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iVinArrow [24]
4 years ago
14

What is the kinetic energy in kJ of 1 mole of water molecules (mass=18) if the average velocity is 590 m/s (1300 mph)

Chemistry
1 answer:
zubka84 [21]4 years ago
7 0

Answer:

= 3132.9 Joules

Explanation:

  • Kinetic energy is the energy possessed by a body when in motion.
  • Kinetic energy is calculated by the formula; K.E = 1/2 mV², where m is the mass of the body or object, and V is the velocity.
  • Therefore kinetic energy depends on the mass and the velocity of the body or the object in motion.

In this case;

Kinetic energy = 0.5 × 0.018 kg × 590²

                        <u>= 3132.9 Joules</u>

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During respiration, energy is retrieved from the high-energy bonds found in certain organic molecules. Which of the following, i
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(A.) CO2, H2O

Explanation:

The chemical equation for respiration process is:

C_{6}H_{12} O_{6} + 6O_{2} → 6CO_{2} + 6H_{2} O + ATP↑

Energy is released during the biochemical process in the organism's cells in form of ATP. Byproducts of the reaction are carbon dioxide and water molecules.

Let me know if you require any further assistance.

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In order to predict the outcome of the reaction, write the molecular, full ionic, and net ionic equations for a mixture of aqueo
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Answer:

See explanation

Explanation:

Full molecular equation;

2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)

Full ionic equation

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Net ionic equation;

2NH3(aq) + Ag^+(aq) -------->  [Ag(NH3)2]^+(aq)

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What is the name of the molecule shown below?
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3 years ago
NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
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Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

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Y = 25700J/mol×\frac{1mol}{80,043g}X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

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