b) three
this is because all integers are sig figs, and all numbers between integers are sig figs. This makes the 40,5 part of 40,500 significant. Place holder zeroes that are not after a decimal are not significant, so the last two zeroes of the number are not significant.
Answer:
protons only
Explanation:
proton is present in the nucleus of atom as sub atomic particle ..
proton has + charge in the it's top as p+..
Ok thanks for the valuble info.
<u>Answer:</u> The percent yield of the compound is 30.86 %.
<u>Explanation:</u>
To calculate the percentage yield of a compound, we use the equation:
Experimental yield of compound = 25 g
Theoretical yield of compound = 81 g
Putting values in above equation, we get:
Hence, the percent yield of the compound is 30.86 %.
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
