Is it inches millimeters meters ?
Https://quizlet.com/18404690/chemistry-final-flash-cards/
here is a Quizlet I have for that
Answer:
286.55K
Explanation:
To convert to kelvin , add 237 .15

Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>
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.