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MAXImum [283]
3 years ago
13

Sea water contains roughly 28.0 g of nacl per liter. What is the molarity of sodium chloride in sea water?

Chemistry
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

The molarity of sodium chloride in sea water is 0.479 M

Explanation:

Step 1: Data given

Mass of NaCl = 28.0 grams

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 28.0 grams / 58.44 g/mol

Moles NaCl = 0.479 moles

Step 3: Calculate molarity NaCl in sea water

Molarity = moles / volume

Molarity NaCl = 0.479 moles / 1L

Molarity of NaCl in sea water = 0.479 mol/L = 0.479 M

The molarity of sodium chloride in sea water is 0.479 M

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Answer:

d making models.

Explanation:

When scientists create a representation of a complex process, they are inferring that they are making models.

A model is an abstraction of the real world or a complex process. Models are very useful in developing solutions to processes that are not easily simplified.

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In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a
igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

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Explanation:

                       

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