All categories

3 years ago

How many molecules are in 0.26 mol of CO2

1 answer:

7 0

<span>1 mole of CO2 = 6.023*10^23 molecules.3 moles of CO2 = 3*6.023*10^23 moleculestherefore, 3 moles of CO2 = 18069*10^20 molecules.
</span>

You might be interested in

Convert 6.42 moles of H2O to molecules

Ivahew [28]

Answer:

3.87 x 10^24

Explanation:

Simply multiply the moles by avogadros number

6.42 moles of H2O x 6.022 x 10^23 molecules/1 mole of H2O = 3.87 x 10^24 molecules of H2O

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20%2B%20%20%5Cfrac%7B2%7D%7B5%7D%20" id="TexFormula1" title=" \frac{1

Travka [436]

3/4 cause its addition

8 0

3 years ago

Which of the following would label the boxes correctly?

choli [55]

The last one 1) exothermic; 2) exothermic

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

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.

7 0

3 years ago

If 2.0 g of water at 0.00°c is to be vaporized, how much heat must be added to it? the specific heat of water is 1.0 cal/g·k, it

krok68 [10]

45 hope it helps hope you get a good grade

6 0

3 years ago