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Leno4ka [110]
3 years ago
7

What is the HCl concentration if 52.0 mL of 0.350 M NaOH is required to titrate a 30.0 mL sample of the acid?

Chemistry
1 answer:
In-s [12.5K]3 years ago
8 0
Volume HCl = 52.0 mL in liters: 52.0 / 1000 => 0.052 L

Molarity HCl = ?

Number of moles NaOH :

V = 30,0 mL / 1000 => 0.03 L 

Molarity = 0.350 M

n = M x V

n = 0.350 x 0.03 => 0.0105 moles of NaOH

Mole ratio :

<span>HCl + NaOH = NaCl + H2O
</span>
1 mole HCl ---------- 1 mole NaOH
? moles HCl --------- 0.0105 moles NaOH

0.0105 x 1 / 1 => 0.0105 moles of HCl

Therefore:

M = n / V

M = 0.0105 /  0.052

= 0.201 M 
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b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
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3.-4.
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The power for the first tank is 2kW  and the second tank is 150 kw for the theoretical pumping power lifts a solution with a density of 1200 kg/m³ from a tank to a reservoir.

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