<h3><u>Answer;</u></h3>
Cations are much smaller than their corresponding parent
<h3><u>Explanation;</u></h3>
- Parent atom has more electrons and thus the effective nuclear charge on each electron is less.
- When a cation is formed electron(s) is/are lost. Thus the effective nuclear charge or simply put, the attraction of the nucleus towards the electrons increases. Therefore, due to greater pull, the nucleus pulls the shells towards it, there by reducing the size, which makes cations smaller than their corresponding parent.
Answer:
the answer to that question is d
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%
m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water
%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
Answer:
Rb2CO3(aq)+Fe(C2H3O2)2(aq)--> 2Rb(C2H3O2)(aq) + FeCO3(s)
Explanation:
The reaction shown in the answer is the reaction of rubidium carbonate and iron II acetate. Rubidium is far more reducing than Fe II hence it can displace Fe II from its salt as shown.
The reducing property of metals depends on the value of their individual electrode potential values. For rubidium, its standard reduction potential is -2.98 V while that of Fe II is -0.44V. Hence rubidium can displace Fe II from its salt as shown above.
