Answer:
There will be 143,67g CO2 produced
Explanation:
2 C6H6 + 15 O2 → 12 CO2 + 6 H2O
(42,5 g C6H6) / (78.1124 g C6H6/mol) = 0.54408775 mole C6H6
(113.1 g O2) / (31.9989 g O2/mol) = 3.534496 moles O2
0.54408775 mole of C6H6 would react completely with 0.54408775 x (15/2) = 4.080658 mole O2, but there is more O2 present than that, so O2 is in excess and C6H6 is the limiting reactant.
(0.54408775 mol C6H6) x (12/2) x (44.0096 g/mol) = 143.67 g CO2
Answer:
3.1% is the fraction of the sample after 28650 years
Explanation:
The isotope decay follows the equation:
Ln[A] = -kt + Ln[A]₀
<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>
<em />
k could be obtained from Half-Life as follows:
K = Ln 2 / Half-life
K = ln 2 / 5730 years
K = 1.2097x10⁻⁴ years⁻¹
Replacing in isotope decay equation:
Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]
Ln[A] = -3.4657
[A] = 0.0313 =
<h3>3.1% is the fraction of the sample after 28650 years</h3>
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Answer:
a) 0
b) 4
c) 4
d) 4
e) 2
f) 4
<u>Explanation: No. unpaired electrons:</u>
a) square planar, strong field
d⁴ 0
b) tetrahedral, strong field
d⁴ 4
c) Octahedral, weak field
d⁴ 4
d) square planar, weak field
d⁴ 4
e) Octahedral, strong field
d⁴ 2
f) tetrahedral, weak field
d⁴ 4
Answer:
26.73 mg.
Explanation:
- Firstly, we can calculate the no. of moles of magnesium chlorate (Mg(ClO₃)₂):
no. of moles of magnesium chlorate (Mg(ClO₃)₂) = mass/molar mass = (72.03 mg)/(191.21 g/mol) = 0.377 mmol.
<em>Every 1.0 mole of magnesium chlorate (Mg(ClO₃)₂) contains 2.0 moles of Cl.</em>
<em></em>
∴ The no. of moles of Cl in magnesium chlorate (Mg(ClO₃)₂) = 2(0.377 mmol) = 0.754 mmol.
∴ The mass of Cl are found in 72.03 mg of magnesium chlorate (Mg(ClO₃)₂) = (no. of moles of Cl)(atomic mass of Cl) = (0.754 mmol)(35.453 g/mol) = 26.73 mg.