The ability to attract an electron for bonding is called (option B) Electronegativity.
100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................
After six half-lives would be found 1.5625% of readioactive species.
the molar mass of the element is 81.36 g/mol
<u><em>calculation</em></u>
step 1 : multiply each %abundance of the isotope by its mass number
that is 79.95 x 29.9 =2391
81.95 x 70.1 = 5745
Step 2: add them together
2390.5 + 5744.7 =8136
Step 3: divide by 100
= 8136/100 = 81.36 g/mol
Answer:
194.6 mL of SO₂
Explanation:
The reaction that takes place is:
P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)
<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:
- 23.8 °C → 23.8 + 273.15 = 296.95 K
- 747 torr → 747/760 = 0.983 atm
We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):
0.576 g P₄S₃ * 
= 7.85 * 10⁻³ mol SO₂ = n
PV=nRT
0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K
V = 0.1946 L
- Finally we convert L into mL:
0.1946 * 1000 = 194.6 mL
Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
