Answer:
51.79g Li₃P.
Explanation:
Li has a molar mass of 6.94 g (since there are 3, you multiply it 3 times) and P has a molar mass of 30.97 g. 6.94(3) + 30.97 = 51.79g.
Answer:
138.96kJ is the maximum electrical work
Explanation:
The maximum electrical work that can be obtained from a cell is obtained from the equation:
W = -nFE
<em>Where W is work in Joules,</em>
<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>
Zn(s) → Zn²⁺(aq) + 2e⁻
F is faraday constant = 96500Coulombs/mol
E is cell potential = 0.72V
Replacing:
W = -2mol*96500Coulombs/mol*0.72V
W = - 138960J =
<h3>138.96kJ is the maximum electrical work</h3>
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Answer:
i cant see it very much i dont think no one can
Explanation:
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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