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Anon25 [30]
3 years ago
11

Calculate the number of mg of Mn2+ left

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.

Explanation:

The reaction that takes place is:

Mn⁺² + S⁻² ⇄ MnS(s)  

ksp = [Mn⁺²] [S⁻²]

If the pksp of MnS is 13.500, then the ksp is:

ksp=10^{-13.500}=3.1623*10^{-14}

From the problem we know that [S⁻²] = 0.0900 M

We use the ksp to calculate [Mn⁺²]:

3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M

[Mn⁺²] = 3.514 * 10⁻¹³ M.

Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:

3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.

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Answer:

When Epsom salt (magnesium sulfate) dissolves, it separates into its ions: a magnesium ion (Mg2+) and a sulfate ion (SO4 2-), which results in hard water. When hard water and soap are mixed, the magnesium ion reacts with soap molecules and forms a solid material called a precipitate, which does not dissolve.

Explanation:

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3 years ago
Which of the following symbols represents a chlorine ion with a stable arrangement of eight valence electrons?
frozen [14]

A stable arrangement of eight valence electrons : ³⁵Cl⁻¹

<h3>Further explanation</h3>

Chlorine is a halogen gas, located in group 17, p block

Chlorine has an atomic number of 17 and an atomic mass of 35

Electron configuration: [Ne] 3s²3p⁵

If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)

So by binding this one electron, chlorine forms negative ions (anions)

³⁵Cl⁻¹

B. Cl⁻² binds 2 electrons, exceeding the octet rule

C. Cl⁺¹, releases 1 electron, remains unstable

D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration

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3 years ago
Can someone help me with this
Trava [24]

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Significant Digits and Scientific Notation Part 1: Determine the number of significant digits in each number and write out the s
MA_775_DIABLO [31]

Answer:

1. 6 significant digits. 2. 2 significant digits. 3. 8 significant digits. 4. 4 significant digits. 5. 8 significant digits. 6. 7 significant digits. 7. 2 significant digits.

Explanation:

All nonzero digits are significant. All zeros before nonzeros are insignificant. All zeros after nonzeros are insignificant unless they are made for precision of some data. So 405000 may have 3 or 6 significant figures. But usually all zero figures after significant digits are significant.

8 0
3 years ago
How many moles of ammonium nitrate will be produced 110.0g of ammonium carbonate
marshall27 [118]

Answer:

                     2.288 Moles of NH₄NO₃

Explanation:

The Balance chemical equation is as follow:

                  (NH₄)₂CO₃ + 2 HNO₃  →  2 NH₄NO₃ + H₂O + CO₂

To solve this problem we will do following steps:

Finding moles of Ammonium Carbonate:

As we know,

                        Moles  =  Mass / M.Mass

So,

                        Moles  =  110 g  /  96.08 g/mol

                        Moles  =  1.144 moles

Calculating moles of Ammonium Nitrate:

According to balance chemical equation;

                      1 mole of (NH₄)₂CO₃ produces  =  2 moles of NH₄NO₃

So,

          1.144 moles of (NH₄)₂CO₃ will produce  =  X moles of NH₄NO₃

Solving for X,

                     X =  2 moles × 1.144 moles ÷ 1 mole

                     X  =  2.288 moles of NH₄NO₃

8 0
3 years ago
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