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finlep [7]
2 years ago
13

The pendulum has the greatest kinetic energy and no potential energy when it is _____.

Chemistry
1 answer:
Bad White [126]2 years ago
3 0

Answer:

At the top of the swing

Explanation:

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Which state of matter expands when heated and is easy to compress
AlladinOne [14]
That would be gases
6 0
3 years ago
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You find a ring with a mass of 107g . You fill a graduated cylinder up with 10ml of water and put the ring into the cylinder . T
Korvikt [17]

Answer:

5 mL

Explanation:

Given data:

mass of ring = 107 g

volume of water = 10 mL

increase in volume = 15 mL

How much water displace = ?

Solution:

V (ring) = V (water + ring) - V (water)

V (ring) = 15 mL -  10 mL

V (ring) = 5 mL

when the ring is put into cylinder, volume is increased by 15 mL. The volume of water was 10 mL so water is displaced by 5 mL and the volume 5mL is the voulme of ring.

3 0
3 years ago
In which part of the cell is the majority of the energy released from the breakdown of glucose
Vladimir [108]

<em>Answer</em><em>:</em>

<em>Glycolysis</em>

<em>E</em><em>xplanation</em><em> </em><em>:</em>

Glycolysis is the first step in the breakdown of glucose to extract energy for cell metabolism.Many living organisms carry out glycolysis as part of their metabolism. Glycolysis takes place in the cytoplasm of most prokaryotic and all eukaryotic cells.

7 0
3 years ago
Read 2 more answers
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

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

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

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

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 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'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
A nickle is about 2 millimeters thick, or 2/1000 of a meter.how many nanometers is this?
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You know that you can just google it right? It has this converter.
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