Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
Answer:
1.64g
Explanation:
The reaction scheme is given as;
2-bromocyclohexanol --> 1,2-epoxycyclohexane + HBr
From the reaction above,
1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane
3.0 grams of trans-2-bromocyclohexanol.
Molar mass = 179.05 g/mol
Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol
This means 0.016755 mol of 1,2-epoxycyclohexane would be produced.
Molar mass = 98.143 g/mol
Theoretical yield = Number of moles * Molar mass
Theoretical yield = 0.016755 * 98.143 ≈ 1.64g
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
Answer:
The answer is A. Lithium and Sodium
Answer:
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. ( B )
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not. ( C )
Explanation:
<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. and
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.
Given the relationship below,
Δ G = -nFE
E = emf of cell , G = free energy.
This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable
