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3 years ago

Given the following equation: 8Fe+S8=8FeS

Chemistry

1 answer:

maxonik [38]3 years ago

5 0

Answer:

a. 2954g of FeS

b. 1143g of FeS

Explanation:

Based on the reaction:

8 Fe + S₈ → 8 FeS

8 moles of Fe reacts with 1 mole of S₈ to produce 8 moles of FeS

a. 4.2mol sulfur produce:

4.2mol S₈ × ( 8 mol FeS / 1 mol S₈) = 33.6mol FeS. As molar mass of FeS is 87.92g/mol, mass is:

33.6mol FeS × ( 87.92g / 1 mol FeS) = 2954g of FeS

b. 13mol Fe produce:

13mol Fe × ( 8 mol FeS / 8 mol Fe) = 13mol FeS. In mass:

13 mol FeS × ( 87.92g / 1 mol FeS) = 1143g of FeS

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

What is the only planet in our solar system that rotates in a different direction from the other planets??

aksik [14]

Venus so that the sun rises in the west

4 0

3 years ago

at standard pressure how do the boiling point and the freezing point of NaCl compare to the boiling point and freezing point of

Ede4ka [16]

Usually in this context you would be referring to the boiling and freezing point of a NaCl solution (saltwater) compared to pure H_{2}O. Sematics would be different for NaCl compound itself, you would say melting and boiling point for a solid substance- and the temperatures would be very, very radical (high).
The boiling point of pure water is 100 degrees C (212 F), and the freezing/melting point is below 0 degrees C (32 F). For a salt water solution, the boiling point is raised and the melting point is lowered. This means that water will stay liquid for an increased range of temperature. Depending on the amount of NaCl solute in the water, the boiling and melting points may change a few degrees.

7 0

3 years ago

If 5.0 g of copper metal reacts with a solution of silver nitrate, how many grams of silver metal are recovered?

sukhopar [10]

Answer:

16.9g

Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2

Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.

Then to calculate the theoretical yield you need to compare moles of the reactants:

m(Cu)=5g

M(Cu)=63.55

n(Cu)=5/63.55=0.0787

By comparing coefficients you require twice as much silver: 0.157mol

n(Ag)=0.157

M(Ag)=107.86

m(Ag)=0.157x107.86=16.9g

Hence, the theoretical yield of this reaction would be 16.9g

3 0

3 years ago

Write the formula for potassium oxide. why do you not need prefixes in the name

Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

Potassium- K- on the left end of period four.
Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

Potassium is a metal,
Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.

Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion $\text{O}^{2-}$ when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each $\text{O}^{2-}$ ion is twice that on a $\text{K}^{+}$ ion. Each $\text{K}^{+}$ would pair up with two $\text{O}^{2-}$ . Hence the subscript in the formula: $\text{K}_{\bf 2}\text{O}$ .

There are two classes of compounds:

Covalent compounds, which need prefixes, and
Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide $\text{CO}_2$ , the prefix di- indicates that there are two oxygen atoms in the formula $\text{CO}_2$ . However, there's no need for prefix in ionic compounds such as $\text{K}_2\text{O}$ .

7 0

3 years ago