Answer:
Two moles of KClO3 decompose to form 5 moles of product.
Volume is the amount of space the matter takes up, density is mass OF the volume
Answer:
3/2a
Explanation:
The complete step by step answer is found in the attachment
Answer:
The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C
Explanation:
Here we make use of the Clausius-Clapeyron equation;

Where:
P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K
P₂ = 0.2 atm = The substance vapor pressure at temperature T₂
= The heat of vaporization = 28.5 kJ/mol
R = The universal gas constant = 8.314 J/K·mol
Plugging in the above values in the Clausius-Clapeyron equation, we have;


T₂ = 440.37 K
To convert to Celsius degree temperature, we subtract 273.15 as follows
T₂ in °C = 440.37 - 273.15 = 167.22 °C
Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.
Answer:
B
Explanation:
CH4 + O2 →H2O + CO2
Left side
C = 1
H = 4
O = 2
Right side
C = 1
H = 2
O = 3
So find common denominator in this case would be 6 O
CH4 + 3O2 →2H2O + CO2
Left side
C = 1
H = 4
O = 6
Right side
C = 1
H = 4
O = 6