No−3(aq)+4h+(aq)+3e−→no(g)+2h2o(l)e ∘=0.96v clo2(g)+e−→clo2−(aq)e∘=0.95v cu2+(aq)+2e−→cu(s)e∘=0.34v 2h+(aq)+2e−→h2(g)e∘=0.00v pb2+(aq)+2e−→pb(s)e∘=−0.13v fe2+(aq)+2e−→fe(s)e∘=−0.45v part a use appropriate data to calculate e∘cell for the reaction. 3cu(s)+2no−3(aq)+8h+(aq)→3cu2+(aq)+ 2no(g)+4h2o(l)

The standard electrode potential of the cell is 0.62 V.

Explanation:

For the given chemical equation:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Oxidation half reaction:       ( × 3 )

Reduction half reaction:      ( × 2 )

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the of the reaction, we use the equation:

Hence, the standard electrode potential of the cell is 0.62 V.

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms

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