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Maksim231197 [3]
3 years ago
8

What property do atoms of these elements have that helps make the molecules polar

Chemistry
1 answer:
DENIUS [597]3 years ago
4 0
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
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Heat transfer between two substances is affected by specific heat and the chemical composition of the substances. state of matte
lesya [120]

Answer:

the answer is D

Explanation:

7 0
3 years ago
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How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
3 years ago
If the acceleration is 600 m/s/s and the mass is 300 kg, what force is acting on the object?
vampirchik [111]

Explanation:

Acceleration (a) = 600 m/s²

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Sugar is made of molecules that are bonded together based on the positively and negatively charged areas.  They will slowly dissolve in water.  Pebbles are solids.  They will sit in water for a long time.  Though shale pebbles will break apart or fall apart.
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3 years ago
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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