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notsponge [240]
3 years ago
7

Three of the adaptations would be very important to help a temperate forest plant disperse its population and spread out. One ad

aptation does not aid in population dispersal. That adapatation is
Chemistry
2 answers:
yan [13]3 years ago
8 0

Answer:

The correct answer is "the presence of secondary xylem as wood tissue".

Explanation:

The missing options of this question are:

A. runner roots that produce new plants by vegetative propagation

B. the use of large sugary fruits to attract animals

C. the presence of secondary xylem as wood tissue

D. the use of pollinating insects, such as bees

The correct answer is option C. "the presence of secondary xylem as wood tissue".

Temperate forest plants have a few adaptations that help them to disperse its population, however, the presence of secondary xylem as wood tissue is not one of them. The adaptation of developing a secondary xylem as wood tissue serves the plant as a defense system. Numerous studies report that trees with secondary xylem as wood tissue have less propensity of developing fungal and bacterial infections.

Likurg_2 [28]3 years ago
3 0

Answer:

C, <u>The presence of secondary xylem as wood tissue.</u>

Explanation:

The presence of secondary xylem as wood tissue. While this strengthens a plant and allows it to grow taller, it does not directly help a plant to spread.

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For the reaction 3h2(g) + n2(g) 2nh3(g), kc = 9.0 at 350°c. calculate g° at 350°c.
miss Akunina [59]
When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° =  - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc 

and Kc = 9 

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

           = - 4536 
4 0
3 years ago
Is snails growing shell conserve mass
postnew [5]

The thing that two changes have in common that snails growing shells and  rust forming on a bicycle frame is option D. Both are caused by cooling.

<h3>How come snails develop shells?</h3>

Calcium carbonate is said to be the material that makes up the shell. The snail's shell expands as it grows to accommodate its growing body. Snails and slugs are also members of the mollusc family of creatures.

Therefore, note that  air that has been mixed with the metal can make rust to develop. and as such, option D. Both are caused by cooling. is correct.

Learn more about snails growing shells from
brainly.com/question/15758489

#SPJ1

See full question below

What do these two changes have in common? snails growing shells rust forming on a bicycle frame Select all that apply.

A. Both are only physical changes.

B. Both are caused by heating.

C. Both are chemical changes.

D. Both are caused by cooling

5 0
1 year ago
Which of the following most influenced the changing nature of classification systems
Lyrx [107]
Hope I helped you enjoy your day

5 0
3 years ago
Read 2 more answers
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order
yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)      

log .75 / ( log 1/2) = 40 / h

<u>h = half life =   96.37683 min</u>

8 0
2 years ago
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