Answer: 362,07 cm3
To answer this question you need to convert the lb into gram first. One lb equal to 453.592g, so: 3.6lb x 453.592gram/lb= 1632.9312gram.
Now we have mass(1632.9312g), density (4.51g/cm3). Volume is mass divided by density. The equation would be:
Volume= mass/density
Volume = 1632.9312gram / (4.51g/cm3)= 362,07 cm3
Answer:
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In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.
Explanation:
I think the correct answer from the choices listed above is option A. The three components of air are all <span>classified as pure substances since they are not chemically bonded so they can be separated by certain processes and be present as a pure substance. Hope this answers the question.</span>
Answer:
d. 8 moles of H2O on the product side
Explanation:
Hello,
In this case, we need to balance the given redox reaction in acidic media as shown below:
![MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\](https://tex.z-dn.net/?f=MnO_4%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%20%2B%20Cl%5E%7B1-%7D%20%28aq%29%20%5Crightarrow%20%20Mn%5E%7B2%2B%7D%20%28aq%29%20%2B%20Cl_2%20%28g%29%5C%5C%5C%5C%5C%5C%5C%5C%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5C%5C%5C%5C2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5C%5C%5C%5C2%2A%5B%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B8H%5E%2B%2B5e%5E-%20%5Crightarrow%20Mn%5E%7B2%2B%7D%2B4H_2O%5D%5C%5C%5C%5C5%2A%5B2Cl%5E%7B1-%7D%5Crightarrow%20Cl_2%5E0%2B2e%5E-%5D%5C%5C%5C%5C%5C%5C%5C%5C2%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B16H%5E%2B%2B10e%5E-%20%5Crightarrow%202Mn%5E%7B2%2B%7D%2B8H_2O%5C%5C%5C%5C10Cl%5E%7B1-%7D%5Crightarrow%205Cl_2%5E0%2B10e%5E-%5C%5C)
Then, we add the half reactions:
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.
