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castortr0y [4]
3 years ago
9

if 334.6 g of phosphoric acid is reacted with excess potassium hydroxide. the final mass K3PO4 produced is found to be 248g. wha

t is the percent yield?
Chemistry
1 answer:
Orlov [11]3 years ago
3 0

Answer:

                     %age Yield  =  34.21 %

Explanation:

                   The balance chemical equation for the decomposition of KClO₃ is as follow;

                            3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

Step 1: Calculate moles of H₃PO₄ as;

Moles = Mass / M/Mass

Moles = 334.6 g / 97.99 g/mol

Moles = 3.414 moles

Step 2: Find moles of K₃PO₄ as;

According to equation,

                 1 moles of H₃PO₄ produces  =  1 moles of K₃PO₄

So,

              3.414 moles of H₃PO₄ will produce  =  X moles of K₃PO₄

Solving for X,

                      X = 1 mol × 3.414 mol / 1 mol

                      X = 3.414 mol of K₃PO₄

Step 3: Calculate Theoretical yield of K₃PO₄ as,

Mass = Moles × M.Mass

Mass = 3.414 mol × 212.26 g/mol

Mass = 724.79 g of K₃PO₄

Also,

%age Yield  =  Actual Yield / Theoretical Yield × 100

%age Yield  =  248 g / 724.79 × 100

%age Yield  =  34.21 %

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Is selenium tetrafluoride an ionic or covalent bond?
nasty-shy [4]

Answer:

Synthesis

The first reported synthesis of selenium tetrafluoride was by Paul Lebeau in 1907, who treated selenium with fluorine:[1]

Se + 2 F2 → SeF4

A synthesis involving more easily handled reagents entails the fluorination of selenium dioxide with sulfur tetrafluoride:[2]

SF4 + SeO2 → SeF4 + SO2

An intermediate in this reaction is seleninyl fluoride (SeOF2).

Other methods of preparation include fluorinating elemental selenium with chlorine trifluoride:

3 Se + 4 ClF3 → 3 SeF4 + 2 Cl2

Structure and bonding

Selenium in SeF4 has an oxidation state of +4. Its shape in the gaseous phase is similar to that of SF4, having a see-saw shape. VSEPR theory predicts a pseudo-trigonal pyramidal disposition of the five electron pairs around the selenium atom. The axial Se-F bonds are 177 pm with an F-Se-F bond angle of 169.2°. The two other fluorine atoms are attached by shorter bonds (168 pm), with an F-Se-F bond angle of 100.6°. In solution at low concentrations this monomeric structure predominates, but at higher concentrations evidence suggests weak association between SeF4 molecules leading to a distorted octahedral coordination around the selenium atom. In the solid the selenium center also has a distorted octahedral environment.

Reactions

In HF, SeF4 behaves as a weak base, weaker than sulfur tetrafluoride, SF4 (Kb= 2 X 10−2):

SeF4 + HF → SeF3+ + HF2−; (Kb = 4 X 10−4)

Ionic adducts containing the SeF3+ cation are formed with SbF5, AsF5, NbF5, TaF5, and BF3.[3] With caesium fluoride, CsF, the SeF5− anion is formed, which has a square pyramidal structure similar to the isoelectronic chlorine pentafluoride, ClF5 and bromine pentafluoride, BrF5.[4] With 1,1,3,3,5,5-hexamethylpiperidinium fluoride or 1,2-dimethylpropyltrimethylammonium fluoride, the SeF62− anion is formed. This has a distorted octahedral shape which contrasts to the regular octahedral shape of the analogous SeCl62−. [5]

Explanation:

4 0
2 years ago
How much heat do you need to raise the temperature of 100g of aluminum from 30 C to 150 C
SVEN [57.7K]

Answer:

Q = 10.8 KJ

Explanation:

Given data:

Mass of Al= 100g

Initial temperature = 30°C

Final temperature = 150°C

Heat required = ?

Solution:

Specific heat of Al = 0.90 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30°C

ΔT = 120°C

Q = 100g×0.90 J/g.°C× 120°C

Q = 10800 J       (10800j×1KJ/1000 j)

Q = 10.8 KJ

4 0
3 years ago
Read 2 more answers
PLEASE HELP!! <br> this is on USAtestprep <br> a)<br> b)<br> c)<br> d)
igomit [66]

Answer:

B

Explanation:

the fluorine has an high tendency to gain electrons from other elements with lower electronegativities

3 0
3 years ago
Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+
Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest E^o is I_2+Cu\rightarrow Cu^{2+}+2I^-

<u>Explanation:</u>

We are given:

E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

E^o_{cell}=E^o_{oxidation}+E^o_{reduction}

The combination of the cell reactions follows:

  • <u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)

Oxidation half reaction:  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-   E^o_{oxidation}=0.45V

Reduction half reaction:  I_2(s)+2e^-\rightarrow 2I_-(aq.)   E^o_{reduction}=0.54V

E^o_{cell}=0.45+0.54=0.99V

Thus, this cell will not give the spontaneous cell reaction with smallest E^o_{cell}

  • <u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)

Oxidation half reaction:  Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-   E^o_{oxidation}=-0.34V

Reduction half reaction: I_2(s)+2e^-\rightarrow 2I_-(aq.)   E^o_{reduction}=0.54V

E^o_{cell}=-0.34+0.54=0.20V

Thus, this cell will give the spontaneous cell reaction with smallest E^o_{cell}

  • <u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)

Oxidation half reaction:  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-   E^o_{oxidation}=0.45V

Reduction half reaction:  Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)   E^o_{reduction}=0.34V

E^o_{cell}=0.45+0.34=0.79V

Thus, this cell will not give the spontaneous cell reaction with smallest E^o_{cell}

Hence, the spontaneous cell reaction having smallest E^o is I_2+Cu\rightarrow Cu^{2+}+2I^-

7 0
3 years ago
Declare the volume of the yeast alcohol as a fraction of the total amount of ethanol liquor liquefied alcohol in 30cm of water p
bogdanovich [222]

alcohol = A/A+B

= 30/100

= 0.3

5 0
3 years ago
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