Answer:
172.385 g/mol
Explanation:
Magnesium Tartrate is C4H4MgO6
C - 12.01 g/mol
H - 1.01 g/mol
Mg - 24.305 g/mol
O - 16.00 g/mol
12.01(4) + 1.01(4) + 24.305 + 16(6) = 172.385 g/mol
Answer:
the first energy level is closest to nuclear the second energy level is a little farther away than the first
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:
Atomic mass of E is 128.24
Explanation:
- The percentage composition by mass of an element in a compound is given by dividing the mass of the element by the total mass of the compound and expressing it as a percentage.
- In this case; the compound Bi₂E₃
Percentage composition of bismuth = 52.07%
Percentage composition of E = 47.93%
Mass Bismuth in the compound is (2×208.9804) = 417.96 g
Therefore,
To calculate the atomic mass of E
52.07% = 417.96 g
47.93% = ?
= (47.93 × 417.96 ) ÷ 52.07 %
= 384.729
E₃ = 384.729
Therefore; E = 384.729 ÷ 3
= 128.24
The atomic mass of E is 128.24
