Answer:
Freezing point = 1.25
Explanation:
If we increase the concentration of the solution, the concentration of H+ does not change.
Convert 2.5% in to decimal
2.5% = 2.5 ÷100
= 0.025
The freezing point = 0.025 × 50
= 1.25
1. temperature is dependent
2. energy is independent
3. the graph looks like a line sloping upward
4. the line means that as energy increases the temperature also increases
5. a straight line would mean that as energy increases temperature remains constant
sorry that's all I can do
El número de Avogadro es 6,022 x 10^23, y es el número de átomos que hay en un mol de dicho elemento. A su vez un mol es la cantidad de un elemento cuya masa en gramos coincide con el peso atómico.
Por tanto, 6,022 x 10^23 átomos del elemento tienen una masa en gramos igual a su peso atómico. Hacemos una regla de tres:
1 gramo -------- 1,5 x 10^22 átomos
x ------------------ 6,022 x 10^23 átomos
=> x = 40,1 gramos por mol del elemento.
De modo que su peso atómico es 40,1. Se trata del calcio.
Saludos.
Answer:
Approximately 6.81 × 10⁵ Pa.
Assumption: carbon dioxide behaves like an ideal gas.
Explanation:
Look up the relative atomic mass of carbon and oxygen on a modern periodic table:
Calculate the molar mass of carbon dioxide
:
.
Find the number of moles of molecules in that
sample of
:
.
If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:
,
where
is the pressure inside the container.
is the volume of the container.
is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
is the ideal gas constant.
is the absolute temperature of the gas.
Rearrange the equation to find an expression for
, the pressure inside the container.
.
Look up the ideal gas constant in the appropriate units.
.
Evaluate the expression for
:
.
Apply dimensional analysis to verify the unit of pressure.
Answer:
oxygen
Explanation:
The given isotope has 8 protons and 8 electrons, so the atomic number of the given isotope is 8, which is the atomic number of oxygen.