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s344n2d4d5 [400]
2 years ago
13

Assuming complete precipitation, how many moles of each ion remain in solution? if an ion is no longer in solution, enter a zero

(0) for the number of moles.
SAT
1 answer:
Alex2 years ago
3 0

Based on the information given, it should be noted that there'll be 0.06537 moles of HgS and 0.13074 moles of the remaining compound.

<h3>What is precipitation?</h3>

Precipitation simply means the process of transforming a dissolved substance into an insoluble solid. It's the conversion of a solution into a solid.

From the complete information, it can be inferred from the reaction that was balanced that there'll be 0.06537 moles of HgS and 0.13074 moles of the remaining compound. This shows the moles of each ion that remains.

Learn more about precipitation on:

brainly.com/question/24704707

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A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin
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The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

  • 131.6 N/C
  • 12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}

Which gives;

\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C

\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C

Which gives;

\vec{E}_1 = \mathbf{21.6 \, N/C  \cdot \hat x +  28.8 \, N/C \hat y}

\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C

Therefore;

\vec  {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)

\vec  {E} = \mathbf{\left( -128.4 \, N/C  \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}

The magnitude of the net electric field is therefore;

E = \sqrt{(-128.4^2 + 28.8^2)} ≈ 131.6

  • The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

  • The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis

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