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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.

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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Branches of chemistry with definition and examples in real life

nataly862011 [7]

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Sub-branches of physical chemistry include:

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

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

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2 years ago

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3 years ago

Which fossil fuel is mainly used for heating and cooking?

Anna [14]

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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.

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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