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2 years ago

A tin atom has 50 electrons. Electrons in the ________ subshell experience the lowest effective nuclear charge.

Physics

1 answer:

Alexus [3.1K]2 years ago

5 0

Answer:

Explanation:

We are given that a tin has 50 electrons.

We have to find in which subshell electrons experience the lowest effective nuclear charge.

We know that the electron in outermost shell experience the lowest effective nuclear charge.

Electronic configuration is given by

$50=1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^2$

Outer most sub-shell is 5p. Therefore, 5p subshell experience the lowest effective nuclear charge because the distance of 5p sub-shell is large from nucleus.

Answer: 5p

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Sergeeva-Olga [200]

Answer:

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3 years ago

1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.

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Answer:

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3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

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3 years ago

With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be

Delvig [45]

Answer:

The resistance must be 6.67 $\Omega$

Solution:

Resistance, $R_{1} = 20\Omega$

Resistance, $R_{2} = 10\Omega$

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Thus in parallel:

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}$

$R_{eq} = 6.67\ \Omega$

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3 years ago

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3 years ago