All categories

4 years ago

The ability of atoms to attract electrons is referred to as

Physics

2 answers:

dexar [7]4 years ago

8 0

Answer: electronegativity

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases and thus the electronegativity increases.

IRISSAK [1]4 years ago

4 0

Answer:

Electronegativity

Explanation:

APEX I just took the test.

You might be interested in

An electrochemical cell has a constant potential of 2 V and provides a steady current of 1 A. How many electrons move from the a

myrzilka [38]

Answer:

$n=6.25\times 10^{21}$

Explanation:

Potential across the electrochemical cell, V = 2 V

Current, I = 1 A

Time, t = 1000 s

Let n electrons move from the anode to the cathode. We know that electric current is calculated as :

$q=I\times t$

Also, q = ne

$ne=It$

$n=\dfrac{It}{e}$

$n=\dfrac{1\times 1000}{1.6\times 10^{-19}}$

$n=6.25\times 10^{21}$

So, $6.25\times 10^{21}$ electrons move from the anode to the cathode in 1000 seconds.

8 0

4 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago

A shark travels with an average velocity of 12 m/s s. How long (time) would it take the shark to swim 42 m at that velocity

liq [111]

Answer:

t=3.5 seconds

Explanation:

Moving at Constant Speed

Let's suppose and object travels in a constant direction and covers the same distances x at the same time t, we say it has a constant speed. It can be computed as:

$\displaystyle v=\frac{x}{t}$