Answer:
The potential wrt. calomel is 1.254 V
Explanation:
Given:
Potential wrt. silver chloride 
V
Potential wrt. saturated silver chloride 
V
Potential wrt. SCE 
V
Now potential wrt. hydrogen is given by,
V
And we find for potential wrt. calomel,
potential wrt. hydrogen + potential wrt. SEC
V
Therefore, the potential wrt. calomel is 1.254 V
Answer:
v = 46.5 m/s
Explanation:
Given data:
Mass of car = 1210 kg
Momentum of car = 56250 kg m/s
Velocity of car = ?
Solution:
Formula:
p = mv
p = momentum
m = mass
v = velocity
Now we will put values in formula:
56250 kg m/s = 1210 kg × v
v = 56250 kg m/s / 1210 kg
v = 46.5 m/s
So a car having mass of 1210 kg with momentum 56250 kg m/s having 46.5 m/s velocity.
Answer: During the summer, the sun's rays hit the Earth at a steep angle. ... Also, the long daylight hours allow the Earth plenty of time to reach warm temperatures. During the winter, the sun's rays hit the Earth at a shallow angle. These rays are more spread out, which minimizes the amount of energy that hits any given spot.
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
