Answer:
The atomic mass of phosphorus is 29.864 amu.
Explanation:
Given data:
Atomic mass of phosphorus = ?
Percent abundance of P-29 = 35.5%
percent abundance of P-30 = 42.6%
Percent abundance of P-31 = 21.9%
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100
Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100
Average atomic mass = 1029.5 + 1278 + 678.9/ 100
Average atomic mass = 2986.4 / 100
Average atomic mass = 29.864 amu.
The atomic mass of phosphorus is 29.864 amu.
Answer:
carbon
Explanation:
tbh im not sure just guessing
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
Answer:
V = 4/3 * 3.1416 * (37x10-10)3
V = 2.12x10-25 cm3
d = m/V
d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3
The difference in temperature, let's convert F to ºC:
ºC = -80-32/1.8 = -62.22 ºC
dT = -92.6 + 62.2 = -30.4 ºC
Answer:
4 4 8 4 is the balanced equation
