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oee [108]
3 years ago
15

Copper has a density of 8.96 g/cm3. If 75.0 g of copper is added to 50.0 mL of water in a graduated cylinder, to what volume rea

ding will the water level in the cylinder rise?
Chemistry
1 answer:
Tcecarenko [31]3 years ago
3 0

Answer:

The answer to your question is    Final volume = 58.37 ml

Explanation:

Data

density = 8.96 g/cm³

mass = 75 g

volume of water = 50 ml

Process

1.- Calculate the volume of copper

  Density = mass / volume

Solve for volume

  Volume = mass / density

Substitution

  Volume = 75/8.96

Simplification

  Volume = 8.37cm³    or 8.37 cm³

2.- Calculate the new volume of water in the graduated cylinder

  Final volume = 50 + 8.37

  Final volume = 58.37 ml

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Julli [10]
To solve this problem we can use following equation.

v =u + at

Where v is the final velocity (m/s), u is the initial velocity (m/s), a is the acceleration (m/s²) and t is the time taken (s).

v = 7 m/s
u = 4 m/s
a = ?
t = 5 s

By applying the equation, we can get
    7 m/s = 4 m/s + a x 5 s
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          a = 0.6 m/s²

Hence, the acceleration is 0.6 m/s² towards north.

Answer is "C".
4 0
3 years ago
A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.
8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
  • Mathematically, P\alpha \frac{1}{V}

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

P2=\frac{P1V1}{V2}

Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

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Thus, the pressure of air inside the balloon after squeezing is 812 kPa

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Answer:

"Thermometer C, because it measures accurately in the ones place."

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Explanation:

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