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2 years ago

If a buffer solution is 0.130 M in a weak acid (K_a = 1.7 x 10^-5) and 0.590 M in its conjugate base, what is the pH?

Chemistry

1 answer:

Serggg [28]2 years ago

8 0

Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43

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Does the iconic compound NaCl have the same particles as the individual elements of sodium and chlorine gas

valentina_108 [34]

Answer:It is not the same

Explanation: the normal sodium atom has an atomic number of 11 while Chlorine has an atomic number of 17. For Nacl; the sodium has lose an electron because it want to attain it octet configuration while chlorine accept the electron release to make it attain its octet configuration. At this point the elements are electrically stable.

Therefore, sodium become 10 and chlorine become 18

4 0

2 years ago

Calculate the rate of dissolution (dM/dt) of relatively hydrophobic drug particles with a surface area of 2.5×103 cm2 and satura

Crank

Answer:

$\large \boxed{\text{1.22 mg/s}}$

Explanation:

We can use the Noyes-Whitney equation to calculate the rate of dissolution.

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{DA(C_{s} - C)}{d}$

Data:

D = 1.75 × 10⁻⁷ cm²s⁻¹

A = 2.5 × 10³ cm²

Cₛ = 0.35 mg/mL

C = 2.1 × 10⁻⁴ mg/mL

d = 1.25 µm

Calculations:

Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³

d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{(1.75 \times 10^{-7} \text{cm}^{2}\text{s}^{-1})(2.5 \times 10^{3} \text{ cm}^{2})(0.350\text{ mg$\cdot$cm$^{-3}$})}{1.25 \times 10^{-4} \text{ cm}} = \textbf{1.22 mg/s}\\\\\text{The rate of dissolution is $\large \boxed{\textbf{1.22 mg/s}}$}$

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2 years ago

What is the speed at which molecules or atoms move dependent on temperature and state of matter.

den301095 [7]

The hotter it gets, the faster molecules move, solid form is in low temperature, liquid in medium temperature and gas in high temperature.

3 0

3 years ago

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

Naddika [18.5K]

Imagine we have mass of solvent 1kg (1000g)
According to that: $molality =n(solute) / m(solvent) =\ \textgreater \ 4.8m = 4.8mole / 1kg$
= 4.8 mole * 98 g/mole = 470g
$m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \$ $=\ \textgreater \ Molarity = 4.8 mole / 1.2 L = 4 M$ m(H2SO4) which is =470g

m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =>
$Molarity = n(solute) / V(solution) =\ \textgreater \$

5 0

3 years ago

Both protons and neutrons (and their anti-particles) froze out:

VladimirAG [237]

I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge.

6 0

3 years ago