Answer:
The molar mass of the unknown gas is
Explanation:
Let assume that the gas is O2 gas
O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.
Under the same conditions;
the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.
From Graham's Law of Diffusion;
Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.
i.e
where K = constant
If we compare the rate o diffusion of two gases;
Since the density of a gas d is proportional to its relative molecular mass M. Then;
Rate is the reciprocal of time ; i.e
Thus; replacing the value of R into the above previous equation;we have:
We can equally say:
Answer:
El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:
El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.
Si se continúan colocando átomos de carbono con enlaces sencillos entre ellos e hidrógenos en los enlaces libres, se crean largas cadenas de compuestos. Al etano le sigue el propano (C2H8) y a éste, el butano (C4H10). Todos estos compuestos forman parte de la familia de los alcanos, y sus nombres terminan con el sufijo –ano para indicar que pertenecen a la misma familia.
Answer:
a
Explanation:
euejejejueeijeejejejejejje
Answer: The mass of given amount of copper (II) cyanide is 462.4 g
Explanation:
To calculate the number of moles, we use the equation:
We are given:
Moles of copper (II) cyanide = 4 moles
Molar mass of copper (II) cyanide = 115.6 g/mol
Putting values in above equation, we get:
Hence, the mass of given amount of copper (II) cyanide is 462.4 g
Answer is: gamma emission or gamma decay.
<span>During gamma emission the nucleus emits radiation without changing its composition, if for example have nucleus with six protons and six neutrons (carbon atom) and after gamma decay there is nucleus with six protons and six neutrons.
</span>Gamma rays are the
electromagnetic waves with the shortest wavelengths (1 pm), highest
frequencies (300 EHz) and highest energy (1,24 MeV).