Answer:
A. Position B.
Explanation:
In the summer the northern hemisphere is closer to the Sun so it's Position B.
2.1648 kg of CH4 will generate 119341 KJ of energy.
Explanation:
Write down the values given in the question
CH4(g) +2 O2 → CO2(g) +2 H20 (g)
ΔH1 = - 802 kJ
2 H2O(g)→2 H2O(I)
ΔH2= -88 kJ
The overall chemical reaction is
CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ
CH4 +2 O2 → CO2 +2 H20
(1mol)+(2mol)→(1mol+2mol)
Methane (CH4) = 16 gm/mol
oxygen (O2) =32 gm/mol
Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O
which generate 882 KJ /mol
Therefore to produce 119341 KJ of energy
119341/882 = 135.3 mol
to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require
=135.3 *16
=2164.8 gm
=2.1648 kg of CH4
2.1648 kg of CH4 will generate 119341 KJ of energy
Answer:
Having as wide a range of organisms as possible.
Hope it helps! :)
Rust is classified as Metal Oxide!
Answer:
(a) a = 5.08x10⁻⁸ cm
(b) r = 179.6 pm
Explanation:
(a) The lattice parameter "a" can be calculated using the following equation:
<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>
<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>
Hence the lattice parameter is:
![a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7B1.32%20%5Ccdot%2010%5E%7B-22%7D%20cm%5E%7B3%7D%7D%20%3D%205.08%20%5Ccdot%2010%5E%7B-8%7D%20cm%20)
(b) We know that the lattice parameter of a FCC structure is:
<em>where r: is the atomic radius of Th</em>
Hence, the atomic radius of Th is:
I hope it helps you!
