Low life?- i’m pretty sure
Answer:
Explanation:
1. Miles travelled in an average month
![\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}](https://tex.z-dn.net/?f=%5Ctext%7BMiles%7D%20%3D%20%5Ctext%7B1%20mo%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B52%20wk%7D%7D%7B%5Ctext%7B12%20mo%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B5%20da%7D%7D%7B%5Ctext%7B1%20wk%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B16%20mi%7D%7D%7B%5Ctext%7B1%20da%7D%7D%20%3D%20%5Ctext%7B347%20mi%7D)
2. Using a gasoline powered vehicle
(a) Moles of heptane used
(b) Equation for combustion
C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O
(c) Moles of CO₂ formed
(d) Volume of CO₂ formed
At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.
3. Using an electric vehicle
(a) Theoretical energy used
![\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}](https://tex.z-dn.net/?f=%5Ctext%7BTheor.%20Energy%7D%20%3D%20%5Ctext%7B347%20mi%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20kWh%7D%7D%7B%5Ctext%7B5.2%20mi%7D%7D%20%3D%20%5Ctext%7B66.7%20kWh%20theor.%7D)
(b) Actual energy used
The power station is only 85 % efficient.
![\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\](https://tex.z-dn.net/?f=%5Ctext%7BActual%20energy%20used%7D%20%3D%20%5Ctext%7B66.7%20kWh%20theor.%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B100%20kWh%20actual%7D%7D%7B%5Ctext%7B%2085%20kWh%20theor.%7D%7D%5Ctimes%20%5Cdfrac%7B%5Ctext%7B3600%20kJ%7D%7D%7B%5Ctext%7B1%20kWh%7D%7D%5C%5C%5C%5C%20%3D%202.82%5Ctimes%2010%5E%7B5%7D%20%5Ctext%7B%20kJ%7D%5C)
(c) Combustion of CH₄
CH₄ + 2O₂ ⟶ CO₂ +2 H₂O
(d) Equivalent volume of CO₂
The heat of combustion of methane is -802.3 kJ·mol⁻¹
![V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}](https://tex.z-dn.net/?f=V%3D%202.82%5Ctimes%2010%5E%7B5%7D%5Ctext%7B%20kJ%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20methane%7D%7D%7B%5Ctext%7B802.3%20kJ%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20CO%24_%7B2%7D%24%7D%20%7D%7B%5Ctext%7B1%20mol%20methane%7D%7D%20%5Ctimes%20%5Cdfrac%7B%20%5Ctext%7B24.0%20L%7D%7D%7B%20%5Ctext%7B1%20mol%20CO%24_%7B2%7D%24%7D%7D%5C%5C%5C%5C%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20gal%7D%7D%7B%5Ctext%7B3.875%20L%7D%7D%20%3D%20%5Ctextbf%7B2180%20gal%7D)
4. Comparison
![\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7B%5Ctext%7Bgasoline%7D%7D%7D%7BV_%7B%5Ctext%7Belectric%7D%7D%7D%20%3D%20%5Cdfrac%7B10800%7D%7B2180%7D%20%3D%205.0%5C%5C%5C%5C%20%5Ctext%7BAn%20gasoline-powered%20vehicle%20has%20%7D%5C%5C%20%5Cboxed%7B%5Ctextbf%7B%20five%20times%20the%20carbon%20footprint%7D%7D%5Ctext%7B%20of%20an%20electric%20vehicle.%7D)
Answer:
The specie which is oxidized is:- ![CrO_2^-](https://tex.z-dn.net/?f=CrO_2%5E-)
The specie which is reduced is:- ![ClO^-](https://tex.z-dn.net/?f=ClO%5E-)
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
For the given chemical reaction:
The half cell reactions for the above reaction follows:
Oxidation half reaction: ![CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-](https://tex.z-dn.net/?f=CrO_2%5E-%20%2B%202H_2O%20%2B%204OH%5E-%5Crightarrow%20CrO_4%5E%7B2-%7D%20%2B%204H_2O%20%2B%203e%5E-)
Reduction half reaction: ![2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-](https://tex.z-dn.net/?f=2ClO%5E-%20%2B%204H_2O%20%2B%202e%5E-%5Crightarrow%20Cl_2%20%2B%202H_2O%20%2B%204OH%5E-)
Thus, the specie which is oxidized is:- ![CrO_2^-](https://tex.z-dn.net/?f=CrO_2%5E-)
The specie which is reduced is:- ![ClO^-](https://tex.z-dn.net/?f=ClO%5E-)
1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO
solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16
24.6 g is the answer.
if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.
Answer:
<h3>no it is not allowed</h3>
Explanation:
<h3>Liwis structure shows the elements symbol with dots thet represents valance electrons ; in second row elements their atomic number is 3 up to 10 , from Li up to Ne from their electron configuration their valance electron will be from 1 up to 8 respectivelly ,if lewis structure represents the element with it is symbol and dots that represents valance electron the second row elements cannot have more than an octet of valance electrons surrounding it.</h3>
<h3>I think it is help ful for you </h3>