How can models demonstrate theories of our expanding universe?

Chemistry

1 answer:

Scilla [17]2 years ago

7 0

Answer:

Explanation:Puesto que la energía causante de la aceleración del espacio-tiempo no ha podido ser observada en forma directa, se ha dado en llamarla energía oscura. Dos candidatos teóricos que podrían hacer las veces de esta energía son una constante cosmológica no igual a cero (que pudo haber causado la inflación cósmica) y una energía repulsiva más general llamada quintaesencia.

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Why copper should not be disposed of in landfill sites?

Dafna1 [17]

Copper is a heavy metal that can harm many people. This process of heavy metals building in your body is called bioaccumulation and the process of this building in the food chain is called biomagnification. Its hard to rid of this heavy metal because it is very dense, therefore, cannot breakdown in our bodies. Heavy metals get consumed by producers and then reach to the top of the pyramid (us) so we have the highest risk of getting diseases and so on. Make sure you explain that heavy metals get passed from industrial purposes where companies do not dispose of waste properly.
Source: Exam as well
Good luck!

4 0

3 years ago

Read 2 more answers

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M