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vovikov84 [41]
3 years ago
9

In the titration between potassium iodate and the sodium thiosulfate solution, if the titration is not performed immediately aft

er the addition of the sulfuric acid, how would this likely affect the calculated concentration of your diluted sodium thiosulfate solution
Chemistry
1 answer:
N76 [4]3 years ago
5 0

Answer:

The calculated concentration of sodium thiosulphate solution will be less than the actual value.

Explanation:

When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.

If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation

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If 26.2 grams of a pure compound contain 8.77 × 1022 molecules, what is the molecular weight of this compound? Answer in units o
Nataly [62]

Answer:

Mw = 179.845 g/mol

Explanation:

  • Mw [=] g/mol

∴ w = 26.2 g

∴ 1 mol = 6.02 E23 molecules.......Avogadro's number

⇒N° moles = 8.77 E22 molecules * ( mol / 6.02 E23 molecules ) = 0.146 mol

⇒ Mw = 26.2 g / 0.146 mol = 179.845 g/mol

3 0
3 years ago
A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
Why do ionic and polar compounds dissolve in water?
ioda
Because they are salty
3 0
2 years ago
Automobile catalytic converters use a platinum catalyst to reduce air pollution by changing emissions such as carbon monoxide, C
julia-pushkina [17]

Answer: 448 g of O_2 will be required to completely react with 784g moles of CO(g) during this reaction.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} CO=\frac{784g}{28g/mol}=28moles

The balanced chemical equation is:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)  

According to stoichiometry :

2 moles of CO require  = 1 mole of O_2

Thus 28 moles of CO will require=\frac{1}{2}\times 28=14moles  of O_2

Mass of O_2=moles\times {\text {Molar mass}}=14moles\times 32g/mol=448g

Thus 448g of O_2 will be required to completely react with 784g moles of CO(g) during this reaction.

6 0
3 years ago
Read 2 more answers
Which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?
forsale [732]
K₂CrO₄(aq) + Ba(NO₃)₂(aq) = BaCrO₄(s) + 2KNO₃(aq)

2K⁺ + CrO₄²⁻ + Ba²⁺ + 2NO₃⁻ = BaCrO₄ + 2K⁺ + 2NO₃⁻

spectator ions: K⁺, NO₃⁻
3 0
3 years ago
Read 2 more answers
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