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aleksandr82 [10.1K]
3 years ago
9

Qué clase de alimentos podemos encontrar que contengan oligoelementos como Cromo (Cr), Litio (Li), Selenio (Se), Manganeso (Mn),

Molibdeno(Mo) y cuáles son sus funciones principales
Chemistry
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer: There are many foods who have trace elements like seafood, legume, vegetables, tomatoes, bananas, eggs, fishes, meat.

Their five mainly functions are accept or receive electrons in redox reactions, act like cathalytic centers which are necessaries to live, control biologics process like the hormon activation and gen expression, provide stability, 3d structures in important molecules.

Explanation: Trace elements participate in redox reactions because they have acid-base components which alkalize o acidify the way, protecting the organism from free radicals.

Biological sources which are controlled by trace elements are:

  • assimilation of ATP
  • methabolic sources relationed with insuline
  • oxygen transport in the organism
  • stomach acid formation
  • 3d structures from important moleculess are mantained by balance of electrons which make up the atoms from the molecule.
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The two naturally occurring isotopes of copper are copper-63 and copper-65. how many neutrons are in one atom of copper-65? 29 3
vampirchik [111]
It's 36. Since copper has 29 protons you do 65-29=36
7 0
4 years ago
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Calculate the amount of time required to raise temperature of 0.50 liter of water from
Burka [1]
<h3>Answer:</h3>

19 seconds

<h3>Explanation:</h3>

Heat energy was supplied and used to heat 0.50 liters of water from 0°C to 10°C.

This means water gained heat energy that was supplied.

Therefore;

Heat supplied = Heat gained by water

We are required to calculate the time taken to raise the temperature of water from 0°C to 10°C.

Step 1: Calculate the heat gained by water

Quantity of heat = Mass × specific heat capacity × change in temperature

Density of water 1 g/ml

Volume of water is 500 mL

Therefore, since Mass = density × volume

The mass of water = 500 g

Change in temperature = 10° C

Specific heat capacity of water = 4.18 J/g/°C

Thus;

Quantity of heat = 500 g × 4.18 J/g/°C × 10

                           = 20,900 Joules

Step 2: Heat supplied by a heater

Heat supplied = Power × time

Power = 1100 J/s

Assuming the time required is x

Heat supplied = 1100x Joules

Step 3: Time required

Remember; heat supplied = heat gained by water

Therefore;

1100x joules = 20,900 Joules

      x = 20,900/1100

        = 19 seconds

Therefore, the time required to raise the temperature of water from 0°C to 10°C is 19 seconds.

6 0
3 years ago
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
3 years ago
A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/
Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0
3 years ago
Explain how heat would change the density of a parcel of air
BartSMP [9]
<span>Heat would make the molecules move faster so they would spread out therefore making air less dense.
Less dense means less packed. Heat increase the kinetic energy</span>
6 0
4 years ago
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