Answer:
Most common insulation materials work by slowing conductive heat flow and--to a lesser extent--convective heat flow. Radiant barriers and reflective insulation systems work by reducing radiant heat gain. To be effective, the reflective surface must face an air space.
Explanation:
To be effective, the reflective surface must face an air space.
Answer:
the atomic packing factor of Sn is 0.24
Explanation:
a = b = 5.83A and c = 3.18A.
Volume of unit cell = a²c
= (5.83)² * 3.18 * 10⁻²⁴ cm³
= 1.08 * 10⁻²²cm³
Volume of atoms =
![2 \times \frac{4}{3} \pi r^3](https://tex.z-dn.net/?f=2%20%5Ctimes%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E3)
(∴ BCC, effective number of atom is 2)
Volume of atoms =
![2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3](https://tex.z-dn.net/?f=2%20%2A%20%5Cfrac%7B4%7D%7B3%7D%20%2A3.14%2A%280.145%2A10%5E-%5E7cm%29%5E3)
= 2.55*10⁻²³cm³
![\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}](https://tex.z-dn.net/?f=%5Ctext%20%7BAtomic%20packing%20factor%7D%3D%5Cfrac%7B%5Ctext%20%7Bvolume%20occupied%20by%20atom%7D%7D%7B%5Ctext%20%7Bvolume%20of%20unit%20cell%20%7D%7D)
![=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.55%2A10%5E-%5E2%5E3%7D%7B1.08%2A10%5E-%5E2%5E2%7D%20%5C%5C%5C%5C%3D0.24)
<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
C is the answer hope this helps
Answer:
c
Explanation:
because the fire gives off radiation
Answer:
The activation energy for the decomposition = 33813.28 J/mol
Explanation:
Using the expression,
Wherem
is the activation energy
R is Gas constant having value = 8.314 J / K mol
Thus, given that,
= ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
T = (25 + 273.15) K = 298.15 K
So,
![E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}](https://tex.z-dn.net/?f=E_a%3D-%5Cln%20%5Cfrac%7B1.5%5Ctimes%20%5C%3A10%5E3%7D%7B4.0%5Ctimes%20%5C%3A10%5E3%7D%5C%3A%5Ctimes%20%5Cfrac%7B8.314%7D%7B%5Cleft%28%5Cfrac%7B1%7D%7B278.15%7D-%5Cfrac%7B1%7D%7B298.15%7D%5Cright%29%7D)
![E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}](https://tex.z-dn.net/?f=E_a%3D-%5Cfrac%7B8.314%5Cln%20%5Cleft%28%5Cfrac%7B1.5%5Ctimes%20%5C%3A10%5E3%7D%7B4%5Ctimes%20%5C%3A10%5E3%7D%5Cright%29%7D%7B%5Cfrac%7B1%7D%7B278.15%7D-%5Cfrac%7B1%7D%7B298.15%7D%7D)
![E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}](https://tex.z-dn.net/?f=E_a%3D-%5Cfrac%7B689483.53266%20%5Cln%20%5Cleft%28%5Cfrac%7B1.5%7D%7B4%7D%5Cright%29%7D%7B20%7D)
![E_a=33813.28\ J/mol](https://tex.z-dn.net/?f=E_a%3D33813.28%5C%20J%2Fmol)
<u>The activation energy for the decomposition = 33813.28 J/mol</u>