Equation: MgOH2 (s) --> MgO (s) + H2O (g)
And it's already balanced.
I had chemistry this semester too.
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7
Answer:
The elements in__Group_ 0 of the Periodic Table are called the_noble__gases. They are generally __unreactive_. because they have a__full_outer shell of electrons. So they do not need to gain__lose_or share _electrons_ with other atoms.
75% would be your answer since the lower letter represents a shorter plant
