Answer:
Blooms are alluring and show up in various hues and shapes to draw in pollinators who help in dust move. Most blooms have four primary parts: sepals, petals, stamens, and carpels. The stamens are the male part though the carpels are the female piece of the blossom.
Sepal: The outer parts of the flower (often green and leaf-like) that enclose a developing bud.
Petal: The parts of a flower that are often conspicuously colored.
Stamen: The pollen producing part of a flower, usually with a slender filament supporting the anther.
Anther: The part of the stamen where pollen is produced.
Pistil: The ovule producing part of a flower. The ovary often supports a long style, topped by a stigma. The mature ovary is a fruit, and the mature ovule is a seed.
Stigma: The part of the pistil where pollen germinates.
Ovary: The enlarged basal portion of the pistil where ovules are produced.
Receptacle: The part of a flower stalk where the parts of the flower are attached
Answer:
C2H4 + 3O2 --> 2CO2 + 2H2O
Explanation:
For an equation to be balanced, it must have an equal number of atoms on each side. C2H4 + 3O2 --> 2CO2 + 2H2O is the only equation out of these four the fulfills this requirement.
Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
Answer:
see explanations
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
Ci(NH₃) = 3.5mole/4L = 0.875M
Cf(NH₃) = 1.6mole/4L = 0.400M
Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s
Rate-2 => 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s
Rate-3 => 5(Δ[NH₃]/Δt) = 4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s
_________________________________________________________
NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values. Then solve for rate of interest and substitute givens.
example: for NH₃ and H₂O
- set rates expressions equal => Δ[NH₃]/Δt = Δ[H₂O]/Δt
- then swap and insert coefficients from given rxn ...
- solve for rate of interest ...
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
=> 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt)
=> Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s
Answer:
The 
for ![[Fe(NCS)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28NCS%29%5D%5E%7B2%2B%7D)
formation is 
.
Explanation:
![[Fe(NO_3)_3]=0.02 M=[Fe^{3+}]](https://tex.z-dn.net/?f=%5BFe%28NO_3%29_3%5D%3D0.02%20M%3D%5BFe%5E%7B3%2B%7D%5D)
Concentration of ferric ion = ![[Fe^{3+}]=0.02 M](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%3D0.02%20M)
Volume of ferric solution = 3.0 mL = 0.003 L
Moles of ferric ion 
1 mL = 0.001 L
![[NaNCS]=0.002 M=[NCS^-]](https://tex.z-dn.net/?f=%5BNaNCS%5D%3D0.002%20M%3D%5BNCS%5E-%5D)
Concentration of 
ion = ![[NCS^{-}]=0.002 M](https://tex.z-dn.net/?f=%5BNCS%5E%7B-%7D%5D%3D0.002%20M)
Volume of ion solution = 3.0 mL = 0.003 L
Moles of ion= 
Volume of nitric acid solution = 10 mL = 0.010 L
After mixing all the solution the concentration of ferric ion and ion will change
Total volume of solution = 0.003 L + 0.003 L + 0.010 L = 0.016 L
Initial concentration of ferric ion before reaching equilibrium :
= 
Initial concentration of ion before reaching equilibrium :
= 
![Fe^{3+}+NCS^-\rightleftharpoons [Fe(NCS)]^{2+}](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%2BNCS%5E-%5Crightleftharpoons%20%5BFe%28NCS%29%5D%5E%7B2%2B%7D)
Initially:
0.00375 M 0.000375 M 0
At equilibrium :
(0.00375-x) (0.000375-x) x
Equilibrium concentration of ![[Fe(NCS)]^{2+}=x=2.5\times 10^{-4} M](https://tex.z-dn.net/?f=%5BFe%28NCS%29%5D%5E%7B2%2B%7D%3Dx%3D2.5%5Ctimes%2010%5E%7B-4%7D%20M)
The expression of equilibrium constant for formation is given by :
![K_c=\frac{[[Fe(NCS)]^{2+}]}{[Fe^{3+}][NCS^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28NCS%29%5D%5E%7B2%2B%7D%5D%7D%7B%5BFe%5E%7B3%2B%7D%5D%5BNCS%5E-%5D%7D)
The for formation is .
