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zhannawk [14.2K]
3 years ago
13

Approximately how many formula units of NaCl are in 116.88g of table salt (NaCl), knowing that the molar mass of NaCl is 58.44g/

mol?
Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

116.88g of table salt (NaCl) contains two formula units

Explanation:

Now,

We know that 1 formula unit of sodium chloride has a molar mass of 58.44g/mol

Hence;

Mass of 1 formula unit = 58.44g

Mass of x formula units = 116.88g

x = 116.88g * 1 formula unit/58.44g

x = 2 formula units

Therefore;

116.88g of table salt (NaCl) contains two formula units

Gnesinka [82]3 years ago
4 0

Answer:

There are  1.2044 × 10²⁴ formula units of NaCl in 116.88 g of table salt (NaCl)

Explanation:

A formula unit is an empirical formula of the smallest collection or number of atoms in an ionic or covalent combination from which a compounds formula can be established and which are used to represent the compound stoichiometrically

Sodium chloride is an ionic compound and is represented by the formula unit NaCl as it composed of ions and is not therefoe represented by a molecular formula

The given mass of the table salt, NaCl = 116.88 g

The molar mass of NaCl = 58.44 g/mol = The mass of 1 mole of NaCl

1 mole of NaCl contains one Avogadro's number or 6.022 × 10²³ formula units of NaCl,

∴ 58.44 g of NaCl contains 6.022 × 10²³ formula units of NaCl

116.88 g of NaCl will have (116.88/58.44) × 6.022 × 10²³ = 1.2044 × 10²⁴ formula units of NaCl

The number of formula units of NaCl in 116.88 g of table salt (NaCl) =  1.2044 × 10²⁴ formula units of NaCl.

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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Explanation:

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The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
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Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

Now put all the given values in this expression, we get:

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s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

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Answer:

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C = 0.737M

7 0
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