Which of the following substances would complete a circuit and allow electricity to flow?
i. potassium chloride lattice
ii. lithium bromide in a solution
iii. diamond
iv. nitric oxide in a solution
v. graphite

The substances which allow the current to flow in the circuit are:  

Lithium bromide in a solution and Graphite.  

Lithium bromide can be pass the current in the circuit, due to the ability to form ions in the solution Li⁺ and Br⁻ which are mobile and can transport the electrical charge.

In the graphite because of the presence of delocalised electrons, each carbon atom forms three covalent bonds with three other carbon atoms, it conducts electrical current.

In the case of potassium chloride lattice there are in a solid form and the atoms are not mobile so they can not form ions.  In the diamond all the carbon valences are satisfied so the electrical charge can not be transported.

Nitric oxide in a solution is isolated and can not transport the electrical current due to its inability of the to form ions .

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

po-215 i got it right on the quiz