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bija089 [108]
3 years ago
7

The melting point of copper is 1084°C. How does the energy of the particles in a certain amount of liquid copper compare to the

Energy of the molecules in the same amount of liquid water? Why?
Chemistry
2 answers:
Kobotan [32]3 years ago
8 0

Explanation:

As it is given that melting point of copper is 1084^{o}C. So, it means liquid state of copper will exist at a temperature greater than 1084^{o}C.

Whereas, it is known that liquid state of water exists at above zero degree celsius.

And, kinetic energy of the particles of a substance is directly related to the temperature as follows.

                    K.E = \frac{3}{2} kT

Hence, more is the temperature of a substance more will be the kinetic energy of its molecules.

Therefore, temperature of liquid copper is more than the temperature of liquid water.

Thus, we can conclude that the energy of the particles in a certain amount of liquid copper is greater than the energy of the molecules in the same amount of liquid water.

Fed [463]3 years ago
5 0
Solids' particles move a lot faster and closer together. Liquids' are more spread apart and move slower. So there is more energy in the solid
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Wittaler [7]

Answer:

Re=2094,76

Explanation:

For a fluid that circulates inside a straight circular pipe, the Reynolds number is given by:

Re=\frac{pvD}{u}

where (using the international measurement system):

  • ρ: density of the fluid [kg/m3]
  • v: velocity of the fluid [m/s]
  • D: diameter of the pipe through which the fluid circulates [m]
  • μ: dynamic viscosity [Pa.s]

To solve the probelm, we just need to replace our data using <u>THE CORRECT UNITS</u> in the Reynolds number equation. So we have:

ρ=1051 kg/m3,

v=34,3 cm/s=0,343 m/s

D=2,15 cm = 0,0215 m

μ = 3,7 cp * 10^-3 Pa.s/1 cp = 3,7*10^-3 Pa.s

Replacing in the main equation:

Re=\frac{1051\frac{kg}{m^{3} }*0,343\frac{m}{s}*0,0215m  }{3,7*10^{-3}Pa.s } =2094,76

So the Reynolds number is 2094,76 (note that the Reynolds number is a dimensionless quantity).

3 0
3 years ago
The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent
anastassius [24]

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have \frac{1}{3} of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

\frac{1}{3}*(-7198kJ)

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

7 0
3 years ago
When 155 mL of water at 26 C is mixed with 75 mL of water at 85 C, what is the final temperature? (assume that no heat is releas
Ket [755]
Depends if it desolved with the water or not

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The nucleoside adenosine exists in a protonated form with a pKa of 3.8. The percentage of the protonated form at pH 4.8 is close
Natasha_Volkova [10]

Answer:

Ok:

Explanation:

So, you can use the Henderson-Hasselbalch equation for this:

pH = pKa + log(A^-/HA) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.

We can solve that

1 = log(A^-/HA\\) and so 10 = A^-/HA or 10HA = A-.  For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.

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2 years ago
Consider the following voltaic cell:(b) In which half-cell does oxidation occur?
damaskus [11]

At the anode, half-cell oxidation occurs in a voltaic cell.

<h3>Voltaic Cell Principle</h3>

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Two half-cells plus a salt bridge make up the voltaic cell. An electrolyte-immersed metallic electrode is present on each side of the cell. These two half-cells are wired together to form a connection to a voltmeter.

<h3>Voltaic Cell Parts</h3>
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Learn more about Voltaic cells here:-

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