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Nimfa-mama [501]
3 years ago
6

Element X has five valence electrons, element Y has one valence electron, and element Z has one valence electron. Which two of t

hese elements are most likely to have similar properties? Explain your answer.
Physics
2 answers:
Grace [21]3 years ago
5 0
The two elements would be element Y and element Z. They both have one valence electron. This means that their bonding properties would be similar because both of them can donate 1 electron. Since both of them have one valence electron, one can assume that they also belong in the same group. 
Tomtit [17]3 years ago
4 0

Answer:

Element Y and element Z

Explanation:

One of the criteria for grouping the elements in the periodic table is according to their valence electrons.

If several elements have the same amount of valence electrons, they belong to the same group.

If they have the same configuration of valence electrons, then electronegativity, atomic radius, ionization energy, among others. It will be similar.

Element Y and element Z have the same number of valence electrons, then they will be located in the same group, and share similar periodic properties

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Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
Free_Kalibri [48]

Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

So if n= 5 it can take 0,1,2,3,4

i.e it can take 5 values

2)for an electron with l =3

it can be from -3 -2 -1 0 1 2 3

i.e it can take 7 values

3) n = 3 !!

l = 0 , 1 , 2

for l=0 , m = 0 total = 1

for l= 1 ,m = -1,0,1 total = 3

for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

total possible states = 9 * 2 = 18

Answer is 168

4)given l=3 and n=3

orbital quantum number cannot be equal to principal quantum number

its max value is l-1 only

5)L = sqrt(l(l+1))x h'

for it to be max l should be max

for n = 3 max l value is 2

therfore it is sqrt(2(2+1)) x h'

\sqrt(6) \times h'

this is the answer

5 0
3 years ago
A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t
Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, \epsilon=1.5\ V

Magnitude of Earth's field, B=10^{-4}\ T

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}

\omega=159.15\ rad/s

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0
2 years ago
An object travels at a speed of 7500 cm/sec . how far will it travel in kilometers in one day
DiKsa [7]

Answer:

6480 km

Explanation:

The speed of the object is

v = 7500 cm/sec

We need to convert centimetres into kilometers and seconds into days. We have:

1 cm = 1\cdot 10^5 km

1 s = \frac{1}{60\cdot 60 \cdot 24}d

Using these conversion factors, we find:

v=7500 \frac{cm}{s} \cdot 1\cdot 10^{-5} \frac{km}{cm}\cdot (24)(60)(60) \frac{s}{d}=6480 km

3 0
3 years ago
A brick in the shape of a cube with sides 10 cm has a density of 2500 kg/m^3. What is its weight? a.) 250 N
Arisa [49]

Answer:

c.) 25 N

Explanation:

 We find the volume of the brick, knowing that the volume of a cube is given by the formula:

l=0,1 metros \\V=l^{3}\\V=(0,1\: \: m)^{3}=0,001\: \: m^{3}

being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg

We find the weight by multiplying the mass of the object with the gravity constant.

W=m*g=(2.5 \:Kg)*(9,81 \:m/s^{2} )=24,5 N\approx25\: N

8 0
3 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
3 years ago
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