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2 years ago

What type of consumer eats only producers?

2 answers:

8 0

Answer: Primary consumers make up the second trophic level. They are also called herbivores. They eat primary producers—plants or algae—and nothing else. For example, a grasshopper living in the Everglades is a primary consumer

FrozenT [24]2 years ago

5 0

Answer:

primary consumer make up the second trophic level .they are also herbivores they eat primary consumer plants or alger and nothing else .for example a grasshopper living in the everglades is a primary consumer

You might be interested in

600.0 mL of air is at 20.0 Celsius. What is the volume of the same air at 60.0 Celsius

REY [17]

Answer:

0.682L or 682mL

Use Charles Law of V1/T1 = V2/T2

V1 = 0.6L

T1 = 293K

V2=?

T2= 333K

Explanation:

6 0

2 years ago

You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration

vovangra [49]

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM

7 0

3 years ago

When planning a simple experiment, what does it mean to “test one variable”?

SSSSS [86.1K]

Answer:

I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."

Explanation:

I remember from last year but I'm not totally sure. Good luck!

3 0

2 years ago

The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal

bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = $\frac{0.840}{148.2}moles$ of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release $(5539\times 0.00567)kJ$ of heat or 31.41 kJ of heat

6.60 kJ of heat increases $1^{0}\textrm{C}$ temperature of calorimeter.

So, 31.41 kJ of heat increases $(\frac{1}{6.60}\times 31.41)^{0}\textrm{C}$ or $4.76^{0}\textrm{C}$ temperature of calorimeter

So, the final temperature of calorimeter = $(20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}$

3 0

3 years ago

Calculate the frequency of radiation with a wavelength of 1.08 x 10^-6 m. simple explanation please

pishuonlain [190]

C = vf
c stands for the speed of waves (which is a constant that is 3 x 10^8)
v stands for the wavelength (which is given)
f stands for frequency (what we are solving for)
3 x 10^8 = (1.08 x 10^-6)f
Divide both sides by the given wavelength
f = 2.78 * 10^14 seconds

8 0

2 years ago