1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.
2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.
Answer:
Evaporation happens from the any water body (like ponds, rivers, water stored in cup etc) while the transpiration essentially happens in plants only.
The evaporation and transpiration plays a major role in the water cycle as they help in formation of clouds and precipitation
Explanation:
Evaporation:
It is the process of formation and transfer of the water vapor from the surface of the liquid into the atmosphere.
Transpiration:
It is the process of transfer of the water vapors in the atmosphere from the pores present in the leaves of the plants.
The water is carried from the ground through the stems to the leaves.
The major difference between the evaporation and transpiration is that the evaporation happens from the any water body (like ponds, rivers, water stored in cup etc) while the transpiration essentially happens in plants only.
The evaporation and transpiration plays a major role in the water cycle as these process helps in transfer of the surface water into the atmosphere, which helps in formation of clouds and precipitation.
Answer:
After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M.
Explanation:
Based on the reaction of the problem, you have as general kinetic law for a first-order reaction:
ln[HI] = -kt + ln [HI]₀
<em>Where [HI] is actual concentration after time t, </em>
<em>k is rate constant </em>
<em>and [HI]₀ is initial concentration of the reactant.
</em>
Initial concentration of HI is 0.310M,
K is 0.0660s⁻¹,
And the actual concentration is 0.0558M:
ln[HI] = -kt + ln [HI]₀
ln[0.0558M] = -0.0660s⁻¹*t + ln [
0.310M]
-1.7148 = -0.0660s⁻¹*t
26.0s = t
<h3>After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M</h3>
<em />
Answer:
Explanation:
14:
a) Shared pairs = 6
per shared pair between each C-H bond.
and two shared pair between C=C thus total shared pairs are 6
a) Lone pairs = 0
b) Single bonds = 4
Double bonds = 1
Triple bonds = 0
c) Formula = C₂H₄
Name = Ethene
d) Lewis structure = given structure is also Lewis structure.
13:
a) Shared pairs = 3
a) Lone pairs = 2
b) Single bonds = 0
Double bonds = 0
Triple bonds = 1
c) Formula = N₂
Name = Dinitrogen
d) Lewis structure = given structure is also Lewis structure.
