For the answer to the question above, I'll show the solution to my answers
moles C = 82.66 g/ 12.0111 g/mol=6.882
<span>moles H = 17.34 g/ 1.008 g/mol=17.20 </span>
<span>17.20/ 6.882 =2.5 => H </span>
<span>6.882 / 6.882 = 1 => C </span>
<span>to get whole numbers multiply by 2 </span>
<span>C2H5 ( empirical formula Molar mass = 29.062 g/mol) </span>
<span>n = pV/RT = 0.732 x 0.158 L/ 0.08206 x 298 K= 0.00473 </span>
<span>molar mass = 0.275/ 0.00473 =58.1 g/mol </span>
<span>58.1 / 29.062 = 2 </span>
<span>multiply by two the empirical formula </span>
<span>C4H10</span>
Explanation:
⁶⁵₃₀Ca → ⁶⁵₂₉Sc + ⁿₓH
The reaction above is nuclear reaction.
In a nuclear reaction, the mass number and atomic number must be conserved.
The mass number is the superscript before the atom
Atomic number is the subscript before the atom
Conserving mass number:
65 = 65 + n
n = 0
conserving atomic number:
30 = 29 + x
x = 1
The unknown atom is a positron i.e a positively charged electron: ⁰₁e
⁶⁵₃₀Ca → ⁶⁵₂₉Sc + ⁰₁e
learn more:
Transmutation brainly.com/question/3433940
#learnwithBrainly
C) MgO aka magnesium oxide
Explanation:
Many meteorites have iron or nickel in them, so they are heavier (and denser) than Earth rocks. Some meteorites have pits (regmaglypts) on the outside, which look like deep thumbprints. Meteorites are not bubbly, and do not have holes. Meteorite are usually not round.
<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H_f_%7B%28CO_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%28g%29%29%7D%29%5D-%5B%284%5Ctimes%20%5CDelta%20H_f_%7B%28CH_3NH_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28l%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%283%5Ctimes%20%28-74.8%29%29%2B%281%5Ctimes%20%28-393.5%29%29%2B%284%5Ctimes%20%28-46.1%29%29%5D-%5B%284%5Ctimes%20%28-22.97%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-138.82kJ)
Hence, the standard heat for the given reaction is -138.82 kJ
