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Nezavi [6.7K]
3 years ago
10

Bromine atoms usually end up with how many valence electrons?

Chemistry
1 answer:
Nastasia [14]3 years ago
8 0
Bromine atoms usually end up with SEVEN valence electrons.

Bromine belongs to group 17 i.e. halogen group. Its atomic number is 35. The electron configuration of Bromine is 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁵. In the last shell, we see 4s²4p⁵, thus 2+5 = 7 valence electrons are present.
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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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M
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m
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n
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=
M
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m
n

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(
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g
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g
m
o
l
−
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, you can use this formula to determine (C3H8)
n
(
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