Answer: I can suck you di ck
Explanation:
Answer:
9.62 μm
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 31.2 THz
Wavelength (λ) =..?
Next, we shall convert 31.2 THz to Hz.
This can be obtained as follow:
Recall:
1 THz = 1×10¹² Hz
Therefore,
31.2 THz = 31.2 THz × 1×10¹² Hz / 1 THz
31.2 THz = 3.12×10¹³ Hz
Therefore, 31.2 THz is equivalent to 3.12×10¹³ Hz.
Finally, we shall determine the wavelength (λ) infrared radiation as follow:
Frequency (f) = 3.12×10¹³ Hz.
Velocity (v) = 3×10⁸ m/s
Wavelength (λ) =..?
V = λf
3×10⁸ = λ × 3.12×10¹³
Divide both side by 3.12×10¹³
λ = 3×10⁸ / 3.12×10¹³
λ = 9.62×10¯⁶ m
Converting 9.62×10¯⁶ m to micro metre (μm) we have:
1 m = 1×10⁶ μm
Therefore,
9.62×10¯⁶ m = 9.62×10¯⁶ m × 1×10⁶ μm / 1 m
9.62×10¯⁶ m = 9.62 μm
Therefore, the wavelength of the infrared radiation is 9.62 μm
Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
Answer:
Correct option is
A
Explanation:
The number of molecules of CO involves in the slowest step will be 0 because CO is not involve in the slowest step i.e. rate determing step.
Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
