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2 years ago

Why is oxygen more chemically reactive than nitrogen?

Physics

2 answers:

grin007 [14]2 years ago

5 0

Answer:

O2 has two more electrons compared to N2, with extra 2 electrons in the higher energy anti-bonding orbitals known as Diradical. These electrons have higher energy and are unpaired; therefore, O2 is more reactive

Explanation:

Anni [7]2 years ago

5 0

Answer:

O B. Oxygen needs to gain only 2 electrons to have a full octet, while nitrogen needs to gain 3.

Explanation:

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What is the best free energy source?<br> Nuclear<br> Solar<br> Natural Gas

Free_Kalibri [48]

Solar it is the cheapest and widely used energy source

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2 years ago

Một oto đang chuyển động thẳng đều với vận tốc 36km/h thì hãm phanh sau 20s thì ô tô dừng lại quãng đường oto đi được là

saw5 [17]

36km/h=10m/s
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3 years ago

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

8 0

2 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

$v_2 = 5.6 \ m/s$

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$