How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration

Amount of HCL = 0.00318 L  of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH  = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH  = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed =  0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795  / 2.5

Amount of HCL = 0.00318 L  of 3.18 ml

the answer would be b i believe

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72