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Brilliant_brown [7]
3 years ago
5

What is our current microscopic view of this world

Chemistry
1 answer:
Fantom [35]3 years ago
6 0
EVERY YEAR, NIKON <span>selects the most artful, scientifically enlightening and skillfully produced images from thousands of submissions for its </span>Small World<span> microscope photography contest. Tomorrow, another set of impressive winners will be announced for the contest’s 40th year.</span>
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Gress.
MrRissso [65]

Answer:

true

Explanation:

the small car also has gravity making it heavy

8 0
3 years ago
Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.
SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

                HClO   ⇄   ClO-  +  H+

start:          0.05            0         0

change       -x               +x       +x

balance     0.05-x         x         x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

3x10^{-8}=\frac{x^{2} }{0.05}

clearing the x and calculating its value we have:

x=3.87x10^{-5}=[H+]=[ClO-]

the pH can be calculated by:

pH=-log[H+]=-log[3.87x10^{-5}]=4.41

7 0
3 years ago
The removal of silver tarnish from silverware using aluminum foil and a solution of electrolyte is an example of
eimsori [14]

Answer:

The answer is C. An electrochemical cell.

Explanation:

The aluminum ion react with the sulfide to form aluminum sulfide.

8 0
3 years ago
Write the electronic configuration of the Na +, Mg2 +, Ca2 + and Rb + ions. Research the size of these ions in picometers and or
Sunny_sXe [5.5K]

Answer:

pen

Explanation:

3 0
3 years ago
Your veterinarian is administering a sedative to your 50 pound dog. The sedative is mixed in saline solution. Unfortunately the
Olin [163]

Answer:

26.25 mL

Explanation:

This is a dilution problem. First, let us calculate the volume of final solution needed:

The dog weighs 50 pounds and the sedative is administered at 0/7 ml per pound. Hence:

50 x 0.7 = 35 mL

A total volume of 35 mL, 2.5% solution of the sedative will be needed.

But 10% solution is available. There needs to be a dilution with saline water, but what volume of the 10% solution would be diluted?

initial volume = ?

final volume = 35 mL

initial concentration = 10%

final concentration = 2.5%

Using dilution equation:

initial concentration x initial volume = final concentration x final volume

initial volume = \frac{final concentration*final volume}{initial concentration}

                     = 2.5 x 35/10 = 8.75 mL

Hence, 8.75 mL of the 10% pre-mixed sedative will be required.

But 35 mL is needed? The 8.75 mL is marked up to 35 mL with saline water.

35 - 8.75 = 26.25 mL

<em>Therefore, 26.25 mL of saline water will be added to 8.75 mL of the 10% pre-mixed sedative to give 2.5%, 35 mL needed for the dog.</em>

5 0
3 years ago
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