Answer:
1. a = -10 m/s²
2. a = 1.8 × 10³ km/h²; s = 0.31 km
Explanation:
1.
Given:
Initial speed (u): 30 m/s
Final speed (v): 0 m/s (rest)
Time elapsed (t): 3.0 s
Asked:
Acceleration (a)
Solution:
We will use the following expression.
a = (v-u)/t = (0 m/s - 30 m/s)/3.0 s = -10 m/s²
Final Answer:
a = -10 m/s²
2.
Given:
Initial speed (u): 50 km/h
Final speed (v): 60 km/h
Time elapsed (t): 20 s
Asked:
(a) Acceleration (a)
(b) Distance traveled (s)
Solution:
(a) First, we will convert the time to hours.
20 s × 1 h/3600 s = 0.0056 h
Then, we will use the following expression.
a = (v-u)/t = (60 km/h - 50 km/h)/0.0056 h = 1.8 × 10³ km/h²
(b) We will use the following expression.
s = u × t + 1/2 × a × t²
s = 50 km/h × 0.0056 h + 1/2 × 1.8 × 10³ km/h² × (0.0056 h)² = 0.31 km
Final Answer:
a = 1.8 × 10³ km/h²
s = 0.31 km
<span>The cycles would eventually come to an end. If the sun were to burn out, the Earth would lose the energy and heat from the sun and soon lose its own heat and energy and fall apart into space. The magnetism would soon become demagnetized, plates would stop moving, trees would die, life etc. no more O2, CO2, cycle and all others would fade.</span>
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
At a certain concentration of 
and 
, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of
Answer : The initial rate of the reaction will be, 
Explanation :
Rate law expression for the reaction:
As we are given that:
Initial rate = 4.0 × 10⁴ M/s
Expression for rate law for first observation:
![4.0\times 10^4=k[H_2]^2[I_2]^2](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E4%3Dk%5BH_2%5D%5E2%5BI_2%5D%5E2)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[I_2]^2](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B4.0%5Ctimes%2010%5E4%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BI_2%5D%5E2%7D%7Bk%5BH_2%5D%5E2%5BI_2%5D%5E2%7D)
Therefore, the initial rate of the reaction will be,
<span>When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 </span>
<span>pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) </span>
<span>so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: </span>
<span>pH = (7.2076+12.3767) / 2 </span>
<span>pH = 9.79</span>
Answer: The boiling point of kerosene ranges from around 150 to 300 degrees Celsius normally being closer to 300 degrees Celsius.
Explanation:
