Answer:
A) Forward rate = 1.1934 × 10^(-4) M/min
B) I disagree with the claim
Explanation:
A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.
The reversible reaction given to us is;
BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)
From this, we can see that the stoichiometric ratio is 1:1:1:1
Thus, concentration of [BF4-] is now;
[BF4-] = 0.150 - 0.0174
[BF4-] = 0.1326 M
From the rate law, we are told the forward rate is kf [BF4-].
We are given Kf = 9.00 × 10^(-4) /min
Thus;
Forward rate = 9.00 × 10^(-4) /min × (0.1326M)
Forward rate = 1.1934 × 10^(-4) M/min
(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.
Thus, I disagree with the claim.
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
Not 100% sure if its right.
When an object covers equal distances in equal amounts of time, it is moving at a constant speed.
Uranium-235 would be more useful for dating in Cambrian time because Cambrian time was 540 million years ago while the half life of carbon-14 is only 5,730 years
Hope this helps
Answer:
mass/13of molecules .........
