Answer: Option (c) is the correct answer.
Explanation:
The bottom of pencil is placed at the starting point of scale. Whereas the tip of pencil depicts the end point of its length.
The bottom of pencil is at 0 mm and tip of pencil is at 18.73 mm. The appropriate amount of significant figures is 18.73 mm.
Therefore, we can conclude that out of the given options, pencil is 18.73 mm long.
(a)- Time
(b)- Heat produced(i guess)
(c)- Material
this is what I think, hope it helps
Answer:
162 g Fe₂O₃
Explanation:
To find the mass of Fe₂O₃, you need to (1) convert grams C to moles C (via molar mass from periodic table), then (2) convert moles C to moles Fe₂O₃ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe₂O₃ to grams (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.
Molar Mass (C): 12.011 g/mol
2 Fe₂O₃(s) + 3 C(s) ---> 4 Fe(s) + 3 CO₂(g)
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
18.3 g C 1 mole 2 moles Fe₂O₃ 159.684 g
-------------- x ---------------- x ------------------------- x ----------------- = 162 g Fe₂O₃
12.011 g 3 moles C 1 mole
<span>Wave A will have a higher pitch than wave B.. This is not true becasue b has the higher PITCH becasue it's closer together.
Wave B will have a lower pitch than wave A... This basically A but worded differently.
Wave A will have a louder sound than wave B... This is CORRECT becasue it's louder the waver are bigger, so it's louder. But it has a lower pitch, 2 different things.
Wave B will have a louder sound than wave A. False..</span><span />
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
