Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.
Answer:
0.158 moles KMnO4
Explanation:
According to the Periodic Table,
K = 39.10 g/mol
Mn = 54.94 g/mol
O = 16.00 g/mol
KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol
25.0 grams KMnO4 1 mole
----------------------------- x -------------------------- = 0.158 moles KMnO4
158.04 grams
1. Both part of the ecosystem
2. There are biotic objects on abiotic objects ( caterpillars on trees ) and abiotic objects on biotic things ( pollen on bees )
3. Both are made of atoms
Answer:
a) cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
b) 0.068 V.
Explanation:
A) Cu2+ + 2e- euilibrium cu (s)
Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-
Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
B) To calculate the cell voltage
E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+
putting values we get
= 0.339V + (90.05916V/2)log(0.100) = 0.309V
E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.
