Answer:
The new volume after the temperature reduced to -100 °C is 0.894 L
Explanation:
Step 1: Data given
Volume of nitrogen gas = 1.55 L
Temperature = 27.0 °C = 300 K
The temperature reduces to -100 °C = 173 K
The pressure stays constant
Step 2: Calculate the new volume
V1/T1 = V2/T2
⇒with V1 = the initial volume of the gas = 1.55 L
⇒with T1 = the initial temperature = 300 K
⇒with V2 = the new volume = TO BE DETERMINED
⇒with T2 = the reduced temperature = 173 K
1.55 L / 300 K = V2 / 173 K
V2 = (1.55L /300K) * 173 K
V2 = 0.894 L
The new volume after the temperature reduced to -100 °C is 0.894 L
Answer:
27.22 dm³
Explanation:
Given parameters:
number of moles = 1 mole
temperature= 50°C, in K gives 50+ 273 = 323K
Pressure= 98.6kpa in ATM, gives 0.973 ATM
Solution:
Since the unknown is the volume of gas, applying the ideal gas law will be appropriate in solving this problem.
The ideal gas law is mathematically expressed as,
Pv=nRT
where P is the pressure of the gas
V is the volume
n is the number of moles
R is the gas constant
T is the temperature
Input the parameters and solve for V,
0.973 x V = 1 x 0.082 x 323
V= 27.22 dm³
Answer:
Ksp = 8.8x10⁻⁵
Explanation:
<em>Full question is:</em>
<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>
<em />
When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:
PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻
Ksp = [Pb²⁺] [Cl⁻]²
If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):
Ksp = [X] [2X]²
Ksp = 4X³
As X is the amount of Pb²⁺ = 2.8x10⁻²M:
Ksp = 4(2.8x10⁻²)³
<h3>Ksp = 8.8x10⁻⁵</h3>
Answer:
The options <u>(A) -</u>The rate law for a given reaction can be determined from a knowledge of the rate-determining step in that reaction's mechanism. and <u>(C) </u>-The rate laws of bimolecular elementary reactions are second order overall ,<u>is true.</u>
Explanation:
(A) -The rate law can only be calculated from the reaction's slowest or rate-determining phase, according to the first sentence.
(B) -The second statement is not entirely right, since we cannot evaluate an accurate rate law by simply looking at the net equation. It must be decided by experimentation.
(C) -Since there are two reactants, the third statement is correct: most bimolecular reactions are second order overall.
(D)-The fourth argument is incorrect. We must track the rates of and elementary phase that is following the reaction in order to determine the rate.
<u>Therefore , the first and third statement is true.</u>
Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
