Answer:
Letter B.
Explanation:
Changing its direction and climbing a steep incline can't be proven only by the graph info. You can see that it's speed isn't constant in the graph.
We have letter B left.
Answer:
It is expressed as a multiple of one-twelfth the mass of the carbon-12 atom, 1.992646547 × 10−23 gram, which is assigned an atomic mass of 12 units. ... In this scale 1 atomic mass unit (amu) corresponds to 1.660539040 × 10−24 gram.
Answer:
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Explanation:
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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:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
<em></em>
∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>