<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life.
Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>
Answer:
The fraction of energy used to increase the internal energy of the gas is 0.715
Explanation:
Step 1: Data given
Cv for nitrogen gas = 20.8 J/K*mol
Cp for nitrogen gas = 29.1 J/K*mol
Step 2:
At a constant volume, all the heat will increase the internal energy of the gas.
At constant pressure, the gas expands and does work., if the volume changes.
Cp= Cv + R
⇒The value needed to change the internal energy is shown by Cv
⇒The work is given by Cp
To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp
Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol
Cv/Cp = 0.715
The fraction of energy used to increase the internal energy of the gas is 0.715
Answer:
V₂ = 15.6 L
Explanation:
Given data:
Initial volume = 175 mL (0.175 L)
Initial pressure = 1 atm
Initial temperature = 273 K
Final temperature = -5°C (-5+273 = 268 K)
Final volume = ?
Final pressure = 1.16 kpa (1.16/101=0.011 atm)
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm
V₂ = 46.9 L / 3.003
V₂ = 15.6 L
The warmer road surface at the end of a sunny day is the black road because during the day it absorbed more radiation (sunlight) than the withe one
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
