Answer:
Your answer is B, Electrochemistry!
Explanation:
This is the part of chemistry that studies the chemical process in which electrons flow. This flow is called electricity. Electricity is generated by the flow of electrons, from one element to another element. This reaction is called oxidation reduction.
We need to know the value of van't hoff factor.
The van't hoff factor is: 2.66 or 2.7 (approximately)
(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.
From the equation: Δ
=i 
.m, where Δ= elevation of boiling point=102.5 - 100=2.5°C.
m=molality of solute=1.83 m (Given)
= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)
i= Van't Hoff factor
So, 2.5= i X 0.512 X 1.83
i=
i=2.66= 2.7 (approx.)
Answer:
<h3>The answer is 40.96%</h3>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
actual density = 2.49g/mL
error = 2.49 - 1.47 = 1.02
We have
We have the final answer as
<h3>40.96 %</h3>
Hope this helps you
Answer:
Option (D)
Explanation:
Phosphorylation can be simply defined as the addition of a phosphate group to an organic and inorganic molecule. This process helps in regulating the processes that occur in the cells. It leads to the growth and development of cells and this process is efficiently carried out with the help of enzymes like kinase. It also plays an important role in transferring the signals within the cells, synthesis, and functioning of proteins within the cells, and storing as well as releasing of energy.
Thus, the correct answer is option (D).
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
