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skad [1K]
3 years ago
13

What is the pOH of a solution with a pH of 8? 14 6 8 0

Chemistry
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

6.

Explanation:

<em>∵ pH + pOH = 14.</em>

<em />

<em>∴ pOH = 14 - pH = 14 - 8 = 6.</em>

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Assuming the metals lose all their valence electrons and the nonmetals gain electrons to complete the s-p subshells, which listi
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Answer: Option (c) is the correct answer.

Explanation:

Atomic number of sodium is 11 and its electronic configuration is 1s^{2}2s^{2}2p^{6}3s^{1}. When sodium loses one electron then it will attain +1 charge and its electronic configuration will be as follows.

Na^{+} : 1s^{2}2s^{2}2p^{6}

Atomic number of fluorine is 9 and its electronic configuration is 1s^{2}2s^{2}2p^{5}. When fluorine gains an electron then it acquires -1 charge and its electronic configuration is as follows.

F^{-} : 1s^{2}2s^{2}2p^{6}

Atomic number of aluminium is 13 and its electronic configuration is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}. When aluminium loses its valence electrons then it acquires +3 charge and its electronic configuration is as follows.

Al^{3+} : 1s^{2}2s^{2}2p^{6}

Thus, we can conclude that the listing for aluminum is correct.

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3 years ago
Three moles of an ideal gas undergo a reversible isothermal compression at temperature 18.0 ?c. during this compression, an amou
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We calculate the entropy of an ideal gas follows:

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Which one A, B, C,D,E,F
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The component that runs the circuit is C the wire
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Read 2 more answers
The specific heats and densities of several materials are given below: Material Specific Heat (cal/g·°C) Density (g/cm3) Brick 0
abruzzese [7]

<u>Answer:</u> The change in temperature is 84.7°C

<u>Explanation:</u>

To calculate the change in temperature, we use the equation:

q=mc\Delta T

where,

q = heat absorbed = 1 kCal = 1000 Cal    (Conversion factor: 1 kCal = 1000 Cal)

m = mass of steel = 100 g

c = specific heat capacity of steel = 0.118 Cal/g.°C

\Delta T = change in temperature = ?

Putting values in above equation, we get:

1000cal=100g\times 0.118cal/g^oC\times \Delta T\\\\\Delta T=\frac{1000cal}{100g\times 0.118cal/g^oC}\\\\\Delta T=84.7^oC

Hence, the change in temperature is 84.7°C

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3 years ago
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