Answer: They are arranged by atomic number.
Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
Given is the specific heat of water equal to 4.18 Joule per gram per *C.
This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.
Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.
To do so we have formula
Q = m C (temperature difference)
Have a look at pic for answer
Answer:
14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.
Explanation:
The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.
During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.
So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.
In this case, being:
- L= 84

and replacing in the expression Q = m*L you get:
Q=172 g*84 
Q=14,448 J
<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>