Answer:
D. How several food chains are related
Explanation:
Answer:
Hey,
So what you would do here is first choose a biome, such as the desert for example, then choose an organism that lives there, or an animal. Then think of what it is like where it lives, is it dry, cold, windy, and so force, just describing the place by its looks and climate. How would this population increase, maybe it would be if there was suddenly more food there for them, or if their ecosystem got better for them to live, then, if they populated more, how would that affect the ecosystem. Maybe some animals would die off more, or some would have to much food. Like what if the fly population suddenly doubled, for example, how would that affect the ecosystem? Now what would happen if that organism was removed? If flies were removed, spiders and a lot of other organisms would not have food, and some organisms would overpopulate because nothing would eat them. That is just an example.
Explanation:
I hope this helps, sorry if not. Have a great day!
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
brainly.com/question/23580857
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Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)
