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3 years ago

What is berylium phosphide anion?

Chemistry

1 answer:

Novosadov [1.4K]3 years ago

4 0

Answer:

Molecular Formula, Be3P2. Synonyms. Beryllium phosphide. Beryllium phosphide (BeP2). EINECS 261-137-1. 57620-29-8. 58127-61-0. Molecular Weight.

Molecular Formula: Be3P2

Molecular Weight: 88.98407 g/mol

Structure: Find Similar Structures

PubChem CID: 6453515

Explanation:

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

Which is a diatomic molecule? A)Ar B)CO C)CO2 D)NaCl

LUCKY_DIMON [66]

Answer:

C. Co2

Explanation:

Oxygen is a diatomic molecule because 2 atoms are conisistebof a normal molecule.

5 0

3 years ago

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How many grams of CaF2 would be needed to produce 1.12 moles of F2?

zlopas [31]

Step 1 : Write balanced chemical equation.

CaF₂ can be converted to F₂ in 2 steps. The reactions are mentioned below.

I] $CaF_{2} + H_{2} SO_{4} -------> 2HF + CaSO_{4}$

II] $2HF -------> H_{2} + F_{2}$

The final balanced equation for this reaction can be written as

$CaF_{2} + H_{2} SO_{4} ---------> CaSO_{4} + H_{2} + F_{2}$

Step 2: Find moles of CaF₂ Using balanced equation

We have 1.12 mol F₂

The mole ratio of CaF₂ and F₂ is 1:1

$1.12mol F_{2} * \frac{1molCaF_{2}}{1molF_{2}} = 1.12molCaF_{2}$

Step 3 : Calculate molar mass of CaF2.

Molar mass of CaF₂ can be calculated by adding atomic masses of Ca and F

Molar mass of CaF₂ = Ca + 2 (F)

Molar mass of CaF₂ = 40.08 + 18.998 = 78.08 g

Step 4 : Find grams of CaF₂

Grams of CaF₂ = $1.12molCaF_{2} *\frac{78.08gCaF_{2}}{1mol CaF_{2}}$

Grams of CaF₂ = 87.45 g

87.45 grams of CaF2 would be needed to produce 1.12 moles of F2.

4 0

3 years ago

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Do you know what this is ??

Sonja [21]

Answer:

is there a pic or anything?

8 0

3 years ago

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Pedro adds 3.23 moles of helium to a balloon that already contained 4.51 moles of helium creating a balloon with a volume of 9.8

sashaice [31]

Answer:

7.The answer is 2.13 because if u add 3.23 and 4.51 and the answer is 7.74 and u add 2.13 to get 9.87. 3.23 + 4.51 + 2.13 is 9.87

Explanation:

5 0

3 years ago