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4 years ago

Does the lattice energy of an ionic solid increase or decrease as the charges or sizes of the ions increase?

Physics

1 answer:

alexandr1967 [171]4 years ago

7 0

Answer:

Explanation:

Lattice energy is the energy required to separate one mole of an ionic solid compound into its components gaseous cations and anions.

Due to increase in size of the ions, the lattice energy decreases while the lattice energy increases as the charge of the ions increases.

When the size increase, the distance between the nuclei also increase leading a decrease the force of attraction between the nuclei

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During which period did the first large herbivores and carnivores appear?

natali 33 [55]

The first large herbivores and carnivores appeared during the Mesozoic Era, which is divided into three periods-- Triassic, Jurassic and Cretaceous. The herbivores and and carnivores increased in size during the Jurassic Period. And some of the largest dinosaurs emerged during this period as well.

7 0

4 years ago

(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0

Sergio039 [100]

<h2>Answer:</h2>

(a) 6.95 x 10⁻⁸ C

(b) 6.25N/C

<h2>Explanation:</h2>

The electric field (E) on a point charge, Q, is given by;

E = k x Q / r² ---------------(i)

Where;

k = constant = 8.99 x 10⁹ N m²/C²

r = distance of the charge from a reference point.

Given from the question;

E = 10000N/C

r = 0.250m

Substitute these values into equation(i) as follows;

10000 = 8.99 x 10⁹ x Q / (0.25)²

10000 = 8.99 x 10⁹ x Q / (0.0625)

10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000/(143.84 x 10⁹)

Q = 0.00695 x 10⁻⁵C

Q = 6.95 x 10⁻⁸ C

The magnitude of the charge is 6.95 x 10⁻⁸ C

(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;

E = k x Q / r²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²

E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100

E = 6.25N/C

Therefore, at 10.0m, the electric field will be just 6.25N/C

3 0

3 years ago

Describing motion with velocity-time graphs.

irina1246 [14]

This link should help you out!!

http://mszopiak.weebly.com/uploads/5/5/8/1/55815541/p_vtgraphskey_mszopiak.pdf

5 0

3 years ago

5. A construction worker on a high-rise building is on a platform suspended between two cables as illustrated below. The constru

Natalka [10]

Answer:

a) Tc = 750 [N] ;b) See the explanation below.

Explanation:

To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.

The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.

In this way we can find by means of a sum of forces on the y axis equal to zero:

- 850 - 450 + 550 + Tc = 0

Tc = 750 [N]

4 0

3 years ago

The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

payback period = $\frac{32 + 66}{520.34}$

payback period = 0.188 year

8 0

4 years ago