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
The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They <span>are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.</span>
Answer:
In a chemical reaction, <u>products</u> are the substances present after the reaction
Answer:
The level of toxins in solutions
Answer:
lattice parameter = 5.3355x10^-8 cm
atomic radius = 2.3103x10^-8 cm
Explanation:
known data:
p=0.855 g/cm^3
atomic mass = 39.09 g/mol
atoms/cell = 2 atoms
Avogadro number = 6.02x10^23 atom/mol
a) the lattice parameter:
Since potassium has a cubic structure, its volume is equal to:
v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]
substituting values:
v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3
but as the cell volume is
a^3 =v
![a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7Bv%7D%3D%5Csqrt%5B3%5D%7B1.5189x10%5E%7B-22%7D%20%7D%20%3D%205.3355x10%5E-8)
cm
for a BCC structure, the atomic radius is equal to
