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OLga [1]
3 years ago
9

Do metal atoms tend to gain or lose electrons?

Physics
2 answers:
german3 years ago
6 0
<span>The elements of the periodic table are classified into three categories: inert gases, nonmetals and metals. Generally, inert gases do not readily gain nor lose electrons, while nonmetals are more likely to acquire electrons. Metals contain low ionization energies, which refer to the amount of energy required to free or remove an electron. These elements also have low electron affinities, or the attractive forces between an incoming electron and the nucleus of an atom. The lower the ionization energies and electron affinities of an atom are, the greater the tendency to lose electrons.</span>
Solnce55 [7]3 years ago
3 0
<span>Metal atoms have the tendency to lose electrons. The type of chemical bonding formed by atoms of metallic elements is called metallic bonding.</span><span>In terms of gaining or losing electrons, the elements of the periodic table are classified into three categories: inert gases, nonmetals and metals. Generally, inert gases do not readily gain nor lose electrons, while nonmetals are more likely to acquire electrons. Metals contain low ionization energies, which refer to the amount of energy required to free or remove an electron. These elements also have low electron affinities, or the attractive forces between an incoming electron and the nucleus of an atom. The lower the ionization energies and electron affinities of an atom are, the greater the tendency to lose electrons.</span> 
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If an airplane undergoes a displace-
PIT_PIT [208]
The formula for velocity is distance divided by time, or d/t. The distance is 500 km and the time is 1.2 hours. 500/1.2 is 416.6 km/hr.
4 0
3 years ago
14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar
Zolol [24]

Answer:

T= 11.0003s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally  the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

 \omega ^2 r = 9.8

  \omega ^2= \frac{9.8}{30}

 \omega = \sqrt{\frac{9.8}{30}}

 \omega= 0.571 rad/s

Therefore solving for  Period T

Generally the equation for solving Period T is mathematically given as

 T= \frac{2\pi}{\omega}

 T= \frac{2\pi}{0.571 rad/s}

 T= 11.0003s

5 0
3 years ago
Please answer D in the image with an explanation
puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0
3 years ago
Place the following types of electromagnetic radiation in order of increasing wavelength. ultraviolet light gamma rays radio wav
kvv77 [185]

Answer: gamma rays < ultraviolet light < radio waves

Explanation: We have to explain the ordering we take into account a relationship for electromagnetic wave, which is given by:

c=λ*ν c is the speed of light.  λ and ν are the wavelegth and the frequency of the waves.

Then we also know that the high frequency of these radiation correspond to the gamma ray (10^18 1/s) while the lowest correspond to radio wave (10^8 1/s) so:

The ultraviolet ligth is ranged at frequencies of 10^16 1/s located bewteen gamma and radio radiation.

as large frequency lower wavelength we can order these radiations from short to long wavelengths as follows:

gamma rays < ultraviolet light < radio waves

3 0
3 years ago
A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

8 0
3 years ago
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