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3 years ago

How many molecular orbitals are present in the valence band of a sodium crystal with a mass of 7.90 g?

1 answer:

8 0

Number of atoms equal the number of molecular orbitals as each atom has 1 atomic orbital and 1 electron.

Therefore,
Number of atoms = (mass/molar mass)*Avogadro's number = (7.90/22.99)*1.36*10^23 = 4.67*10^22 atoms.

Then,
Number of molecular orbitals = 4.67*10^22

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Which description best explains why the view within the rectangle lens is different than the view outside the lens?

Viktor [21]

Answer:

A) Light waves are restricted to a single plane through the lens.

Explanation:

Took the quiz and got it right

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3 years ago

A car and a truck collide in an intersection and the merged wreck continues along. During the collision. both kinetic energy and

Vilka [71]

Answer:

C. Momentum is conserved but not kinetic energy.

Explanation:

This case represents an entirely inelastic collision, that is, a collision between the car and the truck that reduces total kinetic energy of the entire system, whereas linear momentum is conserved. Hence, correct answer is C.

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3 years ago

Which of these objects must have the greatest force acting on it? pls help fast

8_murik_8 [283]

Let's look at Newton's second law

F=ma

Force is directly proportional towards mass

If mass is more force will be more.

Between baseball and bowling ball Bowling ball has higher mass

So it would expert most force

Option D

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2 years ago

A= v50cm-v20cm<br> __________________<br> t

andrezito [222]

Answer:

.....?

Explanation:

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3 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

$F = -1.07 *10^{-8} \ N$

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

4 years ago