The answer is the third one because density is mass divided by volume so 80 divided by 10 is 8
I found a presentation of Food web of a pond that will greatly connect with the above problem. http://www.eduweb.com/portfolio/earthsystems/food/foodweb4.html
Normal setting:
Species Pop. size
Blue heron Medium
Perch Medium
Bass Medium
Minnows Medium
Inverts Medium
Algae Medium
If PERCH population size decreases,
Species Pop. sizeBlue heron LowPerch LowBass MediumMinnows HighInverts HighAlgae Low
As you can see, 4 other species are affected when the Perch population size decreased. Bass is not affected.
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
H₂S
<h3>Further explanation</h3>
Given
ΔH fusion and ΔH vaporization of different substances
Required
The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes
Solution
We can use the formula :
Q=heat/energy absorbed
n = moles
The heat absorbed : 58.16 kJ
moles = 3.11
so ΔH vaporization :
The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S
(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)
