Answer:
The Kelvin Scale has the highest value fir the boiling point of water. The Kelvin degree is the same size as the Celsius degree; hence the two reference temperatures for Celsius, the freezing point of water (0°C), and the boiling point of water (100°C), correspond to 273.15°K and 373.15°K, respectively.
Answer:
1. The pressure will be 32 atm, twice the initial pressure.
2. The pressure will be 1.83 atm, one third of the initial pressure.
Explanation:
Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.
This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:
P1*V1=P2*V2
1. In this case:
- P1= 16 atm
- V1
- P2= ?
- V2= V1÷2=
because the volume is halved.
So:
16 atm*V1= P2* 
Solving:
=P2
16 atm*2= P2
32 atm= P2
<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>
2. Now
- P1= 5.5 atm
- V1
- P2= ?
- V2= V1*3 because the volume is tripled.
So:
5.5 atm*V1= P2* V1*3
Solving:
=P2
= P2
1.83 atm= P2
<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>
Answer:
element
Explanation:
an element is something made up of only one type of atom
Answer:
54g
Explanation:
Given parameters:
Number of moles of H₂O = 3 moles
Unknown:
mass of water = ?
Solution:
To solve this problem, we use the expression below:
mass = number of moles x molar mass
Molar mass of H₂O = 2(1) + 16 = 18g/mol
Mass of water = 3 x 18 = 54g
Answer:
Explanation:
It is volume-volume problems that does not require the use of molar mass.