For the gas phase decomposition of phosphine at 120 °c, the rate of the reaction is determined by measuring the appearance of h2. 4 ph3(g)p4(g) + 6 h2(g) at the beginning of the reaction, the concentration of h2 is 0 m. after 93.0 s the concentration has increased to 0.101 m. what is the rate of the reaction? (mol h2/l) /s

Rate = 1.09*10^-3 (mol H2/L)/s

Explanation:

Given:

Initial concentration of H2, C1 = 0 M

Final concentration of H2, C2 = 0.101 M

Time taken, t = 93.0 s

To determine:

The rate of the given reaction

Calculation:

The decomposition of PH3 is represented by the following chemical reaction

Reaction rate in terms of the appearance of H2 is given as:

Here C1(H2) = 0 M and C2(H2) = 0.101 M

Δt = 93.0 s

Since molarity M = mole/L

rate = 1.09*10^-3 (mol H2/L)/s

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

the answer is   c.) 1575 j.

explanation: