I don’t think I could answer this sorry.........
Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:

Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:

78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
Answer:
c) a readily corroded silvery solid that fizzes with water ... (e) an orange high melting solid and good electrical conductor. (f) soft low melting ...
Explanation:
Answer:

Explanation:
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is:

CaCO₃(s) ⟶ CaO(s) + CO₂(g)
ΔH°f/kJ·mol⁻¹: -1207.1 -157.3 -393.5
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-157.3 + (-393.5)] - (-1207.1)\\& = & -550.8 +1207.1\\& = & \textbf{656.3 kJ/mol}\\\end{array}\\\\\text{The enthalpy of decomposition is } \boxed{\textbf{656.3 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B-157.3%20%2B%20%28-393.5%29%5D%20-%20%28-1207.1%29%5C%5C%26%20%3D%20%26%20-550.8%20%2B1207.1%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B656.3%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5C%5C%5Ctext%7BThe%20enthalpy%20of%20decomposition%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B656.3%20kJ%2Fmol%7D%7D)