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Margaret [11]
2 years ago
13

Mohr's salt is a pale green crystalline solid which is soluble in water. It is a 'double sulfate' which contains two cations, on

e of which is Fe2+.The identity of the second cation was determined by heating solid mohr's salt with solid sodium hydroxide and a colourless gas was evolved. The gas readily dissolved in watdr giving an alkaline solution. A grey green solid residue was also formed which wasd insoluble in water. Whate are the identities of the gas and solid residue? A. Gas: H2 Residue: FeSO4 B.NH3 Na2SO4 C. NH3 Fe(OH)2 D. SO2 Fe(OH)2
Chemistry
1 answer:
Veronika [31]2 years ago
3 0
The gas is NH₃.
H₂ doesn't dissolve readily in water, SO₂ gives an acidic solution in water.

The solid residue is Fe(OH)₂.
FeSO₄ and Na₂SO₄ are soluble in water.

The answer is C.
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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0
3 years ago
If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for t
nikklg [1K]

<u>Answer:</u> The Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{O_2}=K_H\times p_{O_2}

where,

K_H = Henry's constant = ?

C_{O_2} = solubility of oxygen gas = 0.0043g/120.3mL

p_{O_2 = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm

Hence, the Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

7 0
2 years ago
How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
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0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
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Hello!

The reduction in particle size results in an increased rate of solution. It occur because is harder for a a solvent to surround bigger molecules.

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