A 5.0-gram sample of zinc and a 50.-milliliter sample of hydrochloric acid are used in a chemical reaction. Which combination of these samples has the fastest reaction rate?(1) a zinc strip and 1.0 M HCl(aq)
(2) a zinc strip and 3.0 M HCl(aq)
(3) zinc powder and 1.0 M HCl(aq)
(4) zinc powder and 3.0 M HCl(aq)
Answers
Answer:
(4) zinc powder and 3.0 M HCl(aq)
Explanation:
First consider the equation of reaction:
1 mole of Zinc requires 2 moles of HCl for reaction.
mole of zinc = mass/molar mass = 5.0/65.38 = 0.0765 mole
0.0765 mole of zinc will require: 0.0765 x 2 = 0.1530 mole HCl
Mole of 1.0 M HCl = Molarity x volume = 1 x 0.05 = 0.05 mole
Mole of 3.0 M HCl = 3 x 0.05 = 0.15
Hence, the 3.0 M HCl is what will give the required number of mole.
Zinc in powder form will react faster compared to in strip because the former has higher surface area for reaction.
Therefore, zinc powder and 3.0 M HCl will give the fastest reaction rate.
The correct option is option 4.
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Answer: The increase in force will directly effect the pressure exerted by the gas that is high pressure of the gas.
Explanation:
The pressure is defined as force applied per unit area.
If the gas particles colliding with the walls of container with increase in force then the pressure of the gas exerted by the particles of the gas will also get increased.
Answers
Answer:
1.74845
Explanation:
We have the following reaction:
I2 + H2 => 2 HI
Now, the constant Kc, has the following formula:
Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b
In this case I2 is A, H2 is B and C is HI
We know that the values are:
H2 = 1 × 10 ^ -3 at 448 ° C
I2 = 2 × 10 ^ -3 at 448 ° C
HI = 1.87 × 10 ^ -3 at 448 ° C
Replacing:
Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}
Kc = 1.87 ^ 2/2 * 1
Kc = 1.74845
Which means that at 448 ° C, Kc is equal to 1.74845
Answer:
Explanation:
[H2] = 10^-3
[I2] = 2*10^-3
[HI] = 0
in equilbiirum
[H2] = 10^-3 - x
[I2] = 2*10^-3 -x
[HI] = 0 + 2x
and we know
[HI] = 0 + 2x = 1.87*10^-3
x = ( 1.87*10^-3)/2 = 0.000935
then
[H2] = 10^-3 - 0.000935 = 0.000065
[I2] = 2*10^-3 -0.000935 = 0.001065
H₂ + I ⇄ 2 HI
Initially 1 × 10⁻³ 2 × 10⁻³
Change -9.35 × 10⁻⁴ -9.35 × 10⁻⁴ +1.87 × 10⁻³
At equil 6.5 × 10⁻⁵ 1.06 5 × 10⁻³ 1.87 × 10⁻³
HI increase by 1.87 × 10⁻³M
- When [S] << Km, the reaction is second order and V0 depends on [S] and [Et].
- Their kcat is a second order rate constant.
- The lower their Km, the better they recognize their substrate, but the lower their reaction rate.
- When [S] << Km, V0 depends on [S] and [Et].
Answers
Answer:
1. True. 2. True. 3. Not true. 4. True. 5. True
Explanation:
1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).
2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.
3. No, because it is measured in sec-1 and that means it is 1 rate constant.
4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.
5. The same as question two.
Answers
Now to find number of particles we take number of moles x avogadros number.
1.5 x (6.02x10^23) = 9.03x10^23 molecules
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C) the boiling point of water is 100’C
D) helium does not tend to react with anything
Answers
I think the answer is c) the boiling point of water is 100’c
I hope this helped xxx