Answer:
Man-made resources
Explanation:
Humanized resources are items or substances that have value to human lives that do not occur in the natural world. Examples of man-made resources include plastic, paper, soda, sheet metal, rubber and brass. These contrast with natural resources, such as water, crops, sunlight, crude oil, wood and gold.
Answer:
Two electrons each!
Explanation:
The question pretty much requires us to find the oxidation number of Zinc in the compound.
Zn3P2
Following oxidation number rules;
O.N of Zn3P2 = 0
.Since phosphorus has valence of 5, it needs three more electrons to achieve its octet state. Hence;
Oxidation number of P = -3
Let oxidation number of Zn = x
We have;
3x + 2 (-3) = 0
3x + (-6) = 0
3x = 6
x = 6/3 = 2
This means each zn electrons lost 2 electrons.
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.
200 is the answer let me know if you need anything
Answer:
857mL de la solución al 7%
Explanation:
Una solución de agua oxigenada (H₂O₂) al 7% contiene 7mL de peróxido de hidrógeno por cada 100mL de solución = 7mL H₂O₂ / 100mL Sln
Para preparar un litro = 1000mL de disolución al 6% de peróxido de hidrógeno se requieren:
1000mL Sln * (6mL H₂O₂ / 100mL Sln) = 60mL H₂O₂
Para obtener 60mL de H₂O₂ a partir de la solución al 6% se requieren:
60mL H₂O₂ * (100mL Sln / 7mL H₂O₂) =
<h3>857mL de la solución al 7%</h3>
