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JulijaS [17]
3 years ago
10

Determine whether each practice is recommended to keep a micropipette clean and functioning properly.

Chemistry
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer:

1. A

2. B

3. B

4. A

5. B

Explanation:

1. Recommended. That way cross-contamination is prevented.

2. Not recommended. The inner mechanism of the micropipette should not come in contact with any liquid whatsoever.

3. Not recommended. It contradicts statement 1.

4. Recommended. That way the micropipette remains completely vertical, minimizing the risks of being decalibrated.

5. Not recommended. If it lays on the bench it will be horizontal, thus contradicting statement 4.

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The combination of the +vely charged nitrogen and the electronegative oxygen atom leads to delocalization causing the resonance effect.

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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,

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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

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3 years ago
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Ilya [14]

Answer:

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Explanation:

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