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Snowcat [4.5K]
2 years ago
12

Copper(l) sulfide can react with oxygen to produce copper metal by the reactionCu2S + O2 => 2Cu + SO2.If 5.00 g of Cu2S is us

ed, then what is the theoretical yield of Cu?a. 10.0 gb. 1.99 gc. 1.00 gd. 5.00 ge. 3.99 g
Chemistry
1 answer:
Vlad1618 [11]2 years ago
6 0

Answer:

The answer to your question is e. 3.99 g

Explanation:

Data

Cu₂S = 5g

theoretical yield = ?

Chemical reaction

                             Cu₂S + O₂ ⇒ 2Cu + SO₂

Process

1.- Calculate the molecular mass of Cu₂S and Cu

Cu₂S = (2 x 64) + 32 = 160 g

Cu = 2 x 64 = 128 g

2.- Calculate the theoretical production of Cu

                            160 g of Cu₂S  --------------- 128 g of Cu

                                5 g of Cu₂S  --------------   x

                                x = (5 x 128) / 160

                                x = 640 / 160

                               x = 4 g of Cu

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A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa
bogdanovich [222]

Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

6 0
2 years ago
HELP
agasfer [191]

Answer:

3.02× 10²⁴ atoms

Explanation:

Given data:

Number of nitrogen atoms = ?

Number of moles of N₂O = 2.51 mol

Solution:

1 mole contain 2 mole of nitrogen atoms.

2.51 × 2 = 5.02  mol

According to Avogadro number,

1 mole = 6.022 × 10²³ atoms

5.02  mol ×  6.022 × 10²³ atoms / 1 mol

30.2 × 10²³ atoms

3.02× 10²⁴ atoms

5 0
1 year ago
Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K
DanielleElmas [232]

Answer:

Explanation:

  The following equation relates to Born-Haber cycle

\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L

Where

\triangle H_f is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

IE_M is ionisation enthalpy of metal

EA_X is electron affinity of non metal atom

U_L is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

U_L = - 707 KJ / mol

3 0
2 years ago
Show that the Newton has the units of mass times acceleration
Dimas [21]

F = ma = (kg)(m/s2) = kg ´ m/s2 N

hope this helps :D

7 0
2 years ago
Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2
Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has SP^{2}  O has SP^{2} .

Explanation : The orbitals of oxygen and carbon which are involved in SP^{2} hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and  O also has π  orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b:  O SP^{3}  H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen SP^{3} hydrogen atoms in it. It has SP^{3} hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c:   C is SP^{3}  O is also SP^{3}

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of SP^{3} hybridized bonds.

Answer 5) Bond d:   C atom has SP^{3}  C atom has SP^{3}

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be SP^{3}. Therefore, the orbitals that undergo SP^{3} hybridization are SP^{3}.

Answer 6) Bond e : C1 containing O SP^{2}    

C2 is SP^{3}

Explanation : The carbon atom which contains oxygen along with a double bond has SP^{2} hybridized orbitals involved in the bonding process; whereas the carbon at C2 has SP^{3} hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0
2 years ago
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