Answer:
234 KJ ≡ 55.887 Kcal
Explanation:
∴ J ≡ Kg/m².s² ≡ N.m = 0.2389 cal
∴ cal = 4.187 J
⇒ 234 KJ * ( 1000 J / KJ ) * ( cal / 4.187 J ) * ( Kcal / 1000 cal ) = 55.887 Kcal
Hey there!:
Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.
Molar mass of Mn= 54.938 g/mol
Moles of Mn = mass / molar mass
59.1 /54.938 => 1.07 ≈ 1 mol.
and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.
Molar mass of F=18.998 g/mol
Moles of F :
40.9 / 18.999 => 2.15 mol ≈ 2 mol.
The mole ratio between Mn:F= 1 : 2
Therefore the empirical formula of manganese fluoride:
=> MnF2=Mn1F2
Hope that helps!
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
A solution is a homogenous mixture meaning it has a uniform appearance throughout. (example: salt water)
heterogeneous mixture is visibly composed of different things. (example: chicken noodle soup)
Tin to Fluorine mass ratios:
1) For compound A:
38.5/12.3
= 3.13
2) For compound B:
56.5/36.2
= 1.56
The lowest whole number mass ratio is 2. It cannot be 1 because it is less than that required for compound B.
