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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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Explain why two thin sweaters feel warmer then one thick sweater​
jeyben [28]

Answer:

Two slender woolen sweaters are hotter than a thick woolen sweater in light of the fact that there is a layer of air between them that doesn't permit our body warmth to get away yet it likewise it doesn't retains heat from the climate and fleece is additionally a protector that itself doesn't permit our body warmth to get away

7 0
4 years ago
You are given the reaction Cu + HNO3 Cu(NO3)2 + NO + H2O complete the final balanced equation based on half-reactions
Evgen [1.6K]

Answer:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Explanation:

Cu + HNO3 → Cu(NO3)2 + NO + H2O

The first step is to write the oxidation numbers for each atoms in the given equation  

Cu0 + H+1N+5O-23 → Cu+2(N+5O-23)2 + N+2O-2 + H+12O

Identify the oxidizing and reducing agent  

OXIDATION --- Cu0 → Cu+2(N+5O-23)2 + 2e-    

REDUCTION---H+1N+5O-23 + 3e- → N+2O

Balance equation in half reaction  

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e-

H+1N+5O-23 + 3e- → N+2O

Now balance the charge

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O

Balance the oxygen atom  

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O-2 + 2H2O

Make electron gain equivalent to electron lost.

3Cu0 + 6HNO3 → 3Cu+2(N+5O-23)2 + 6e- + 6H+

2H+1N+5O-23 + 6e- + 6H+ → 2N+2O-2 + 4H2O

Complete reaction  

3Cu0 + 8H+1N+5O-23 + 6e- + 6H+ → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 6e- + 4H2O + 6H+

Simplify the equation

3Cu0 + 8H+1N+5O-23 → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 4H2O

Final equation  

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

5 0
3 years ago
A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible
RideAnS [48]

Answer:

TS (total solid, mg/L) = 57,174 mg/L

VS (volatile solid, mg/L) = 24,088 mg/L

Explanation:

First step to solve this problem is to know data and questions:

Data:

Sample volume = 50 mL

Sample weight = 50.1 g

P1 (weight empty crucible) = 17.1234 g

P2 (weight after drying a 103º in oven) = 19.9821  g

P3 (weight after incinatrion 550º) = 18.7777 g

Questions:

TS = ?  

VS = ?

Formula:

We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:

TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}

VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}

We have to transform mL to L so we will divide mL by 1,000 the sample volume:

Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L

In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)

Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:

TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L

VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L

We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.

4 0
3 years ago
The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per uni
hjlf

Answer:

Turnover number = 113182s⁻¹

Explanation:

Turnover number is a concept used to know the activity of an enzyme. The higher turnover number, the higher activity of the enzime.

Turnover number is defined as:

Turnover number = Rmax / [E]t

As Rmax is  249 μmol⋅L⁻¹⋅s⁻¹ and [E]t is 2.20 nmol⋅L⁻¹

As 1μmol = 1000nmol, Rmax = 249000 nmol⋅L⁻¹⋅s⁻¹. Replacing:

Turnover number = 249000 nmol⋅L⁻¹⋅s⁻¹ / 2.20 nmol⋅L⁻¹

<h3>Turnover number = 113182s⁻¹</h3>
8 0
4 years ago
A solution is prepared by mixing 25 mL pentane (C3H12, d =
defon
Hope this helps you with yoyrslef

7 0
3 years ago
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