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charle [14.2K]
3 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]3 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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Romashka [77]
The formula is AI203
4 0
4 years ago
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What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg a
yaroslaw [1]

<u>Answer:</u> The Van't Hoff factor for KCl is 1.74

<u>Explanation:</u>

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

\chi_{(solute)}=\frac{1}{56.56}=0.0177

The equation used to calculate relative lowering of vapor pressure follows:

\frac{p^o-p_s}{p^o}=i\times \chi_{solute}

where,

\frac{p^o-p_s}{p^o} = relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

\chi_{solute} = mole fraction of solute = 0.0177

p^o = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74

Hence, the Van't Hoff factor for KCl is 1.74

7 0
3 years ago
What is the weight of a 7.0-kilogram bowling ball on the surface of the moon.
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5 0
3 years ago
An experiment calls for 7.57g of sugar. You have a sugar son that is 5% by weight (for every 100g there is 5g of sugar). How man
Soloha48 [4]

Answer:

mL sugar sold needed = 2.2215 mL sugar

Explanation:

mass sugar = 7.57 g

∴ wt% = 5% = (g sugar/g sln)×100

⇒ 0.05 = g sugar/g sln

∴ g sln = 100 g

⇒ g sugar = 5 g

∴ δ sugar = 1.157 g/mL

⇒ mL sugar = (5 g)×(mL/1.157 g) = 4.322 mL

⇒ mL sugar needed = (7.57 g)/(1.157 g/mL) = 6.543 mL

mL of the sugar sold needed = 6.543 mL - 4.322 mL = 2.2215 mL sugar

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7. How many electrons can be held in the energy level n = 4?
Dahasolnce [82]

Answer:

32 electrons

Explanation:

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