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Kisachek [45]
3 years ago
7

A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,

the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.

You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Chemistry
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

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Ammonium nitrate decomposes to dinitrogen monoxide and water. If given 45.7 grams of ammonium
statuscvo [17]

Answer:

20.54 g of H₂O

Explanation:

Since you already have a balanced equation, the next step is to see the ratio between the ammonium nitrate and water in the equation:

NH4NO3(s)—N2O(g)+2H2O

1 mole of ammonium nitrate produces 2 moles of H2O

So we have the ration:

\dfrac{1\;mole\;of\;NH_4NO_3}{2\;moles\;of\;H_{2}O}

Let's leave that for later use.

Next step is to covert the mass given into moles. We do that by getting the molar mass of the given and using that as a conversion factor:

Element   number of                 molar mass

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N =                  2              x            14.01 g/mole   =      28.02 g/mole

H =                  4              x              1.01 g/mole   =        4.01  g/mole

O =                  3              x            16.00 g/mole =      <u>48.00 g/mole </u>

                                                                                    80 .03 g/mole

Now we can convert:

45.7g\;of\;NH_4NO_3\times\dfrac{1\;of\;NH_4NO_3}{80.03\;g\;of\;NH_4NO_3}=0.57\;moles\;of\;NH_4NO_3

Now we can use this to determine how many moles of H2O this would produce by using the ration we solved for earlier.

0.57\;moles\;of\;NH_4NO_3\times\dfrac{2\;moles\;of\;H_{2}O}{1\;mole\;of\;NH_4NO_3} =1.14\;moles\;of\;H_{2}O

And we convert that by getting the molecular mass of H2O, which is 18.02 g/mole:

1.14\;moles\;of\;H_{2}O\times\dfrac{18.02\;g\;of\;H_{2}O}{1\;\;mole\;of\;H_{2}O} =20.54\;g\;of\;H_{2}O

But this is only if the whole 45.7 g of ammonium nitrate is used up.

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