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frosja888 [35]
3 years ago
14

Lead(2) sulfate yields to lead(2)sulfite and oxygen

Chemistry
1 answer:
PolarNik [594]3 years ago
4 0
A standardized test of what you want to do with your friends and you get to know what you think you want to be tre
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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 $\large \boxed{\textbf{1.22 mg/s}}$}

8 0
3 years ago
Calculate Density:<br> Volume=13cm3, Mass=147.55g<br> Someone please help
almond37 [142]

Answer:

11.35 g/cm³

Explanation:

If your rounding then 11.4. hope this helps :)

5 0
2 years ago
Question 11 (1 point)
erastova [34]

Answer:

remove product

Explanation:

Removing the product will always shift the equilibrium to the right. This is based on the Le Chatelier's principle which states that "if any of the conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".

  • If a system at equilibrium is disturbed, by changing the concentration of one of the substances all the concentrations will change until a new equilibrium point is reached.
  • Removing the product will increase the concentration of the species on the left hand side, the equilibrium will shift to the right.
5 0
3 years ago
Can you help me with my Chemistry homework please? Can you answer the blank ones? And number 20 please.
Anarel [89]

i just cant understand the question

please take a clear pic

7 0
3 years ago
Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?
sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

3 0
3 years ago
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