Answer:
Explanation:
Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.
The integrated rate law for a first-order reaction is
![\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%5Cdfrac%7B%5BA%5D_%7B0%7D%7D%7B%5BA%5D%7D%20%5Cright%20%29%20%3D%20kt)
Data:
[A]₀ = 1.28 mol·L⁻¹
[A] = 0.17 [A]₀
k = 0.0632 s⁻¹
Calculation:
![\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cln%20%5Cleft%20%28%5Cdfrac%7B%5BA%5D_%7B0%7D%7D%7B0.170%5BA%5D_%7B0%7D%7D%20%5Cright%20%29%20%26%20%3D%20%26%200.0632t%5C%5C%5C%5C%5Cln%20%5Cleft%20%285.882%29%20%26%20%3D%20%26%200.0632t%5C%5C1.772%20%26%20%3D%20%26%200.0632t%5C%5C%5C%5Ct%20%26%20%3D%20%26%20%5Cdfrac%7B1.772%7D%7B0.0632%7D%5C%5C%5C%5Ct%20%26%20%3D%20%26%20%5Ctextbf%7B%7B28.0%20s%7D%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20will%20take%20%7D%20%5Cboxed%7B%5Ctextbf%7B28.0%20s%7D%7D%20%5Ctext%7B%20for%20%5BHI%5D%20to%20decrease%20to%2017.0%20%5C%25%20of%20its%20original%20value.%7D)
A compound is a substance composed of 2 or more chemical elements.
The model does not represent a compound because it shows two atoms of the same element combined. To be a compound the model would need to contain atoms of different element, represented by different colored balls. Therefore, the picture does not represent a compound. :)
<span>Errors can come from various sources: the obtainer of the data, the measuring instrument, the setting and etc. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.</span>
Answer:
254.5 K
Explanation:
Data Given
initial volume V1 of neon gas = 12.5 L
final Volume V2 of neon gas = 10.5 L
initial Temperature T1 of neon gas = 30 °C
convert Temperature to Kelvin
T1 = °C +273
T1 = 30°C + 273 = 303 K
final Temperature T2 of neon gas = ?
Solution:
This problem will be solved by using Charles' law equation at constant pressure.
The formula used
V1 / T1 = V2 / T2
As we have to find out Temperature, so rearrange the above equation
T2 = V2 x T1 / V1
Put value from the data given
T2 = 10.5 L x 303 K / 12.5 L
T2 = 254.5K
So the final Temperature of neon gas = 254.5 K
