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kifflom [539]
3 years ago
8

Which of these time measurements is the smallest? 0.02 seconds, 0.02 teraseconds, 2,500 milliseconds, 25,000 nanoseconds

Chemistry
2 answers:
VladimirAG [237]3 years ago
7 0
25,000 nanoseconds :)))
zheka24 [161]3 years ago
4 0

Answer:

25,000 nanoseconds= 2.5e-5 seconds

Therefore25,000 nanoseconds is the smallest

Explanation:

0.02 seconds,

0.02 teraseconds, =20000000000 seconds  

2,500 milliseconds, = 2.5 second  

25,000 nanoseconds= 2.5e-5 seconds

Therefore25,000 nanoseconds is the smallest

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The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
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Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol  

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = \frac{0.590 g}{122.12 g/mol}=0.004831 mol

Energy released by 0.004831 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J

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15,595.5 J=C\times 2.125^oC

C=7,339.05 J/^oC

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose =\frac{1.400 g}{180.16 g/mol}=0.007771 mol

Energy released by the 0.007771 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

21,603.01 J=7,339.05 J/^oC\times \Delta T'

\Delta T'=2.943^oC

2.943 °C temperature change from the combustion of the glucose has been taken place.

8 0
3 years ago
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velikii [3]

Answer:

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Explanation:

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8 0
3 years ago
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Which element has 13 protons and 8 neutrons?
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only thing close I can see would be aluminun

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Consider the reaction 3X + 2Y 5C + 4D
xeze [42]

Answer:

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Explanation:

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