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2 years ago

If the dimensions of a cube-shaped cell with a volume of 1 pm were to

Chemistry

1 answer:

melamori03 [73]2 years ago

5 0

Answer:

Explanation:

Assuming the units are all supposed to be the same

A cube shaped cell with a 1 um³ volume would have a dimensions of 1 x 1 x 1 (because volume = length x width x height and 1 x 1 x 1 =1)

If we double the dimensions, we have a cube with the dimensions 2 x 2 x 2.

The surface area of each face of the cube is the length x width, so 2 x 2. This means the surface area of each face of the cube is 4 um².

There are 6 faces of a cube, so the total surface area is 6 x 4 um² = 24 um².

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In subsequent experiments you determine that the compound is iron (II) chloride. Which of the following properties would you als

Naddika [18.5K]

Answer:

b. Conducts electricity when dissolved in water

Explanation:

Iron(II) chloride, is the chemical compound with formula FeCl2.

It is a solid with a high melting point of about 677 degree Celsius or 950 K when in anhydrous form but have lower melting point in hydrated form.

The compound is often off-white. FeCl2 crystallizes from water as the greenish tetrahydrate, which is the form that is most commonly encountered in the laboratory.

There is also a dihydrate. The compound is highly soluble in water, giving pale green solutions.

7 0

2 years ago

Separation of light into different wavelengths

TiliK225 [7]

Answer:

I'm sure it's Dispersion

Explanation:

4 0

2 years ago

What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

7 0

2 years ago

REALLY NEED HELP!!! Chemistry

kicyunya [14]

Answer:

C.) 2

Explanation:

The pH equation is:

pH = -log[H⁺]

In this equation, [H⁺] is the molarity of the acid. In this case, the acid is HCl. Molarity can be found using the equation:

Molarity (M) = moles / volume (L)

Since you were given moles and volume, you can find the molarity of HCl.

Molarity = moles / volume

Molarity = 0.01 moles / 1.00 L

Molarity = 0.01 M

Now, you can plug the molarity of the acid into the pH equation.

pH = -log[H⁺]

pH = -log[0.01]

pH = 2

3 0

1 year ago

Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their

dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Pure Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For Br and Br,

E.N of Bromine = 2.96

________

E.N Difference 0.00 (Non Polar/Pure Covalent)

For N and O,

E.N of Oxygen = 3.44

E.N of Nitrogen = 3.04

________

E.N Difference 0.40 (Non Polar/Pure Covalent)

For P and H,

E.N of Hydrogen = 2.20

E.N of Phosphorous = 2.19

________

E.N Difference 0.01 (Non Polar/Pure Covalent)

For K and O,

E.N of Oxygen = 3.44

E.N of Potassium = 0.82

________

E.N Difference 2.62 (Ionic)

6 0

3 years ago