When 20.0 ml of a 0.250 m (nh4)2s solution is added to 150.0 ml of a solution of cu(no3)2, a cus precipitate forms. the precipitate is then filtered from the solution, dried, and weighed. if the recovered cus is found to have a mass of 0.3491 g, what was the concentration of copper ions in the original cu(no3)2 solution?

The concentration fo copper in the Cu(NO₃)₂ solution is 0.0243 M.

Explanation:

Hi there!

Let´s write the balanced equation of the reaction:

(NH₄)₂S (aq) + Cu(NO₃)₂ (aq) → CuS (s) + 2NO₃⁻(aq) + 2NH₄⁺ (aq)

We obtain 0.3491 g of CuS. The molar mass of CuS is calculated adding the molar mass of Cu (63.546 g) plus the molar mass of S (32.065 g):

Molar mass CuS = 63.546 g + 32.065 g = 95.611 g

Then, if 1 mol CuS has a mass of 95,611 g, 0.3491 g of CuS will be the mass of (0.3491 g of CuS · 1 mol CuS / 95,611 g CuS) 3.651 × 10 ⁻³ mol CuS.

Looking at the chemical equation, we can notice that 1 mol of Cu(NO₃)₂ is needed to produce 1 mol CuS, then, the number of moles of Cu(NO₃)₂ in the solution will be equal to the number of moles of CuS obtained: 3.651 × 10 ⁻³ mol.

This number of moles is present in 0.1500 l, then, in 1 l there will be

(1.000 l · 3.651 × 10 ⁻³ mol / 0.1500 l) 0.0243 mol Cu(NO₃)₂

The concentration of Cu(NO₃)₂ is 0.0243 M. Since 1 mol of Cu(NO₃)₂ dissociates in 1 mol of Cu and 2 mol of NO₃, the concentration of Cu will be the same as the concentration of Cu(NO₃)₂ :  0.0243 M.

Let´s corroborate if this is possible. From the chemical equation, we know that 1 mol (NH₄)₂S reacts with 1 mol of Cu(NO₃)₂, so there must be at least 3.651 × 10 ⁻³ mol (NH₄)₂S to produce the amount of CuS obtained.

number of moles of (NH₄)₂S = 0.250 mol/l · 0.020 l = 5.00 × 10⁻³ mol

(NH₄)₂S is in excess (5.000 × 10⁻³ mol > 3.651 × 10 ⁻³ mol) , so that all the Cu(NO₃)₂ will react.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

points

explanation: