If you drop a bath bomb into water, then it will fizz because a chemical reaction is taking place.
Answer:
no
Explanation:
Radium is silvery, lustrous, soft, intensely radioactive. It readily oxidizes on exposure to air, turning from almost pure white to black. Radium is luminescent, corrodes in water to form radium hydroxide. Although is the heaviest member of the alkaline-earth group it is the most volatile.
Answer:
The equilibrium shifts to produce more reactants.
Explanation:
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Consider the following chemical reaction.
Chemical reaction:
2SO₂ + O₂ ⇄ 2SO₃
In this reaction the equilibrium is disturb by increasing the concentration of Product.
When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of SO₃ increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.
Answer:
0.145 moles de AlBr3.
Explanation:
¡Hola!
En este caso, al considerar la reacción química dada:
Al(s)+Br2(l)⟶AlBr3(s)
Es claro que primero debemos balancearla como se muestra a continuación:
2Al(s)+3Br2(l)⟶2AlBr3(s)
Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:
![n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3](https://tex.z-dn.net/?f=n_%7BAlBr_3%7D%5E%7Bpor%5C%20Al%7D%3D4.00gAl%2A%5Cfrac%7B1molAl%7D%7B27gAl%7D%20%2A%5Cfrac%7B2molAlBr_3%7D%7B2molAl%7D%3D0.145mol%20AlBr_3%5C%5C%5C%5Cn_%7BAlBr_3%7D%5E%7Bpor%5C%20Br_2%7D%3D42.00gr%2A%5Cfrac%7B1molr%7D%7B160g%20Br_2%7D%20%2A%5Cfrac%7B2molAlBr_3%7D%7B3molBr_2%7D%3D0.175mol%20AlBr_3)
Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.
¡Saludos!
The answer is B. molality