Answer:
A small positively charged nucleus surrounded by revolving negatively charged electrons in fixed orbits
Answer:
Element 2
Explanation:
If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.
The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.
Fifty percent. plz brainliest my answer
Answer:
собак гандонов в гавне))))))))))
Explanation:
hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluсиськижопыссиськасуксиськипись loride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,whichйух reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.
Answer:
15.4%
Explanation:
If Ka = 0.54 mM = 1.51x10⁻⁵
Then;
C₄H₈O₂ --------> C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Recall that x is the percentage degree of dissociation
From the ICE table;
Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]
1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)
1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x
1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2
1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2
Hence;
0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0
x^2 + 0.028x - 0.028 = 0
Solving the quadratic equation here;
x = 0.154 or −0.182
Ignoring the negative result, x = 0.154
Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%
