Answer:Water absorbs different wavelengths of light to different degrees. The longest wavelengths, with the lowest energy, are absorbed first. Red is the first to be absorbed, followed by orange & yellow. The colors disappear underwater in the same order as they appear in the color spectrum. Even water at 5ft depth will have a noticeable loss of red. For this reason, strobes are usually used to add color back to subjects.
Explanation:
<u>Answer:</u> The amount of energy released per gram of 
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of produces -9078.57 kJ of energy.
Or,
If 
of produces -9078.57 kJ of energy
Then, 1 gram of will produce = 
of energy.
Hence, the amount of energy released per gram of is -71.92 kJ
find mol of N2 present using gas law equation
PV = nRT
P = pressure = 688/760 = 0.905 atm.
V = 100mL = 0.1L
n = ???
R = 0.082057
T = 565+273 = 838
Substitute:
0.905*0.1 = n*0.082057*838
n = 0.0905 / 68.76
n = 0.00132 mol N2
Molar mass N2 = 28 g/mol
0.00132 mol = 0.00132*28 = 0.037g N2 gas
Answer:
Halogen / salt-former
Explanation:
Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".
