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Mademuasel [1]
2 years ago
10

Si se tiene 1mol de NO . cuantas moleculas de NO hay

Chemistry
1 answer:
kumpel [21]2 years ago
3 0

if you have 1mol of NO. how many molecules of NO are there

Answer:

6.02 x 10²³ molecules

Explanation:

Given parameters:

Number of moles of NO = 1 mole

Unknown:

Number of molecules in NO;

Solution:

A mole of compound contains the Avogadro's number of particles.

  1 mole of a substance contains 6.02 x 10²³ molecules

So, 1 mole of NO contained 6.02 x 10²³ molecules

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Answer:

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A composition of reflections over parallel lines is the same as a __________.
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Answer : Option A) Translation


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We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.

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When water freezes into ice, sme of the properties have changed <br> What stays the same
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When water freezes into ice, some of the properties have change. What stays the same.




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Have a goood day!
5 0
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

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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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