Answer:
296.1 day.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).
t is the time of the reaction (t = ??? day).
a is the initial concentration of Ir-192 (a = 560.0 dpm).
(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).
<em>∴ kt = lna/(a-x)</em>
(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).
(9.365 x 10⁻³ day⁻¹)(t) = 2.773.
<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>
Answer:
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The <em>freezing-point depression</em> is expressed by:
We put the data given by the problem and <u>solve for m</u>:
- 2.63 °C = 40°C·kg/mol * m
For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We<u> calculate moles of reserpine:</u>
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.
The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.
The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.
The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)
to the proper number of significant figures) to the following? (12.67+19.2)(3.99)/(1.36+ 11.366).
solution:
Hydration is the addition of water; hydrogenation is the addition of hydrogen.
desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??
knowns:
__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole
__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole
___________2(1/2 O2 + H2 -------------> H2O)___________2*dHox = 2(-285.8 kJ/mole)
Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem
