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3 years ago

the main part of the microscope that is used to see and denlarge specimens what do you call this part?

Chemistry

1 answer:

Crazy boy [7]3 years ago

7 0

Answer:

I believe you are asking about the objective lens or eyepiece.

Explanation:

Objective lenses are used to magnify objects enough for you to see them in great detail

Eyepiece/Ocular is what you look through at the top of the microscope.

A tube connects the eyepiece lens to objective lenses, which enhance the magnification power of the eyepiece lens. Both the eyepiece and the objective lens contribute to the magnification of the object.

The eyepiece lens usually magnifies 10x, and a typical objective lens magnifies 40x.

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Suppose 3.52 g of calcium chloride is completely dissolved in a beaker of water. what would be the number of chloride ions that

yaroslaw [1]

Answer:

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3 0

3 years ago

One brand of extra-strength antacid tablets contains 750 mg of calcium carbonate (100 g/mol) in each tablet. Stomach acid is ess

konstantin123 [22]

Answer:

One tablet can neutralize 150 mL of stomach acid at a pH of 1.0

Explanation:

Step 1: Data given

Mass of calcium carbonate = 750 mg = 0.75 grams

Molar mass of CaCO3 = 100 g/mol

pH = 1.0

Step 2: The balanced equation

2HCl + CaCO3 → CaCl2 + H2CO3

Step 3: Calculate molarity of HCl

pH = -log[H+] = 1

[H+] = 0.1 M = 0.10 mol/L

Step 4: Calculate moles of CaCO3

Moles CaCO3 = 0.75 grams / 100g/mol

Moles CaCO3 = 0.0075 mol

Step 5: Calculate moles of HCl

For 2 moles HCl we need 1 mol CaCO3 to produce CaCl2 and 1 mole of H2CO3

For 0.0075 moles of CaCO3 we have 2*0.0075 = 0.015 moles HCl

Step 6: Calculate volume of HCl

Volume = moles /molarity

Volume = 0.015 moles / 0.1 M

Volume = 0.15 L = 150 mL

One tablet can neutralize 150 mL of stomach acid at a pH of 1.0.

6 0

3 years ago

To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i

enyata [817]

Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$ .....(1)

Molarity of acetic acid solution = 0.0880 M

Volume of solution = 30.6 mL

Putting values in equation 1, we get:

$0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol$

Thus there are 0.00269 moles of acetic acid in buffer.

8 0

3 years ago

If an element is lustrous, brittle, and a semi-conductor, how would you classify it?

just olya [345]

Metalloid

Explanation:

If an element is lustrous, brittle and a semi-conductor, it is best classified as a metalloid.

Metalloids shares attributes of metals and non-metals.

They are often described as semi-metals as they do not share the full properties that makes a metal a metal.
Metalloids are lustrous but not malleable like metals.
They do not conduct electricity but they do so on certain conditions.
Examples are silicon, germanium, boron, arsenic e.t.c
They are usually found in the middle of the periodic table.
They are not readily alloyed with metals.

Learn more:

Metalloid brainly.com/question/3023499

#learnwithBrainly

7 0

3 years ago

Please help me fast...

jekas [21]

B) 1 and 3

it's the right answer

6 0

3 years ago

Read 2 more answers