Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Answer:
Number of moles of solute = 0.6 mole
Mass =13.8 g
Explanation:
Given data:
Number of moles of sodium = ?
Volume = 2.0 L
Molarity = 0.30 M
Mass in gram of sodium= ?
Solution:
<em>Number of moles:</em>
Molarity = number of moles of solute / volume in litter
Number of moles of solute = Molarity × volume in litter
Number of moles of solute = 0.30 M × 2.0 L
Number of moles of solute = 0.6 mole
<em>Mass in gram:</em>
Mass = Number of moles × molar mass
Mass = 0.6 mole× 23 g/mol
Mass =13.8 g
Answer:
7.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.
Thus, 7.5 moles of O₂ were obtained from the reaction.
374u
187u
C₁₄H₂₂N₄O₈
Explanation:
To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.
The empirical weight is determined by using the simplest ratio of the elements involved in the compound;
Molecular weight of C₁₄H₂₂N₄O₈;
atomic mass of C = 12g/mol
H = 1g/mol
N = 14g/mol
O = 16g/mol
Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)
= 168 + 22 + 56 + 128
= 374u
Empirical weight:
Empirical formula:
C₁₄ H₂₂ N₄ O₈
14 : 22 : 4 : 8
divide by 2:
7 : 11 : 2 : 4
empirical formula C₇H₁₁N₂O₄
empirical weight = 
= 
= 187u
The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈
learn more:
Molecular mass brainly.com/question/5546238
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