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2 years ago

10

Which of the following lists the elements in order, from those having the least protons to those having the most protons in the

atoms?
A. O, N, B, Li

B. Na, S, Al, Cl

C. O, S, Se, Te

D. Rb, K, Na, Li

2 answers:

Tema [17]2 years ago

6 0

Answer:

C

Explanation:

O has 8 protones,S tiene 16, Se 34 y Te 52.

Mandarinka [93]2 years ago

3 0

The answer would be C o,s,se,te

You might be interested in

Properties such as luster conductivity and flexibility are used to describe what type of elements

pickupchik [31]

Answer: metals.

Justification:

There are 118 elements which you can find in the periodic table ordered by atomic number. Those elements my be classified into metals, non-metals and metalloids.

The metals are placed on the left side of the periodic table. The metals share the properties of luster, conductivity and flexibility.

The properties of non-metals (which are on the right side of the periodic table) are opposite to those of metals: opaque, low conductivity and brittle.

Metalloids have in between properties.

Copper, for example is a metal: it has luster, is flexible and is highly conductive of the electricity (and temperature).

4 0

3 years ago

Nikitich [7]

Answer:

Protons and Neutrons are found in the nucleus and the electrons are found on the outside of the nucleus

Explanation:

Hope this helps

7 0

3 years ago

A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw

8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0

2 years ago

Electrons are added into the outermost what in groups 1 & 2

aivan3 [116]

S orbital.

Group 1 elements have a general configuration $ns^{1}$ , where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration $2s^{1}$ whereas Cesium has $6s^{1}$ . Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of $ns^{2}$ . For example, Magnesium has $3s^{2}$ as its outer shell configuration while Strontium has the same as $5s^{2}$ .

We see that in both the cases, the outermost S orbital is being filled.

3 0

3 years ago

1. If 800 C of electric charge passed through a light bulb in 4 min; what is the mag

9966 [12]

The electric current passing through the bulb would be 3.3A

<u>Explanation:</u>

Given:

Electric charge, q = 800C

Time, t = 4 min

= 4 X 60 sec

= 240 sec

Electric current, I = ?

We know,

$Current = \frac{Charge}{Time}$

On substituting the value we get:

$I = \frac{800}{4 X 60} \\\\I = 3.3A$

Thus, the electric current passing through the bulb would be 3.3A

5 0

2 years ago