Answer: The standard enthalpy of formation of 
is -252.1 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_{reactantss}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactantss%7D%5D)
Putting the values we get :
![\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%5Ctimes%20H_f%7BFe%7D%2B3%5Ctimes%20H_f%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f%7BFe_2O_3%7D%2B3%5Ctimes%20H_f%7BH_2%7D%5D)
![67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]](https://tex.z-dn.net/?f=67.9kJ%3D%5B%282%5Ctimes%200%29%2B%283%5Ctimes%20H_f%7BH_2O%7D%29%5D-%5B%281%5Ctimes%20-824.2kJ%2Fmol%29%2B3%5Ctimes%200kJ%2Fmol%29%5D)
Thus standard enthalpy of formation of is -252.1 kJ/mol.
Answer:
Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.
Explanation:
Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:
→ 
I hope this explanation is clear and explanatory.
Answer: The answer is 6. Or A on USATestprep.
Explanation: The sum of the coefficients of the balanced chemical equation is 6. The coefficients are the numbers in front of the compounds in the reaction equation.
<span>The representative particle for silicon is atom of silicon.
</span>Representative particles can be atoms, molecules, formula units or ions. Representative particles depend on the nature of the substance. Silicon is chemical element made of atoms.
For sodium chloride (NaCl) for example, representative particles are ions.
Answer:
A. 1.54 mole.
B. 55.39g of carbon
Explanation:
A. Determination of the number of mole in 67.7g of C3H8.
Mass of C3H8 = 67.7g
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Number of mole of C3H8 =..?
Number of mole = Mass/Molar Mass
Number of mole of C3H8 = 67.7/44
Number of mole of C3H8 = 1.54 mole
B. Determination of the mass of carbon in the compound.
This is illustrated below:
The mass of C in compound can be obtained as follow:
=> 3C/C3H8 x 67. 7
=> 3x12 / 44 x 67.7
=> 36/44 x 67.7
=> 55.39g
Therefore, 55.39g of carbon is present in the compound.
