You can split the process in two parts:
1) heating the liquid water from 10.1 °C to 25.0 °C , and
2) vaporization of liquid water at constant temperature of 25.0 °C.
For the first part, you use the formula ΔH = m*Cs*ΔT
ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J
For the second part, you use the formula ΔH = n*ΔHvap
Where n is the number of moles, which is calculated using the mass and the molar mass of the water:
n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol
=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J
3) The enthalpy change of the process is the sum of both changes:
ΔH total = 1,874 J + 73,480 J = 75,354 J
Answer: 75,354 J
Answer:
The first thing we have to do is change and state all the units so that we can use our ideal gas law equation (
).
650 mmHg is a pressure unit, we have to convert this to kiloPascals. We know that 760 mmHg gives us 101 kPa.
P = 86kPa
T = 15°C + 273K = 288K
R (Gas constant) = 8.31 kj/mol
Molar mass of Ammonia (
) = (1 x 3) + (14) = 17g/mol
n (moles) = 
3.34 mol
V = ?
Rearrange the equation to solve for Volume:
Substitute the values inside:
V = 
<u>Therefore 93 L of volume is occupied by the ammonia gas.</u>
<u></u>
Answer:
H3C—CH3
Explanation:
The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.
Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.
The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.
Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.
