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OlgaM077 [116]
4 years ago
15

The density of aluminum is 2.70 g/cm3. an irregular shaped piece of aluminum weighing 40.0 is added to a 100mL graduated cylinde

r containing exactly 50.0mL of water. To what height in the cylinder will the water level rise? SHOW STEPS PLZ.
Chemistry
1 answer:
hram777 [196]4 years ago
4 0
To determine the change in the water level after the aluminum is added, we sum up the forces acting on the piece aluminum. In this case, the weight of the piece is equal to the buoyant force. We do as follows:

W = Fb
mg = m(water)g
m = m(water)
40 = density of water ( volume )
40 = (1) V
V = 40 mL

Therefore, 40 mL of volume is being displaced by the aluminum and the level would be at about 90 mL.
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A sample of gas occupies 6. 8 L at 327 c what will be it volume at 27 c if the pressure and number of particles do not change
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A sample of gas occupies 6. 8 L at 327 c and the volume at 27 c if the pressure and number of particles do not change is 3.401 L.

<h3>What is volume?</h3>

Volume is a measurement of the three-dimensional space that is occupied. Numerous imperial units or the SI-derived units, such as the cubic meter and liter, are frequently used to quantify it numerically (such as the gallon, quart, cubic inch). Volume and the length (cubed) have a symbiotic relationship. The volume of a container is typically thought of as its capacity, not as the amount of the space it takes up. In the other words, the volume is the amount of fluid (liquid or gas) that the container may hold.

Volume was initially measured using naturally occurring vessels of a comparable shape and then with standardized containers. Arithmetic formulas can be used to quickly calculate the volume of the several straightforward three-dimensional shapes. If a formula for shape's boundary is known, it is possible to use integral calculus to determine the volumes of more complex shapes. No object in dimensions of zero, one, or two has volume; in the dimensions of four and above, the hypervolume is a concept similar to the standard volume.

In solving the any gas law problems, kelvin temperature will be used.

kelvin = degree celsius + 273.15

Kelvin = 327 + 273.15

                  = 600.15 K

Kelvin = 27 + 273.15

                  =  300.15 K

Given Data;

V_{1}(the initial volume) = 6.8 L                V_{2} (the final volume) = ? unknown

T_{1}(the initial temperature = 327 °C      T_{2}(the final temperature) = 27 °C

                                   =( 600.15 K)                                      = (300.15 K)

This problem can be solve using the Charles' Law which states that volume is directly proportional to kelvin temperature of gases if amount of gas and pressure are constant. If the temperature decreases, volume decreases.

Formula of the Charles' Law is \frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }

To solve for V2, we can used formula V_{2} =\frac{V_{1}T_{2}  }{T_{1} }

V_{2} =\frac{V_{1}T_{2}  }{T_{1} }

    = 6.8 L × 300.15 K / 600.15 K       (cancel the unit kelvin)

    = 2041.02 L / 600.15

V_{2}= 3.401 L

Therefore, volume is 3.401 L

To know more about volume visit: brainly.com/question/1578538

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