For Ethanol:
D = m / V
D = 3.9 g / 5 mL
D = 0.78 g/mL
For benzene:
D = 4.4 g / 5 mL
D = 0.88 g/mL
benzene has a <span>higher density.
</span>
hope this helps!
I think this is what you're after:
Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1]
Convert to J and divide by the Avogadro Const to give E in J per photon
E = 375700/6.022×10^23 = 6.239×10^-19 J
Plank relationship E = h×ν E in J ν = frequency (Hz s-1)
Planck constant h = 6.626×10^-34 J s
6.239×10^-19 = (6.626×10^-34)ν
ν = 9.42×10^14 s^-1 (Hz)
IP are usually given in ev Cs 3.894 eV
<span>E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon </span>
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Ah is gold and Rh is rhodium and Os is osmium and Ir iridium
Answer:
B. It would lead to an increase of global temperatures.
Explanation:
Gas burns when it is hot. Therefore the more gas that our cars consume, the hotter the air will be when the car releases those gas vapors. Causing the air to become hotter.
Hope this helped <3 :) brainliest?
