Answer:
We will not know immediately
Explanation:
The distance of star Betelgeuse = 325 light years
Therefore, if it where to explode, at the location, we would have a supernovae such that the luminosity of the star multiplies several multiple times
However, due to the distance of Betelgeuse from the Earth, it will take the light of the explosion, 325 years to reach Earth and the explosion as it happens will not be noticed here on Earth immediately
Answer:
s<p<q<r
Explanation:
Given
Element s displaces p from its oxide. Thus, s is more reactive that p
p reacts with cold water but element q cannot react with cold water. Thus,
p is more reactive than q
Element q is able to react with weak or dilute acids but element r in unable to react with weak or dilute acids. Thus, q is more reactive that r
The order of four elements on the basis of their reactivity in descending order is as follows
s<p<q<r
Answer:
1.66 * 10^-23 moles
Explanation:
If you follow dimensional analysis, molecules to moles is 10 molecules divided by 6.022 times 10 to the 23rd and that gives you your answer
<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
<span>Number of moles of Fe(CO)5 = mass/molar mass = (6.0 g) / (195.8955 g/mol) = 0.03 mol
Number of moles of PF3 = (4.0 g)/ (87.97 g/mol) = 0.0455 mol
Number of moles of H2 = (4.0 g) / (2.016 g/mol) = 1.984 mol
From the balanced equation
1 mol of Fe(CO)5 reacts with 2 mol of PF3 and 1 mol of H2
then 0.03 mol of Fe(CO)5 will need 0.03 (2/1) = 0.06 mol of PF3 and 0.03 mol of H2
Since we have less PF3 ( 0.0455 mol instead of 0.06 mol), that is the limiting agent</span>
