12.2 C
It has 3 significant figures now.
Answer:
Thermal energy is transferred between particles that are in direct contact with each other.
Thermal energy is transferred between objects of different temperatures.
Thermal energy is transferred from fast-moving particles to slow-moving particles.
Explanation:
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:
Part A: 47.8 mi/h
Part B: 0.072 M/s
Part C: 0.144 M/s
Explanation:
Part A
The average speed or velocity (V) is the variation of the space divided by the variation of the time:
V = (241 - 2)/(8 -3)
V = 47.8 mi/h
Part B
As Part A, the average rate (r) of formation of I2 is the variation of the concentration divided by the variation of time:
r = (1.83 - 1.11)/(15 - 5)
r = 0.072 M/s
Part C
The rates of the substances are proportional of their number of moles (n) which are their coefficient, so:
rI2/nI2 = rHCl/nHCl
0.072/1 = rHCl/2
rHCl = 2*0.072
rHCl = 0.144 M/s
When the balanced reaction equation is:
2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2
2:1
∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) * (1L ca(OH)2/0.585 mol Ca(OH)2
= 20.4 L
