To balance this equation, first we should consider balancing C because it only presents in one reactant and one product. Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2. Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O. So the total number of O in the products is 12+3 = 15. O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
The answer is: a relatively large electron cloud.
Atom is composed of the nucleus and the electron cloud.
Protons (positive particles) and neutrons (neutral particles) are in the nucleus of an atom and electrons (negative particles) are in the electron cloud.
Nucleus is in the centar of the atom atom and electron cloud is surrounding it.
Atoms have their mass concentrated in a very small nucleus.
A neutral carbon doesn't lack any electrons. It has exactly the same number of electrons as it has neutrons.
However, it has four electrons in its outer shell in comparison with eight electrons for a noble gas.
In that sense, it needs four electrons to complete its second shell.<span />