Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles
Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles
Here, the limiting reagent is NaOH
The reaction is represented as:
HCOOH + NaOH ↔HCOONa + H2O
Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles
Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles
Total final volume is given as 1 L
Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M
[HCOONa] = 0.015/1 = 0.015 M
pKa of HCOOH = 3.74
As per Henderson-Hasselbalch equation
pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22
Therefore, pH of the final solution = 4.22
Answer : The correct option is, (C) 6
Explanation :
Oxidation-reduction reaction : It is a reaction in which oxidation and reduction reaction occur simultaneously.
Oxidation reaction : It is the reaction in which a substance looses its electrons. In this oxidation state increases.
Reduction reaction : It is the reaction in which a substance gains electrons. In this oxidation state decreases.
The given unbalanced chemical reaction is,
Half reactions of oxidation and reduction are :
Oxidation : 
......(1)
Reduction : 
.......(2)
In order to balance the electrons, we multiply equation 1 by 2 and equation 2 by 3, we get:
Oxidation : 
......(1)
Reduction : 
.......(2)
The overall balanced chemical reaction will be:
From this reaction we conclude that the electrons are getting transferred from iron to iodine and the number of electrons transferred are 6 electrons.
Hence, the correct option is, (C) 6
Answer: D. a structure that provides a specific service within a cell
Explanation:
Answer:
Mass, m = 1.51 grams
Explanation:
It is given that,
The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm
Length of the cylinder, h = 4.2 cm
We know that the density of the Aluminium is 2.7 g/cm³
Circumference, C = 2πr
Density is equal to mass per unit volume.
m is mass of the cylinder
V is the volume of the cylinder
So,
So, the mass of the cylinder is 1.51 grams.
Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) S₈
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.
(b) C₂H₁₂
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Carbon , C = 12 g/mol
Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol
(c) Sc₂(SO₄)₃
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
scandium , Sc = 45 g/mol.
Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol
(d) CH₃COCH₃ (acetone)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol
(e) C₆H₁₂O₆ (glucose)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.
