The energy would be 3.31*10-19J
M{(NH₄)₂Cr₂O₇}=2A(N)+8A(H)+2A(Cr)+7A(O)
M{(NH₄)₂Cr₂O₇}=252.065 g/mol
M(Cr)=51.996 g/mol
m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}=m(Cr)/2M(Cr)
m(Cr)=2M(Cr)m{(NH₄)₂Cr₂O₇}/M{(NH₄)₂Cr₂O₇}
m(Cr)=2*51.996*35.8/252.065=14.770 g
m(Cr)=14.770 g
<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
A I think but idrk I just guess sorry if you get it wrong cause of me just be careful I could be wrong
Answer:
Option B. 4.25×10¯¹⁹ J
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 6.42×10¹⁴ Hz
Energy (E) =?
Energy and frequency are related by the following equation:
Energy (E) = Planck's constant (h) × frequency (f)
E = hf
With the above formula, we can obtain the energy of the photon as follow:
Frequency (f) = 6.42×10¹⁴ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =?
E = hf
E = 6.63×10¯³⁴ × 6.42×10¹⁴
E = 4.25×10¯¹⁹ J
Thus, the energy of the photon is 4.25×10¯¹⁹ J
