Look up and compare the ionic radii of In3+, Mn3+, and Y3+ for a specific coordination number. Why do you think Mn3+ forms a solid substitution on the In3+ rather than the Y3+ site?

See explanation

Explanation:

Substitution implies that one thing is repalced by the other.

Substitution of one ion by another largely depends on the sizes of the ions in question.

The ionic radius of Mn3+ is about 0.65 Angstroms while that of In3+ is about 0.94 Angstroms. On the other hand, the ionic radius of Y3+ is about 1.032 Angstroms.

It is clear that the sizes of Mn3+ and In3+ are closer to each other hence Mn3+ can substitute for In3+ easily than Y3+ in a lattice site.

The larger size of Y3+ explains why it is not easily replaced by Mn3+.

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol