The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
Answer: X could represent the element of oxidation state (+2) such as (Mg2+, Pb2+, Ba2+, Ca2+, Ba2+, Zn2+, ....etc)
Explanation:
- The formula of the compound XSO4 is a neutral compound that the algebraic summation of the oxidation states of different elements in it must be zero.
- The group SO4 has the oxidation state (2-), that S has (6+) oxidation state and O has (2-) oxidation state, so the oxidation of SO4 = (6+) + (-2*4) = -2.
- It is clear that X must have the oxidation state 2+.
- So, X could be represents by many different elements such as (Mg2+, Pb2+, Ba2+, Ca2+, Ba2+, Zn2+, Fe2+, ....etc)
<span>The symbol on a line that indicates a cold front on a map is a triangle. The point of the triangle will indicate the direction of movement of the cold front.</span>
Answer:
phosphodiester bond
Explanation:
<em>Phosphodiester linkage/bond is found in deoxyribonucleic and ribonucleic acids. It is formed from a reaction involving the elimination of water from a reaction involving the hydroxyl groups of two different 5-carbon (pentose) sugars and a phosphate group.</em>
The elimination of water, also known as condensation reaction occur twice, resulting in the formation of two ester bonds which then bind the phosphate group to the pentose sugars to become a phosphodiester bond.
The bond links the 3'-hydroxyl group of one of the pentose sugars and the 5'-hydroxyl group of the other pentose sugar in the nucleotides that make up nucleic acids.
Answer: Temperature final = 103 °C
Explanation: To solve for final temperature we use the equation of heat:
Q= mc∆T
Next derive the equation to find final temperature
Q = mc(T final - T initial)
Q / mc = T final - T initial
Transpose T initial and change the sign so that T final will be left.
T final = Q / mc + T initial
Substitute the values:
T final = 305 J / 28.8 g x 0.128 J/(g°C)
= 305 J / 3.6864 J/°C
= 82.7 + 20.0°C
= 103 °C final temperature.
