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3 years ago

5

How do you determine the chemical reactivity for metals by using the periodic table?

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

Chemical reactivity increases down a group and decreases from left to right of a period.

Explanation:

The higher the ionization energy is, the lower the reactivity is. Since the ionization energy is highest in the top right corner of the periodic table, we can assume that the most reactive elements are in the opposite bottom left corner. This is because the electrons that react are farther away from the nucleus thus experience less attraction to the nucleus (called nuclear shielding). Therefore their electrons are more easily removed than elements that don't ecperience nuclear shielding.

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2 years ago

Read 2 more answers

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago