I suppose it would be forest because in order to have organic matter the soil needs to be rich and fertile,therefore it is forest.
Answer:
H₂SO₄
Explanation:
Given data:
Number of moles of H₂SO₄ = 15 mol
Number of moles of Fe = 13 mol
Which reactant is limiting reactant = ?
Solution:
Chemical equation:
3H₂SO₄ + 2Fe → Fe₂(SO₄)₃ + 3H₂
now we will compare the moles reactant with product.
H₂SO₄ : Fe₂(SO₄)₃
3 : 1
15 : 1/3×15 = 5
H₂SO₄ : H₂
3 : 3
15 : 15
Fe : Fe₂(SO₄)₃
2 : 1
13 : 1/2×13 = 6.5
Fe : H₂
2 : 3
13 : 3/2×13 = 19.5
Number of moles of product formed by H₂SO₄ are less thus it will act as limiting reactant.
Answer:
<h2>
0.50 m/s</h2><h2>
</h2>
Explanation:
Velocity = distance over time
where distance = 5.20 m
time = 10.4 s.
velocity = <u> 5.20 m </u>
10.4 s.
= 0.50 m/s
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>
Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
