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Alex777 [14]
3 years ago
13

Please if you just want point TELL ME I can’t fail this.

Chemistry
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

here you go. I took the test :)

Explanation:

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Explain the difference between commensalism,parasitism,and mutualism and give examples of each
vaieri [72.5K]
Commercialism is when one benefits while the other gets no harm nor benefit.
Example: the Remora rides attached to bigger types of fish or shark and eat the remains of meals left behind from the organism they are attached to. 

Parasitism is when benefits and the other is harmed. 
Example: A tape worm feeds off the intestines and eat what the host eats, killing the host because they are deprived of nutrients. 

Mutualism is when both benefit.
Example: when a bird is allowed to peck the food from a rhinos mouth to help the rhinos teeth and allows the birth food. 
6 0
4 years ago
Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2
abruzzese [7]

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

<u>Step 1:</u> Data given

2S (s)+3O2 (g)→2SO3 (g)     ΔH = -790.4 kJ  

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

<u>Step 2</u>: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ  

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

4 0
4 years ago
What is the molar mass of sodium phosphate, Na3PO4?
baherus [9]

Answer:

164g

Explanation:

3×23+31+16×4

≈69+31+64

≈164.0 g

8 0
3 years ago
Which causes genetic variations and can result in different alleles? O predation rate O random mutations O competition o environ
alekssr [168]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.
Maslowich
Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
                                                     = 10 moles
Mole of hydrogen in the compound = 20.2/1.01
                                                         = 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
So
The empirical formula of the compound is H2O. I hope the answer has come to your help.
3 0
4 years ago
Read 2 more answers
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