Answer: Biodiversity is important to humans for many reasons. ... Ecological life support— biodiversity provides functioning ecosystems that supply oxygen, clean air and water, pollination of plants, pest control, wastewater treatment and many ecosystem services.
Explanation:
Answer:
a= <em>In scientific notation</em>
6.96000×10⁵ Km
b =<em>In expanded notation</em>
0.00019 mm
Explanation:
Given data:
Radius of sun = 696000 Km
size of bacterial cell = 1.9 ×10⁻⁴ mm
Radius of sun in scientific notation = ?
Size of bacterial cell in expanded notation = ?
Solution:
Scientific notation is the way to express the large value in short form.
The number in scientific notation have two parts.
. The digits (decimal point will place after first digit)
× 10 ( the power which put the decimal point where it should be)
for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³
Radius of sun:
696000 Km
<em>In scientific notation</em>
6.96000 × 10⁵ Km
The expanded notation is standard notation of writing the numerical values which is normal way. The numbers are written as they are, without the power of 10.
Size of bacterial cell:
1.9 ×10⁻⁴ mm
<em>In expanded notation</em>
1.9/ 10000 = 0.00019 mm
when carbon dioxide gas is collected down ward of water wet gas is collected by the downward displacement of water . This is used for gases that are not very soluble in water . ... In water , carbon dioxide produces a weakly acidic solution , carbonic acid .
Answer:
Kc = 3.90
Explanation:
CO reacts with
to form
and
. balanced reaction is:
![CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)](https://tex.z-dn.net/?f=CO%28g%29%20%2B%203H_2%20%28g%29%20%5Cleftrightharpoons%20CH_4%28g%29%20%20%2B%20%20H_2O%28g%29)
No. of moles of CO = 0.800 mol
No. of moles of
= 2.40 mol
Volume = 8.00 L
Concentration = ![\frac{Moles}{Volume\ in\ L}](https://tex.z-dn.net/?f=%5Cfrac%7BMoles%7D%7BVolume%5C%20in%5C%20L%7D)
Concentration of CO = ![\frac{0.800}{8.00} = 0.100\ mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B0.800%7D%7B8.00%7D%20%3D%200.100%5C%20mol%2FL)
Concentration of
= ![\frac{2.40}{8.00} = 0.300\ mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B2.40%7D%7B8.00%7D%20%3D%200.300%5C%20mol%2FL)
![CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)](https://tex.z-dn.net/?f=CO%28g%29%20%2B%203H_2%20%28g%29%20%5Cleftrightharpoons%20CH_4%28g%29%20%20%2B%20%20H_2O%28g%29)
Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium
= 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium
= 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)
![Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B0.0386%20%5Ctimes%200.0386%7D%7B%280.184%29%5E3%20%5Ctimes%200.0614%7D%20%3D3.90)