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Dennis_Churaev [7]
3 years ago
12

What does a banana have that a human does not?

Chemistry
2 answers:
NARA [144]3 years ago
8 0

Answer:

appealing taste

ba dump shhhh

um i dont know if this was a real question

Explanation:

Vikentia [17]3 years ago
7 0
It would be a face or toes
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Viruses are not considered organisms, because they lack many of the characteristics of living things. A virus is essentially a s
tensa zangetsu [6.8K]

Answer:

well, I think the answer is C

7 0
3 years ago
Help please <br> What percentage of the world does not have a reliable source for clean water?
disa [49]

Answer:9%

Explanation:

7 0
2 years ago
In a laboratory experiment, John uses a mesh to separate soil particles from water. Which technique of separation is he using?
m_a_m_a [10]
Correct answer -->filtration
8 0
3 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0
3 years ago
40.0L of N₂ gas are in a sealed container at STP.How many moles of N₂ are present?9 mol
Vinvika [58]

Explanation:

We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.

STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.

1 mol of N₂ = 22.4 L

moles of N₂ = 40.0 L * 1 mol/(22.4 L)

moles of N₂ = 1.79 mol

Answer: 1.79 moles of nitrogen are present.

8 0
1 year ago
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