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Bad White [126]
3 years ago
11

If a graduated cylinder has 28ml of water in it and you add an object which causes the water level to rise to 47ml, what is the

volume of the object you placed in the water?
Chemistry
1 answer:
grin007 [14]3 years ago
7 0

Answer:

The object placed in the water has a volume of 19 cm³

Explanation:

<u>Step 1: </u>Data given

volume of the cylinder before adding the object = 28 mL = 28 cm³

After adding an object with volume X the volume rises to 47 mL = 47 cm³

<u>Step 2:</u> Calculate the volume of the object

Volume of the object = Final volume - initial volume

Volume of the object = 47 cm³ (or 47 mL) - 28 cm³ ( or 28 mL) = 19 cm³ (or 19 mL)

The object placed in the water has a volume of 19 cm³

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The strong nuclear force is always smaller then the electrostatic force true or false
Y_Kistochka [10]

Answer: I think that the answer was False.

3 0
3 years ago
Read 2 more answers
A sample of neon occupies a volume of 478 mL at STP. What will be the volume of the neon when the pressure is reduced to 93.3 kP
Andre45 [30]

The volume of neon when the pressure is reduced to 93.3 kPa is 519 mL.

Explanation:

The kinetic theory of gases is mostly based on Boyle's law. From the Boyle's law, the pressure experienced by any gas molecules is inversely proportional to volume of the gas molecules. Also this inverse relation is obeyed if and only if the number of moles and temperature of the gas molecules remained constant.

So,P=\frac{1}{V}

So if there is a change in pressure then there will be inverse change in volume. That means if there is decrease in the pressure of gas molecules then there will be increase in the volume and vice versa.

So the Boyle's law is combined as P_{1} V_{1} = P_{2} V_{2}

As here the initial pressure or P_{1} is 1 atm or 101.3 kPa and the initial volume is 478 mL. Similarly, the final pressure is 93.3 kPa and the final volume will be

101.3*10^{3}*478*10^{-3}  = 93.3*10^{3} * V_{2}

V_{2} = 519 mL

So, the volume of neon when the pressure is reduced to 93.3 kPa is 519 mL.

5 0
3 years ago
Mass of 60.009 g and radius of 1.02 cm what is the density of the sphere in (g/ml)
alexandr402 [8]
Density is given as mass / volume.

Mass is the sphere is 100 g.

Volume of the sphere = (pi∗r3)∗4/3
(
p
i
∗
r
3
)
∗
4
/
3

=(4∗22∗3∗3∗3)/(7∗3)cm3
=
(
4
∗
22
∗
3
∗
3
∗
3
)
/
(
7
∗
3
)
c
m
3

=792/7
=
792
/
7
cm3
3

Therefore, Density is 100/(792/7)g/cm3
100
/
(
792
/
7
)
g
/
c
m
3

Which gives: density = 0.883838 g/cm3
g
/
c
m
3

If you want to change the units to kg per cubic metres, then we need to divide this value by 1000( for g to kg) and multiply by 100 * 100 * 100 (for cm to m).

This makes the density to be 883.83 kg/m3
4 0
3 years ago
The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc
schepotkina [342]

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

  • PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

4 0
3 years ago
Need the names for these asap please ​
jeyben [28]

Answer:

3_ethylpentane

4_ethylheptane

1_pentene

3_hexene

5 0
3 years ago
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