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3 years ago

NEED URGENT HELP WITH SCIENCE HOMEWORK. TRIED MY BEST TO FIGURE OUT BUT COULDNT. I WILL MARK YOU BRAINLIEST!!

Chemistry

1 answer:

Dennis_Churaev [7]3 years ago

8 0

Answer:

1. a

2. c

3. b

4. d

5. True

6. True

7. Cold Tundra

8. False

Explanation:

These are the answers. I had the same questions on my science quiz.

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A substance that contains only one type of particle is an __________

Dvinal [7]

the answer is a chemical element

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3 years ago

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s

Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.

In order to prepare 10 mL, 10 μM; 4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM; 6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM; 8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

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3 years ago

In the figure below, two nonpolar molecules are interacting.Which interaction would most likely cause these molecules to repel e

Fed [463]

Two non-polar molecules are most likely to interact by
induced dipole-induced dipole interaction.

Non-polar substances do not have a permanently established charge distribution due to similar electron affinities of the atoms that are present. Moreover, due to the absence of a polar hydrogen, they cannot exhibit hydrogen bonding. They interact with one another by induced dipole-induced dipole interactions which arise from the molecules of the substances coming into close vicinity of one another.

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What sublevels are contained in the hydrogen atoms first four energy levels what orbitals are related to each?

erik [133]

1st level = s
2nd level = s,p 
3rd level = s,p,d 
4th level = s,p,d,f

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3 years ago

. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub

elena-s [515]

The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use are:

v = v(f)-v(i)
d = m/v

Where:

d= density
m= mass
v= volume
v(f) = final volume
v(i) = initial volume

Information about the problem:

m = 132.6 g
v(i) = 48.6 ml
v(f) = 61.2 ml
v = ?
d =?

Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

Learn more about density in: brainly.com/question/1354972

#SPJ4

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1 year ago