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Sholpan [36]
3 years ago
9

Calculate the number of molecules in 120 grams of water

Chemistry
1 answer:
marissa [1.9K]3 years ago
4 0

7.02x<u><em>10^{24}</em></u>

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Why didn't mendeleev use atomic number to arrange the elements?
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The primary reason for this was that Mendeleev didn't know that atomic numbers actually existed. Atomic numbers were only discovered a period after Mendeleev's time. The use of X-rays made it possible to find the atomic number, and those had not been discovered yet. <span>
<span>The periodic table was then arranged in 1913 by Henry Moseley in an arrangement according to atomic number.</span></span>

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How do the metals in group 1 compare with the transition metals?
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Group 1 metals and transition metals are different from each other, mainly based on the colour of the chemical compounds that they form. The key difference between group 1 metals and transition metals is that the group 1 metals form colourless compounds, whereas the transition metals form colourful compounds.

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3 years ago
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How many valence electron does sulfur
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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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6. How are Fisher projections used to represent three-dimensional perspective?
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