Answer:
Explanation:
It is malleable, ductile, and a good conductor of electricity and heat.
A: Trial 1, because the average rate of the reaction is lower.
The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.
If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.
Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.
Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.
Explanation:
a) when zinc burnt in oxygen.
2Zn + O2 -----∆-----> 2ZnO(black residue)
b) when carbon burnt in oxygen.
C+O2----∆---> CO2.
c) when sulphur burnt in oxygen.
S+O2-----∆-----> SO2.
d) when Calcium burnt in oxygen.
2Ca+O2-----∆-----> 2CaO(black residue)
e) when Magnesium burnt in oxygen.
2Mg+O2-----∆----> 2MgO.
f) when sodium burnt in oxygen.
4Na+O2----∆-----> 2Na2O.
hope all these reactions help you.
Physical change alters a given material without changing its chemical.
Answer : The equilibrium concentration of
at
is,
.
Solution : Given,
Equilibrium constant, 
Initial concentration of
= 0.260 m
Let, the 'x' mol/L of
are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,

Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get

By rearranging the terms, we get the value of 'x'.

Therefore, the equilibrium concentration of
at
is,
.