what volume of a 3.5 m hcl is required to completely neutralize 50.0 ml of a 2.0 m naoh? 58 ml 29 ml 15 ml 50 ml

Hey there!:

Molarity HCl = 3.5 M

Volume HCl = ?

Molarity NaOH = 2.0 M

Volume NaOH in liters = 50.0 mL / 1000 => 0.05 L

Number of moles NaOH:

n = M * V

n = 2.0 * 0.05

n = 0.1 moles of NaOH

Given the reaction:

HCl + NaOH = NaCl + H2O

1 mole HCl 1 mole NaOH

1 mol HCl reacts with 1 mol NaOH , so moles NaOH = moles HCl

0.1 moles of NaOH = 0.1 moles of HCl

Therefore:

M( HCl ) = n / V

3.5 = 0.1 / V

V = 0.1 / 3.5

V = 0.029 L in mL :  0.029 * 1000 => 29.0 mL

Answer B

hope that helps!

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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