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Marianna [84]
3 years ago
7

What is the maximum number of bonds that can form in a molecule with a central atom containing five electrons in sp 3 hybrid orb

itals?
Physics
1 answer:
nikitadnepr [17]3 years ago
7 0
2 single bonds form <span>in a molecule with a central atom containing five electrons in sp 3 hybrid orbitals.</span>
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disa [49]

displacement of the plane is given as

d = 1995 miles

time taken by the plane

t = 5.00 hours

now the velocity is given as

v = \frac{displacement}{time}

v = \frac{1995}{5}

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4 0
3 years ago
A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy
SashulF [63]

Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

Hence,Kp/Ktot is 5.444x10^(-4)

4 1
3 years ago
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Answer:

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Explanation:

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Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

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3 years ago
Which of the following is NOT a symptom associated with hypertension?
zepelin [54]
I would say a or d hope this helps haha ( also if you could do brainliest for this that would be AMASINF bc I really want to rank up, u don’t have to tho)
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