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ra1l [238]
3 years ago
8

Which type of substance is soft has a low melting point is a poor conductor of heat and electricity

Chemistry
1 answer:
stealth61 [152]3 years ago
4 0
Covalent solid, such as candle wax.
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What is the atomic weight of a hypothetical element consisting of two isotopes, one with mass = 64.23 amu (26.0%), and one with
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I think it should be A or D
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3 years ago
A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centr
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<h3><u>Answer;</u></h3>

<em><u> = 48,828.125 mi/hr²</u></em>

<h3><u>Explanation and solution</u>;</h3>
  • <em><u>Centripetal acceleration is the rate of change of angular velocity. Centripetal acceleration occurs towards the center of the circular path along the radius of the circular path</u></em>.
  • Centripetal acceleration is given by; <em>V²/r ; </em>

<em>V = 125 mi/h and r = 0.320 miles </em>

  • <em>Thus; centripetal acceleration = 125²/0.320 </em>

                                                  =15625/0.320

                                                 <em><u> = 48,828.125 mi/hr²</u></em>

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3 years ago
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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.
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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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Infrared (IR) and Nuclear Magnetic Resonance (NMR) are two spectroscopic techniques you've encountered in organic chemistry I, C
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Answer:

The solution to this question can be defined as follows:

Explanation:

Please find the attached file for the solution:

8 0
3 years ago
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