figure 13-2 represents the half-life of 238U. Color in the area of circles B, C, and D that represents the amount of 238U remaining in the rock layer as each half-life passes. 3. If Figure 13-2 represented the half-life of 238U, which circle that you colored would represent a rock layer with the greatest concentration of lead? _A__ 4. Which circle that you colored shows the amount of the original isotope remaining after two half-life periods have expired? 5. IT Figure 13-2 represented the half-life of 14C, how many years would have passea to reach letter D? years 70 um ens ghe ting Fie ni boe nibs D rlt to got no oA teB Lgoc velFIGURE 13-2. Half-Life of a Radioactive Isotope C teob
Answer:
H+(aq) + OH-(aq) ⟶H2O(l)
Explanation:
Step 1: The balanced equation
HCN(aq) + KOH(aq) ⟶ H2O (l) + KCN (aq)
H+(aq) + CN-(aq) + K+(aq) + OH-(aq) ⟶H2O(l) + K+(aq) + CN-(aq)
Step 2: The net ionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:
H+(aq) + OH-(aq) ⟶H2O(l)
Answer:
After 18 hours, the amount of pure technetium that will be remaining is 12.5 grams
Explanation:
To solve the question, we note that the equation for half life is as follows;
Where:
N(t) = Quantity of the remaining substance = Required quantity
N₀ = Initial radioactive substance quantity = 100 g
t = Time duration = 18 hours
= Half life of the radioactive substance = 6 hours
Therefore, plugging in the values, we have;
Therefore, after 18 hours, the amount of pure technetium that will be remaining = 12.5 grams.
Answer:
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Explanation:
