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MAVERICK [17]
3 years ago
12

Which characteristic is typical of metals? A. They are brittle. B. They do not conduct electricity. C. They are highly soluble i

n water. D. They are malleable.
Chemistry
2 answers:
kkurt [141]3 years ago
8 0

Answer: Option (D) is the correct answer.

Explanation:

Metals are the species which have excess of electrons. Metals are placed from left side to more than middle side of the periodic table.

Some properties of metals are as follows.

  • Metals are good conductors of heat and electricity.
  • They are malleable and ductile i.e, they can be drawn into thin sheets and wires.
  • They are lustrous in appearance.
  • Metals generally have high melting point.

Thus, we can conclude that out of the given options, characteristic they are malleable is typical of metals.

shutvik [7]3 years ago
7 0

D. They are malleable.


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A physicist

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What event typically causes upwelling?
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How many atoms are in 3.690 moles of oxygen
sweet [91]

1 mole of any substance contains Avogadro's number.  

So, 1 mole of O2= 6.023x10^23 molecules  

3 mole of O2= 6.023x10^23x3 molecules  

= 1.8069x10^24 molecules  

Each molecule of Oxygen has 2 atoms.  

therefore,  

1.8069x10^24 molecules= 1.8069x10^24 x 2 atoms  

= 3.6138x10^24 atoms.

4 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
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