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Debora [2.8K]
4 years ago
11

Use ionic lewis structure to predict which of these two ionic solids has the higher melting point. use ionic lewis structure to

predict which of these two ionic solids has the higher melting point. kf cao
Physics
2 answers:
son4ous [18]4 years ago
8 0
Best Answer<span>: </span>CaO<span> - the </span>attraction between<span> a </span>2<span>+ and a </span>2<span>- </span>ion<span> is </span>much stronger than between<span> a + and a -. Source(s):. Gervald F · 1 ... </span>yeah<span> i </span>think CaO would have<span> the </span>strongest bond<span>, and </span>because<span> of that </span>im guessing<span> that it </span>would have<span> a </span>higher melting point because<span> it </span>has<span> a </span><span>stronger network structure</span>
Mandarinka [93]4 years ago
3 0

Answer:

CaO

Explanation:

As lewis structure tells us about the charge on the ions in the lattice So by using it we can predict the balanced lattice of CaO And KF after proceeding we will find that CaO has the Ionic Bond of +2 and -2 which is greater than the ionic bond of KF which is +1 and -1 so greater the charge more the attraction and more the attraction more powerful bond becomes so after all of this we can say that Having higher bond strength leads to the much higher melting and boiling points.

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A block with a mass of 1kg moving at a velocity of 3m/s collides and sticks to a block of mass of 4kg initially at rest. What is
Vsevolod [243]

Answer:using Newton third law

Let initial velocity of block be u1=3m/s

Mass of moving block m1 =1kg

Final velocity of block =V

Mass of stationary block m2= 4kg

Since they stick together, their final velocity will be the same.

m1u1 + m2u2=(m1+m2)v

(1*3)+(0*4)=(1+4)v

3=5v

Divide both sides by 5

V=0.6

Final velocity is 0.6m/s

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3 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

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