Molecular structure. Isomers have the same formula, so there is the same elements in both. But they have a different molecular structure and when drawn they look different
Answer:
2812.6 g of H₂SO₄
Explanation:
From the question given above, the following data were obtained:
Mole of H₂SO₄ = 28.7 moles
Mass of H₂SO₄ =?
Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:
Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)
= 2 + 32 + 64
= 98 g/mol
Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:
Mole of H₂SO₄ = 28.7 moles
Molar mass of H₂SO₄ =
Mass of H₂SO₄ =?
Mole = mass / Molar mass
28.7 = Mass of H₂SO₄ / 98
Cross multiply
Mass of H₂SO₄ = 28.7 × 98
Mass of H₂SO₄ = 2812.6 g
Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄
Answer:
Bonds basically differs with each other due to sharing of electrons .
Explanation:
There are majorly three kinds of bonds
1. Ionic bonds which forms due to an element donate an electron to another element completely .
2. covalent bonds which forms with the mutual sharing of electrons b/w two atoms .
3. metallic bonds which forms b/w metals & they share electrons due to electron negativity difference b/w two atoms or elements
Answer:
please mark me as brainlest
Explanation:
carbon fluoride is the answers
<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
