Moles of PF₃ : 4
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)
Limiting reactant : the smallest ratio (mol divide by coefficient)
P₄ : F₂ =

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
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Answer:
205.12 atm
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
R = 0.0821 Latm/perK)
T = temperature (K)
n = number of moles (mol)
According to the information in this question;
P = ?
V = 34.25 mL = 34.25 ÷ 1000 = 0.03425L
n = 0.215 mol
T = 125.0°C = 125 + 273 = 398K
Using PV = nRT
P = nRT ÷ V
P = (0.215 × 0.0821 × 398) ÷ (0.03425)
P = 7.025 ÷ 0.03425
P = 205.12 atm