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Tems11 [23]
3 years ago
10

Which physical property refers to the temperature at which substance in a solid state transforms to a liquid state?

Chemistry
2 answers:
professor190 [17]3 years ago
6 0
Melting, in which heat of fusion is needed.
tester [92]3 years ago
5 0

Answer:

Meltting point

Explanation:

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What is the mass in grams of 6.022×1023 atoms of mass 16.00 amu?
skelet666 [1.2K]
You need the Avogadro's number.  I can't remember exactly the calculation.
6 0
2 years ago
When people conserve electricity, power plants do not need to produce as much electricity because there is less demand for it.
almond37 [142]

True

Explanation:

When people conserve electricity, power plants do not need to produce as much electricity because there is less demand for it.

This very correct.

  • When electricity is used in a sustainable way, there is little of it consumed.
  • Switching off idle electronics would make that amount of electric current to be re-distributed to another place.
  • Less coal and fuel would be used and power plants do not need to produce as much electricity.
  • Most electricity produced in power plants are often wasted due to idle appliances consuming the bulk of them.
  • When these appliances are turned off, less energy is in demand and less will be produced.

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5 0
3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
3 years ago
Two categories of compounds are
crimeas [40]

Answer is: (3) ionic and molecular.

Ionic compounds are made of ions held together with ionic bonds.

Ionic bond forms when a cation transfers its extra electron to an anion who needs it.  

For example compound magnesium chloride (MgCl₂) has ionic bond (the electrostatic attraction between oppositely charged ions).  

Magnesium (metal) transfers two electrons (became positive cation) to chlorine (became negative anion).  

Molecular compounds are made up of molecules whose atoms are connected with covalent bonds.

Covalent bond is bond between nonmetals.  

For example, molecule carbon monoxide CO has covalent bond.

Carbon (C) and oxygen (O) are nonmetals.  

Carbon atom and oxygen atom are connected by a triple bond (six shared electrons in three bonding molecular orbitals) that is formed of two covalent bonds and one dative covalent bond.  

4 0
3 years ago
Read 2 more answers
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
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