Pb + Mg(NO₃)₂ → Pb(NO₃)₂ + Mg
This reaction would NOT occur because Pb is less reactive than Mg and as such Pb cannot displace the Mg in order for the reaction to occur under normal conditions.
Mg + Fe(NO₃)₂ → Fe + Mg(NO₃)₂
This reaction would occur. This is because Mg is more reactive than Fe and as such can displace it in the reaction, thus allowing the reaction to occur under normal conditions.
Cu + Mg(NO₃)₂ → Cu(NO₃)₂ + Mg
This reaction would NOT occur. Mg is more reactive than Cu, and as such copper cannot displace magnesium in order for the reaction to occur under normal conditions.
Answer:
The element is Rubidium . The ion only loss one electron to make the electron 36 . But the proton number(atomic number) is 37. The chemical compound should RbBr
X = Rubidium(Rb)
Explanation:
The ion of element X has a mass number of 85 and electron of 36. An elements is charge if it either loss or receive electron. Base on the charge the ion has loss electron. Bromine usually possess a negatively charge ion. That makes the X electron a positively charge ion.
proton number = Atomic number
Proton number = electron number when not charged
Mass number = proton number + neutron number
Proton is usually same number with the electron to make the element neutral. But when the atom is charge the atom usually loss or gain electron.
In this case, the atom X loss one electron thereby reducing the electron number to 36 . Recall, the electron is equal to proton only if the atom is uncharged. This means the proton number should be 37 as one electron was loss.
Element with proton number(atomic number) of 37 is rubidium. It is an alkali metal that is very reactive.
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
Answer:
draw a square with an errow pointing down
Explanation:
