Question

The standard heat of combustion of solid boron is equal to

Answer 1

I think its 100° C, Hope its correct.........

Answer 2

ANSWER:

The standard heat of combustion of boron is equal to $\frac{1}{2} \Delta H^{\circ} f\left(B_{2} O_{3}\right).$

EXPLANATION:

$4 B+3 O_{2} \rightarrow 2 B_{2} O_{3}$

or $B+\frac{3}{4} O_{2} \rightarrow \frac{1}{2} B_{2} O_{3}$

∴$\Delta H^{\circ} \mathrm{COMb}(\mathrm{B})=\frac{1}{2} \Delta H^{\circ} f\left(B_{2} O_{3}\right)$

The potential taken between two point across a resistor will be the product of the current flowing through it and its resistance.

"Ohm's law states" that the product of the "current flowing" through a resistor and its resistance is the "potential difference" across the resistor. The formula is "V = I × R".  

Ohm's law can also be defined as J = sigma × E, where "J is the current density" in the conductor, sigma is the conductivity and E is the electric field in the conductor.