Some distilled water is added to an empty beaker. a gram of copper(ii) nitrate is added to the beaker with stirring. after a few minutes, what is in the beaker?

a. solid copper(ii) nitrate and water
b. nitrogen gas, copper atoms, electrons, and water
c. a sticky material of copper hydronitrite
d. solid copper, nitrate ions, and water
5. copper ions, nitrate ions, and water

e. copper ions, nitrate ions, and water.

Explanation:

a and b are eliminated because copper and nitrate turn out to be aqueous due to solubility rules.

that leaves you with b, c, and e.

with these types of problems, you'll never end up with electrons as the answer, it'll only ask you what the products are, and in the reactants, 'electrons' wasn't part of a compound.

lastly, you are with c and e.

this is pretty obvious as well, because H2O will stay H2O when you dissociate it, so you won't have copper hydronitrite. (Plus, that would require taking out oxygen and changing nitrate to nitrite, which doesn't follow the law of conservation of mass.

they are both synthetic elements

explanation:

d

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