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a_sh-v [17]
3 years ago
8

How Do Clownfish Catch Their Food?

Chemistry
1 answer:
Kobotan [32]3 years ago
7 0
Clown fish get there food from anemone by eating by eating the algae
You might be interested in
What material is least likely to be recognized as a mixture by looking under a microscope
I am Lyosha [343]
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
6 0
3 years ago
Lithium is a metal,chlorine is a non metal,so LiCl would be a ______bond
ivann1987 [24]
Ionic is the answer. This is because lithium has a positive charge, while chlorine has a negative charge, meaning the compound doesn’t necessarily have an overall charge.
4 0
3 years ago
PLZ HELP!!
Nataly [62]

A and B are experiencing winter. The picture which isn't available here in this question is attached below.

Option C.

<h3><u>Explanation:</u></h3>

The earth is tilt by an angle of 23.2° to the vertical plane. This makes the seasonal variation of earth, because in some time of the year, the northern hemisphere faces the sun directly, experiencing summer and then southern hemisphere is away from summer experiencing winter and vice versa. The summer occurs when the place directly faces the sun. And the winter happens when the place obliquely faces the sun or doesn't face the sun at all.

Here in this diagram, we can see that the points A and B are the north pole and the part in northern hemisphere respectively which aren't facing the sun directly, whereas C and D are facing the sun. Thus the southern hemisphere is experiencing summer and the northern hemisphere the winter.

4 0
2 years ago
Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?
Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0
2 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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