Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - 
logQ
= 0.18 - 
log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
Answer:
metal can hold heat keep it near fire
Answer: The pH of the solution is 11.2
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
moles of 
= 
(1g=1000mg)
Now put all the given values in the formula of molality, we get
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
According to stoichiometry,
1 mole of gives 2 mole of 
Thus 0.0298 moles of gives =
moles of
Putting in the values:
![pOH=-\log[0.0596]=2.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0596%5D%3D2.82)
Thus the pH of the solution is 11.2
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
