Answer:
The value is 
Explanation:
From the question we are told that
The equation is 
The temperature is ![T = 25^oC = 298 K [room \ temperature ]](https://tex.z-dn.net/?f=%20T%20%3D%2025%5EoC%20%3D%20%20298%20K%20%20%20%5Broom%20%20%5C%20temperature%20%5D)
The emf at standard condition is 
Generally at the cathode

At the anode

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

Here n is the no of electron with value n = 6
F is the Faraday's constant with value 96487 J/V
=>
=> 
This Gibbs free energy can also be represented mathematically as

Here R is the cell constant with value 8.314J/K
K is the equilibrium constant
From above
=> 
Generally antilog = 2.718
=>
=> 
Answer:
6.79 g of phosphine can be produced
Explanation:
The reaction is this:
3H₂ + 2P → 2PH₃
We have the mass of the two reactants, so let's find out the limiting reactant, so we can work with the equation. Firstly, we convert the mass to moles (mass / molar mass)
6.2 g / 30.97 g/mol = 0.200 moles of P
4g / 2 g/mol = 2 moles of H₂
Ratio is 3:2.
3 moles of hydrogen react with 2 moles of P
Then, 2 moles of H₂ would react with (2 . 2)/ 3 = 1.3 moles of P.
We have only 0.2 moles of P, so clearly the phosphorous is the limiting reactant.
Ratio is 2:2. So 2 moles of P can produce 2 moles of phosphine. Therefore, 0.2 moles of P must produce the same amount of phosphine.
Let's convert the moles to mass ( mol . molar mass)
0.2 mol . 33.97 g/mol = 6.79 g
Answer:
[Ne] 2s2 2p3
Explanation:
Phosphorus will most likely have an ion that will be 3- because it wants to have a full outer shell. Thus, the elctron configuration is: 1s2 2s2 2p6 3s2 3p3.
Correct answer: Option D, <span>
K = 5.04 × 10^52</span>
Reason:
We know that,
Ecell =

,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>

Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
No they wouldn't. <span>You can't make an </span>ionic compound<span> with these elements.</span>