Answer:
Density of the He atom = 12.69 g/cm³
Explanation:
From the information given:
Since 1 mole of an atom = 6.022x 10²³ atoms)
1 atom of He = 
The volume can be determined as folows:
since the diameter of the He atom is approximately 0.10 nm
the radius of the He = 
= 0.05 nm
Converting it into cm, we have:
Assuming that it is a sphere, the volume of a sphere is
= 
= 
= 
Finally, the density can be calcuated by using the formula :
D = 12.69 g/cm³
Density of the He atom = 12.69 g/cm³
Answer:
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1 0 0 1s 1 2 2
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2 0 0 2s 1 2
2 1 1,0,-1 2p 3 6 8
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3 0 0 3s 1 2
3 1 1,0,-1 3p 3 6
3 2 2,1,0,-1,-2 3d 5 10 18
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4 0 0 4s 1 2
4 1 1,0,-1 4p 3 6
4 2 2,1,0,-1,-2 4d 5 10
4 3 3,2,1,0,-1,-2,-3 4f 7 14 32
Explanation:
The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
<h3>Weight of one gallon of water</h3>
The weight of 1 gal of water is given as 3785 g
Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g
<h3>Volume of the water in cubic meters</h3>
Volume = mass/density
Volume = 3.2 x 10¹² g/1 gmL
Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³
Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
Learn more about volume here: brainly.com/question/1972490
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I’m not sure what the answer is but I hope someone can help you
Answer:
A
Explanation:
Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:
Therefore, from the chemical equation, we have that:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%3D%20%5Cleft%5B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24O%7D%20%20%5Cright%5D%20%20%20-%5Cleft%5B3%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24%7D%2B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24O%7D%5Cright%5D%20%5Cend%7Baligned%7D)
Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%26%20%3D%20%5Cleft%5B%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%28-285.8%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20-%5Cleft%5B%203%280%29%20%2B%20%2882.1%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20%5C%5C%20%5C%5C%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%26%20%3D%20%28-317%20%2B%20285.8%20%2B%2082.1%29%5Ctext%7B%20kJ%2Fmol%7D%20%5C%5C%20%5C%5C%20%26%20%3D%2050.9%5Ctext%7B%20kJ%2Fmol%7D%20%5Cend%7Baligned%7D)
In conclusion, our answer is A.
