20 points

Chemistry

2 answers:

Gemiola [76]3 years ago

7 0

Answer:

Explanation:

I am just saying this is correct! The other answer is wonderful, they just had a typo and meant to put "Ag". :)

allochka39001 [22]3 years ago

5 0

Answer:

Which of these elements would you expect to find a sea of electrons surrounding nuclei is Ar

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions

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2. A destructive, rotating column of air that has very high wind speeds and that is sometimes visible

lesya692 [45]

It should be a tornado

5 0

3 years ago

Read 2 more answers

I need some help with with question:

defon

Answer:

C1V1=C2V2

C1 is 2.0mol/l

V1=?

C2=.4mol/L

V2=100ml or for this 0.1L

V1 is 20ml

Best way to prepare this is to measure out 20ml of the 2 molar solution and add 80mL to it to get to 100mL

Explanation:

8 0

3 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 $\large \boxed{\textbf{1.22 mg/s}}$}$