Answer : The correct option is, (C) 0.675 M
Explanation :
Using neutralization law,

where,
= concentration of
= 13.5 M
= concentration of diluted solution = ?
= volume of
= 25.0 ml = 0.0250 L
conversion used : (1 L = 1000 mL)
= volume of diluted solution = 0.500 L
Now put all the given values in the above law, we get the concentration of the diluted solution.


Therefore, the concentration of the diluted solution is 0.675 M
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
Answer:
The process or result of oxidizing or being oxidized.(Rust)
Explanation:
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