Answer:
C₄H₈O₂.
Explanation:
- Firstly, we can calculate the no. of moles (n) of each component using the relation:
<em>n = mass/atomic mass,</em>
mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.
mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.
mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.
- To get the empirical formula, we divide by the lowest no. of moles (2.26 mol) of O:
∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.
<em>∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) =</em> 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol)<em> = 44.0 g/mol.</em>
∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.
∴ The molecular formula is, 2(C₂H₄O), that is; <em>(C₄H₈O₂)</em>
Answer: a) pH of a 0.1 M vinegar solution is 2.9
b) It is an acid as pH is less than 7
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.
As vinegar is a weak acid, its dissociation is represented as;
cM 0 0
So dissociation constant will be:
Give c= 0.1 M
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.1\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.1%5Ctimes%20%5Calpha)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[1.3\times 10^{-3}]=2.9](https://tex.z-dn.net/?f=pH%3D-log%5B1.3%5Ctimes%2010%5E%7B-3%7D%5D%3D2.9)
Thus pH of a 0.1 M vinegar solution is 2.9
As pH is less than 7, it is an acid.
A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
<h3>What is a first order reaction?</h3>
It is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.
First, we will calculate the rate constant using the following expression.
ln ([A]/[A]₀) = - k × t
ln (1/16[A]₀/[A]₀) = - k × 500 s
k = 5.55 × 10⁻³ s⁻¹
where,
- [A] is the final concentration of the reactant.
- [A]₀ is the initial concentration of the reactant.
- k is the rate constant.
- t is the elapsed time.
Next, we can calculate the half-life (th) using the following expression.
th = ln 2 / k = ln 2 / (5.55 × 10⁻³ s⁻¹) = 125 s
A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
Learn more about first order reactions here: brainly.com/question/518682
Answer:
<em>The correct option is C) All of the genetic information comes from one parent.</em>
Explanation:
The description in the paragraph tells that Astrammina rara are unicellular organisms and their offspring born are identical to the parent cell.
Asexual reproduction can be described as a method of reproduction in which a parent cells produces two offsprings which are identical to the parent cell. Asexual reproduction is the common mode of reproduction in unicellular organisms.
Hence, the description shows that Astrammina rara reproduce by this method and hence all of the information will arise from one, single parent.
