A redox reaction --> a reaction whereby oxidation & reduction occurs
Reduction:
Charge of Cl2 = 0
Charge of Cl- in NaCl = -1
Hence, since charge of Cl2 decreased from 0 in Cl2 to -1 in NaCl, reduction occured.
Oxidation:
Charge of Na = 0
Charge of Na+ in NaCl = +1
Hence, since charge of Na increased from 0 in Na to +1 in NaCl, oxidation occured.
Since both oxidation & reduction occured in the reaction, it is a redox reaction.
Q = mcΔθ
67.5 = m x 0.45 x (28.5 - 21.5)
M = 67.5 / 3.15
= 21.4 g
Answer:
Number of moles = 0.042 mol
Explanation:
Given data:
Number of moles = ?
Mass of calcium carbonate = ?
Solution:
Formula:
Number of moles = mass/ molar mass
now we will calculate the molar mass of calcium carbonate.
atomic mass of Ca = 40 amu
atomic mass of C = 12 amu
atomic mass of O = 16 amu
CaCO₃ = 40 + 12+ 3×16
CaCO₃ = 40 + 12+48
CaCO₃ = 100 g/mol
Now we will calculate the number of moles.
Number of moles = 4.15 g / 100 g/mol
Number of moles = 0.042 mol
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
Answer:
[H₂SO₄] = 6.07 M
Explanation:
Analyse the data given
8.01 m → 8.01 moles of solute in 1kg of solvent.
1.354 g/mL → Solution density
We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g
Mass of solvent = 1kg = 1000 g
Mass of solution = 1000g + 785.4 g = 1785.4 g
We apply density to determine the volume of solution
Density = Mass / volume → Volume = mass / density
1785.4 g / 1.354 g/mL = 1318.6 mL
We need this volume in L, in order to reach molarity:
1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L
Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M
