Answer:
The answer to your question is 49.08 u
Explanation:
Data
Average atomic mass = 45.737 u
Isotope 1 = 40.149 u abundance = 37.46%
Isotope 2 = ?
Process
1.- Calculate the abundance of isotope 2
abundance isotope 1 + abundance isotope 2 = 100%
37.46 + abundance isotope 2 = 100
abundance isotope 2 = 100 - 37.46
abundance isotope 2 = 62.54
2.- Calculate the mass of isotope 2
Average atomic mass = (abundance x mass isotope 1) + (abundance x
mass isotope 2)
45.737 = (0.3746 x 40.149) + (62.54 x mass isotope 2)
mass isotope 2 = [45.737 - (0.3746 x 40.149)]/ 62.54
mass isotope 2 = [45.737 - 15.0398]/ 62.54
mass isotope 2 = 30.6972/62.54
mass isotope 2 = 0.4908 x 100
mass isotope 2 = 49.08 u
Answer:
Density = 25.2 g/20.0 ml = 1.26 g/ml
57 ml x 1.26 g/ml = 71.8 g
Answer:
(1) The two column proof is presented here as follows;
Statement 
Reasons
1. C is the midpoint of 
Given
B is the midpoint of 
D is the midpoint of 
2. AC = CE, AB = BC, CD = DE Definition of midpoint
3. AB + BC = AC, CD + DE = CE Segment addition postulate
4. CD + DE = AC Substitution property of equality
5. AB + BC = CD + DE Substitution property of equality
6. BC + BC = CD + CD Substitution property of equality
7. 2·BC = 2·CD Addition of two identical quantities
8. BC = CD Division property of equality
9. BC ≅ CD Definition of Congruence
(2) The two column proof for the triangular geometric figure is presented here as follows;
Statement Reasons
1. 2·WV = XY; 2·YZ = XW; WV = YZ Given
2. 2·WV = 2·YZ Multiplication property
3. XY = XW Substitution Property
4. XW + WV = XV Segment Addition Property
XY + YZ = XZ
5. XW + WV = XV Substitution Property
XW + WV = XZ
6. XV = XZ Substitution Property
Explanation:
The answer would be B red, violet would be the shortest
Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
