Answer:
they are equal
Explanation:
the Law of Conservation of Mass states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change
Answer:
Some energy is lost as heat
Explanation:
It is correct to say that as the gasoline is converted to mechanical energy in the automobile engine, some of the energy is lost as heat.
Heat energy is on of the ways energy is lost in any system. The movement of mechanical parts and even the combustion of the gasoline produces heat energy.
These energy are usually lost to the environment.
Answer:
Hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system and first aid kit.
Explanation:
Working in laboratories has many risks, therefore, preventive measures that should be incorporated to avoid the occurrence of any laboratory accidents.
Some of the important emergency equipment that should be available in laboratories are: hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system, chemical storage cabinet, first aid kits and fume hood.
Some of the personal protective equipment include lab coats, goggles, safety gloves and face shield.
Answer:
ΔHreaction = 263.15 kJ/mol
Explanation:
The reaction is as follow:
OH + CF₂Cl₂ → HOF + CFCl₂
You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol