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3 years ago

6

What picture of the electron is suggested by the quantum-mechanical model of the hydrogen atom? a. a raisin in pudding c. a plan

etary orbiting body b. a probability cloud d. a light quantum

1 answer:

Ivanshal [37]3 years ago

3 0

D. a light quantum this is the answer .

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The element radon is at the opposite end of the range, with the lowest specific heat of all naturally occurring elements. Radon'

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Explanation:

Q = mc∆T

= (0.34 kg)(94 J/kg-°C)(25°C)

= 799 J

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3 years ago

An atom of a certain element has 36 protons, 36 electrons, and a mass number of 84. At room temperature, this element is a very

Grace [21]

Answer:

12 neutrons

Explanation:

The number of protons also shows the atomic number. Therefore the element in question is Krypton (Kr), which also is a noble gas.

Neutrons = Mass number - protons - electrons

Here neutrons = 84 - 36 - 36 = 12

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3 years ago

Higher levels of co2 are more likely to be found where on the earth? question 15 options: uniformly around the globe as winds bl

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Uniformly around the globe. it is mostly found in earths atmosphere.

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2 years ago

Both formal and informal

Tema [17]

Explanation:

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2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago