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3 years ago

The graphics below shows how photosynthesis and cellular respiration are related.

Chemistry

1 answer:

lisov135 [29]3 years ago

8 0

Options found from another source are:

a. oxygen. b. glucose. c. energy stored as ATP. d. carbon dioxide and water

Answer:

c energy stored as ATP

Explanation:

Cellular respiration converts glucose into energy in the form of ATP (c). The answer cannot be oxygen (a), because this is required for this process as a final electron acceptor. In terms of photosynthesis, oxygen is released as a by-product. The answer cannot be glucose (b) because that is our starting point for respiration, and what is synthesised during photosynthesis. The answer cannot be (d) as carbon dioxide and water are released by cellular respiration, and required by photosynthesis

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Which of the following is NOT true about energy?*

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Answer:

the last one

Explanation:

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What is metallurgy?<br>

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2 years ago

Arrange the molecules in order from strongest intermolecular force to weakest.

motikmotik

Answer:- $C_4H_1_0>C_3H_8>C_2H_6>CH_4$

Explanations:- Alkanes are non polar molecules as these only have carbons and hydrogens. Electron negativity difference of C and H is very low and it makes them non polar. These have weaker London dispersion forces.

The forces of attraction becomes stronger in alkanes as the number of carbon increases because the surface area as well as molecular weight of the alkanes increases with an increase in number of carbons.

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3 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

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2 years ago