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Shtirlitz [24]
3 years ago
12

A student is trying to calculate the density of an ice cream cone. She already knows the mass, but she needs to determine the vo

lume as well. Which of the following formulas can be used to calculate the volume of the cone?
V equals four-thirds times pi times r cubed
V equals one-third times pi times r squared times h
V = s3
V = πr2h
Chemistry
2 answers:
lyudmila [28]3 years ago
8 0

Answer:

V equals one-third times pi times r squared times h

Explanation:

IgorLugansk [536]3 years ago
6 0
<span>V equals one-third times pi times r squared times h</span> 
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10.0 mL of 3.0 M sulfuric acid has been added to 50.0 mL of water.
Cloud [144]

Answer:

the new concentration is 0.60M

Explanation:

The computation of the new concentration is shown below;

We know that

M1V1=M2V2

(3.0M) (10.0 mL) = M2 (50.0mL)

30 = M2 (50.0mL)

So, M2 = 0.60 M

Hence, the new concentration is 0.60M

The same is considered and relevant

3 0
3 years ago
An atom has 18 protons, 18 electrons, and 20 neutrons. This atom has _____.
Lera25 [3.4K]

an atom that has 18 protons, 18 neutrons and 20 neutrons. This atom has mass number of 38

mass number is calculated as number of protons + number on neutrons

therefore the mass number of the atom = 18+20= 38

Since the atom has a electronic configuration of 2.8.8,the atom cannot be negatively or positively charged because it is stable.

The atomic number of the atom is 18 since atomic number is equal to number of protons

4 0
3 years ago
Read 5 more answers
On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

5 0
3 years ago
Read 2 more answers
Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and
Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

            \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

The given data is as follows.

     mass = 500 g,         C_{p} = 24.4 J/mol K

     T_{h} = 500 K,          T_{c} = 250 K               

   Mass number of copper = 63.54 g /mol

Number of moles = \frac{mass}{/text{\molar mass}}

                                 = \frac{500}{63.54}

                                 = 7.86 moles

Now, equating the entropy change for both the substances as follows.

     7.86 \times 24.4 \times [T_{f} - 250] = 7.86 \times 24.4 \times [500 -T_{f}]

       T_{f} - 250 = 500 - T_{f}

          2T_{f} = 750

So,       T_{f} = 375^{o}C

  • For the metal block A,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

              = 24.4 log [\frac{375}{500}]

              = -3.04 J/ K mol

  • For the block B,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

                  = 24.4 log [\frac{375}{250}]

                  = 4.296  J/Kmol

And, total entropy change will be as follows.

                       = 4.296 + (-3.04)

                      = 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol  and change in entropy of block B is 4.296  J/Kmol.

8 0
3 years ago
80 protons, 119 neutron
g100num [7]
Mercury-199 is composed of 80 protons, 119 neutrons, and 80 electrons.
4 0
2 years ago
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