Can metals have covalent radius and van der Waal's radius instead of metallic radius?

Chemistry

1 answer:

romanna [79]3 years ago

4 0

Explanation:

Whether you choose to use van der Waals radii or metallic radii as a measure of the atomic radius, for metals the ionic radius is smaller than either, so the problem doesn't exist to the same extent. It is true that the ionic radius of a metal is less than its atomic radius (however vague you are about defining this).

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Hot water enters a double-pipe counter-flow water-to-oil heat exchanger at 220°7 and leaves a 100°F. Oil enters at 70°F and leav

il63 [147K]

Answer:

Oil has the smaller heat capacity. The effectiveness of the heat exchanger is 0.80.

Explanation:

Part 1:

In order to know which fluid has the smaller heat capacity we need to consider the heat equation below:

Q = CΔT, where Q is the heat exchanged, C is the heat capacity and ΔT is the variation in temperature.

As the heat exchange is the same for both fluids, the smaller the temperature variation, the smaller the heat capacity.

Water: ΔT = 120 °F

Oil: ΔT = 80 °F

Therefore, oil is the fluid with the smallest heat capacity.

Part 2:

The effectiveness of a counter-flow heat exchanger is given by the equation bellow:

$E = \frac{Th1 - Th2}{Th1 - Tc1} \\$

Th1: initial temperature of the hot fluid

Th2: final temperature of the hot

Tc1: initial temperature of the cold fluid

$E = \frac{220 - 100}{220 - 70} \\E = 0.8$

7 0

3 years ago

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Lunna [17]

The answer could be a satellite

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3 years ago

What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?

Pie

$\text{Given that,}\\\\\text{Pressure,}~ P = 1.99~ \text{atm}\\ \\\text{No. of moles,}~ n = 0.182~ \text{mol.}\\\\\text{Temperature,}~ T = 83.4^{\circ}~ C = 356.4~K\\\\ \text{Molar gas constant,}~ R = 0.082~ \text{L}~ \text{atm}~ \text{mol}^{-1}~ \text{K}^{-1}\\\\\text{Volume,}~ V =?\\\\\text{We know that,}~\\\\~~~~~~~PV =nRT\\\\\implies V = \dfrac{nRT}{P}\\\\\\~~~~~~~~~~=\dfrac{0.182 \times0.082 \times 356.4 }{1.99}\\\\\\~~~~~~~~~~\approx 2.68~ \text{L.}\\\\\\$