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charle [14.2K]
2 years ago
5

An object has a density that is greater than the density of water. How could you make it float? Explain by giving a real-life ex

ample.
Chemistry
1 answer:
vladimir1956 [14]2 years ago
5 0

Answer:

Put it on a ship or an aerogel

Explanation:

You can use an aerogel, a synthetic porous ultralight material. The aerogel can  support a mass many times greater than their own.

Even simpler, you can put the object on board a ship. The ship has a smaller density than water, making it float in water.

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Photochemistry — the study of the chemical changes caused by light.

Surface chemistry — the study of chemical reactions at surfaces of substances

8 0
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Help plzzzz only the answer
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Which fossil fuel is mainly used for heating and cooking?
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A is wrong

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4 0
3 years ago
Read 2 more answers
What are the properties of aluminium and their use​
Wewaii [24]

Answer: Answers are in bulleted lists.

Explanation: Aluminum...

  • has a low density
  • is non-toxic
  • has a high thermal conductivity
  • has excellent corrosion resistance
  • can be easily cast, whether it's machined or formed.

Uses:

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Have a great day! :)

6 0
2 years ago
One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume unt
Umnica [9.8K]

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K    0^00C=273K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas  = 1

V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?

Putting values in above equation, we get:

\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K

The final temperature is 900 K

b) The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 6.00 atm

P_2 = final pressure of gas = 2\times 6.00atm=12.0atm

V_1 = initial volume of gas = 4.10 L

V_2 = final volume of gas =  2\times 4.10 L=8.20L

T_1 = initial temperature of gas = 300K

T_2 = final temperature of gas =?

Now put all the given values in the above equation, we get:

\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}

T_2=1200K

The final temperature is 1200 K

5 0
3 years ago
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