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3 years ago

Calculate the volume of a sample of iron that has a density of 6.8 g/mL and a mass of 6780 mg.

Chemistry

1 answer:

Veronika [31]3 years ago

8 0

Answer:

m= 6780mg = 6.78g = 6.8g

d= 6.8g/ml

V= m/d

V= 1ml

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3 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M TRIS base, what will be the resulting pH?

Katyanochek1 [597]

Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For TRIS acid:

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

For TRIS base:

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

4 years ago

To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo

gtnhenbr [62]

V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH

moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

hope this helps!

4 0

3 years ago