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3 years ago

How is energy Associated with food stored

Chemistry

2 answers:

scoundrel [369]3 years ago

8 0

A form of energy associated with the positive and motion of the object.

Flauer [41]3 years ago

7 0

There is molecular atoms that have energy like for example uh a kool aid has a lot of sugar and that sugar produces the energy and it is stored mainly how the inventors make the product.

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Help me. ASAP!!!!! 23 POINTS!!!!!!!

Alona [7]

a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.

b) they lose 2 electrons as they are transferred to the non metal.

c)They obtain this charge as when they are made into an ionic compound the 2 electrons on the outer shell are transferred to the non metal, meaning there are 2 more protons that electrons, giving it a positive charge.

hope this helps! :)

6 0

2 years ago

The reaction 2a → a2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if

olga nikolaevna [1]

According to the second order formula:
1/[At] = K t + 1/[Ao]
and when we have the K constant =0.0265 & we have t = 180 min & we have the initial concentration of A = 4.25 so by substitution:

1/[At] = 0.0265 X 180min + 1/4.25
1/[At] = 5
∴[At] = 1/5 = 0.2 m

7 0

3 years ago

Read 2 more answers

If two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is

Fed [463]

If the gases are at the same temperature and pressure, the ratio of their effusion rates is directly proportional to the ratio of the square roots of their molar masses:

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

Where

R₁ and R₂ are the rates of the two gas
M₁ and M₂ are the molar masses of the two gas

From the Graham's law equation, we can see that the ratio of the rates of effusion of the two gases is directly proportional to the square root of their molar masses

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

#SPJ1

3 0

2 years ago

77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a

BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

Non examples of chemical change<br> List some please

Ivan

Answer:

Crushing a can.

Melting an ice cube.

Boiling water.

Mixing sand and water.

Breaking a glass.

Dissolving sugar and water.

Shredding paper.

Chopping wood.

3 0

2 years ago