Answer:
0.0250 g
Explanation:
Step 1: Determine the molar mass of Vitamin C.
The molar mass is the mass in grams corresponding to 1 mole. In order to calculate the molar mass of vitamin C (C₆H₈O₆) we need to add the molar masses of the elements that compose it.
M(C₆H₈O₆) = 6 × M(C) + 8 × M(H) + 6 × M(O)
M(C₆H₈O₆) = 6 × 12.01 g/mol + 8 × 1.01 g/mol + 6 × 16.00 g/mol
M(C₆H₈O₆) = 176.14 g/mol
Step 2: Calculate the mass corresponding to 0.000142 mol of vitamin C.
![0.000142 mol \times \frac{176.14g}{mol} =0.0250 g](https://tex.z-dn.net/?f=0.000142%20mol%20%5Ctimes%20%5Cfrac%7B176.14g%7D%7Bmol%7D%20%3D0.0250%20g)
Answer:
water is H2O having different structure than alcohol R-OH which means they have different properties too.
Explanation:
In water one oxygen atom is covalently bound with two hydrogen atoms while alcohol is an organic compound having Oh group attached to the carbon chain.
Other than liquid water can occur in solid form that is ice and in gaseous form that is vapors too while alcohol only present in liquid form.
heat of evaporation of alcohol is lower than water means water need more heat to evaporate than alcohol that is why we can say alcohol having more cooling effect than water.
Answer:
Wave energy is the transport and capture of energy by ocean surface waves. The energy captured is then used for all different kinds of useful work, including electricity generation, water desalination, and pumping of water.
Explanation:
Answer:
here's is ur answer in attachment
Explanation:
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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:
![2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0](https://tex.z-dn.net/?f=2%28Mn%5E%7B7%2B%7DO%5E%7B2-%7D_4%29%5E%7B1-%7D%20%28aq%29%2B16H%5E%2B%2B10Cl%5E%7B1-%7D%20%5Crightarrow%202Mn%5E%7B2%2B%7D%2B8H_2O%2B5Cl_2%5E0)
Thereby, we can see d. 8 moles of H2O on the product side.
Best regards.