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Tanzania [10]
3 years ago
6

Elements are arranged in groups by similar atomic structure on the periodic table. This allows for an elements properties to be

predicted based on general periodic trends. One of these trends, atomic radius, increase down a group and to the left along a period and can be defined as _________.
A). One-half the distance between the nuclei of two atoms

B). The distance between the nuclei of two atoms

C). One-half the diameter of an atom

D). The diameter of an atom
Physics
2 answers:
DiKsa [7]3 years ago
5 0

C

Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.

Explanation:

This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has  a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius

The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.

Learn More:

brainly.com/question/1604565

brainly.com/question/13126562

#LearnWithBrainly

cestrela7 [59]3 years ago
3 0

Answer:

the answer is actually A I just took it on usatestprep

Explanation:

Atomic radius can be defined as the atom's ability to one-half the distance between the nuclei of two atoms. Atomic radius is measured between two nuclei because the edge of the electron cloud is too difuse to clearly measure otherwise. Atomic radius increases down a group and to the left along a period.

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A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the
densk [106]

Answer:10.4 times of initial velocity

Explanation:

Given

Diameter reduced by 69 %

it approaches with velocity v_0

suppose its velocity is v during blocked passage

suppose d is the initial diameter and d_2 diameter is

d_2=d(1-0.69)

d_2=0.31 d

A_2=\frac{\pi d_2^2}{4}

As flow is constant

Q_1=Q_2

d^2v_0=d_2^2v

v=10.40 v_0

6 0
3 years ago
If the mass of the products measured 120 g, what would be the mass of the reactants? 30 g 60 g 120 g 240 g
Tanya [424]

Answer:

The correct answer option is C

Explanation:

In a balanced chemical reaction mass of the reactant are always equal to mass of the products. Also known as Law of Conservation of Mass  which states that " mass can nor be created nor be destroyed in a chemical reaction."

So, the mass of the reactant will be equal to the mass of products.That is 120 grams.

Hence, the correct answer option(C).

3 0
3 years ago
Read 2 more answers
A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured
Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}

Put the value into the formula

0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}

v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}

v_{3}=4.98\ m/s

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0
3 years ago
Which of these is not an even locomotor movement? Question 3 options:
antiseptic1488 [7]

Answer:

skipping

Explanation:

your welcome :)

7 0
3 years ago
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Air becomes a conductor when the electric field strength exceeds 3.01 × 106 N/C. Determine the maximum amount of charge that can
Blizzard [7]

Answer:

8.37×10⁻³ C.

Explanation:

The formula of electric field is given as

E = kq/r².......................... Equation 1

Where E = electric field, q = amount of charge, k = coulomb constant, r = radius of the sphere

make q the subject of the equation

q = Er²/k .................... Equation 2

Given: E = 3.01×10⁶ N/C, r = 5.0 m, k = 8.99×10⁹ Nm²/C².

Substitute into equation 2

q = (3.01×10⁶)(5²)/(8.99×10⁹)

q = (75.25/8.99)×10⁻³

q = 8.37×10⁻³ C.

Hence the maximum amount of charge = 8.37×10⁻³ C.

3 0
4 years ago
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