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Sloan [31]
3 years ago
8

2. A solution that contains a small amount of salt and a large amount of water is said to be a _______ solution.

Physics
2 answers:
koban [17]3 years ago
6 0
2. A dilute solution means that the amount of solvent (water, for this particular case) is significantly larger than the solute (salt). Thus, the answer is D. 6. John Dalton's atomic theory states that matter is made up of tiny indestructible objects called matter. The theory also indicates that although same elements have the same atoms, each element have unique set atoms that deinfe them. From this, we can conclude that the wrong assumption is C. 9. Atoms, by default, are electrically neutral. When an atom loses or gains electron/s, then they become ionized atoms or commonly called as ions. Thus, ionized atoms imply unequal number of protons and electrons. This means the answer must be A. 11. Analgesics are commonly used to relieve pain. Thus, the answer is C. 14. Adding up the atomic mass of the individual atoms will give you the molar mass of a compound. Therefore, the answer is B.: 15. The pH scale provides us the alkalinity or acidity of a solution based on the value. A value between 0 to 6 indicates that the solution is acidic. 7 is considered neutral and a value between 7 and 14 indicates that the solution is basic. Thus, the answer is D. 19. An element has consistent properties and can no longer be further identified into having individual properties. Thus, the answer is A. 20. The valence of an element dependeds on the number of electrons on the outermost shell. Thus, it is equal to the number of charge negative or positive charges on the ion. Hence, we have A<span>. :</span>
astra-53 [7]3 years ago
4 0

Which of the following classifications of matter includes materials that can no longer be identified by their individual properties?

A. Element

B. Mixture

C. Atom

D. Compound

THE CORRECT ANSWER IS D. COMPOUND

ANY OTHER QUESTIONS YOU CAN MESSAGE ME FOR THE CORRECT ANSWER. EVERYONE IS WELCOMED :)

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
Enrique is given information about a satellite orbiting Earth. R = 3. 8 Ă— 108 m T = 18 days In order to calculate the tangentia
NISA [10]

The first step that  Enrique must take in order to calculate the tangential speed of the satellite is to convert the period from days to seconds.

We know that the SI unit of speed is meter per second and now, we with to obtain the tangential speed of the satellite.

Since the period is given in days, the first step is to convert the period from days to seconds.

Learn more: brainly.com/question/17638582

6 0
2 years ago
What can electromagnetic waves travel through?
Kisachek [45]
Electromagnetic waves do not require a medium to travel through. They can travel through empty space or matter.
6 0
2 years ago
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
2 years ago
An 8-kg mass is in free fall. What is the velocity of the mass after 9 seconds
zimovet [89]
The velocity of the mass after 9 second is 88 m/s
7 0
2 years ago
Read 2 more answers
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