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kari74 [83]
3 years ago
13

Why is the amount of usable energy lower for anaerobic respiration than for aerobic respiration

Chemistry
1 answer:
photoshop1234 [79]3 years ago
3 0
Anaerobic is when there is not enough oxygen for aerobic respiration, so it uses glucose and lactic acid for example instead of glucose and oxygen for aerobic. Anaerobic takes longer to convert to energy. 
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HELP PLEASE!! (100 points)
MrRa [10]

Answer:

As you cool a matter to absolute zero, their kinetic energy reduces significantly and the molecules slows down and begins to aggregate together. ... As heat is added, the molecules gain more kinetic energy. This shown in their increase motion. When heat is withdrawn, the particles slows down hope this helped

3 0
3 years ago
What is diffusion of water?
andre [41]

Answer:it’s called osmosis

Explanation:Osmosis is a specific type of diffusion; it is the passage of water from a region of high water concentration through a semi-permeable membrane to a region of low water concentration. 

6 0
1 year ago
How many grams of chlorine gas can be produced from the decomposition of 73.4
ser-zykov [4K]
1) first, we have to convert the grams to moles of AuCl3 using the molar mass of the molecule. 

molar mass of AuCl3= 197 + (35.5 x 3)= 303.5 g/mol

73.4 g (1 mol AuCl3/ 303.5 g)= 0.242 moles

2) now. let's convert moles of AuCl3 to moles of chlorine gas (Cl2) using the mole-mole ratio

0.242 mol AuCl3 (3 mol Cl2/ 2 mol AuCl3)= 0.363 mol Cl2

3) finally, we convert moles to grams using the molar mass of Cl2.

molar mass of Cl2 = 35.5 x 2= 71.0 g/mol

0.363 mol Cl2 ( 71.0 g/ 1 mol)= 25.8 grams
6 0
3 years ago
Cu2(s)+O2(g)=Cu2O(g)+SO2(g)<br><br>please help urgent will give brainiest​
AysviL [449]

Answer:

2 Cu2S + 3 O2 = 2 Cu2O + 2 SO2

Explanation:

2 Cu2S + 3 O2 = 2 Cu2O + 2 SO2

3 0
3 years ago
At 25 ∘C only 0.0180 mol of the generic salt AB is soluble (at equilibrium) in 1.00 L of water.What is the value for Ksp for thi
Ghella [55]

Answer:

Ksp = 3.24 x 10⁻⁴

Explanation:

The dissociation equilibrium for a generic salt AB is:

AB(s) ⇄ A⁺(aq) + B⁻(aq)

              s            s

For instance, the expression for the Ksp constant is:

Ksp = [A⁺] [B⁻] = s x s = s²

According to the problem, 0.0180 mol of the salt is soluble in 1.00 L os water. That means that the solubility of the salt (s) is equal to 0.0180 mol per liter.

s = moles of solute/L of solution = 0.0180 mol/L

Thus, we calculate Ksp from the s value as follows:

Ksp = s² = (0.0180)² = 3.24 x 10⁻⁴

6 0
3 years ago
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