(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻
k=2 k=1
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>
Answer: 40.3 L
Explanation:
To calculate the moles :
According to stoichiometry :
1 moles of 
produces = 3 moles of 
Thus 0.600 moles of will produce=
of
Volume of 
Thus 40.3 L of CO is produced.
Answer:
your answer will be C
Explanation:
remember low souunds carry sound waves farther than high pithed sounds.
