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Leya [2.2K]
2 years ago
15

What is the concentration of an alcl3 solution if 150. ml of the solution contains 550. mg of cl- ion?

Chemistry
1 answer:
valina [46]2 years ago
6 0

The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M


calculation


concentration = moles /volume in liters


volume in liters = 150 /1000= 0.15 L


number of moles calculation

write the equation for dissociation of Al2Cl3

that is AlCl3 ⇔ Al^3+ + 3 Cl ^-


find the moles of Cl^- formed

moles =mass/molar mass

mass in grams= 550/ 1000 =0.55 grams

molar mass of Cl^- =35.5 g/mol


moles is therefore= 0.55/35.5 =0.0155 moles


by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles



concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M

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Answer:

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Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

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Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

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Q = mCΔT

m = mass of substance

c = specific heat capacity

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m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

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T2 = T1+ ΔT

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Therefore, final temperature is 91°C

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Explanation :

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Thus, the oxidizing and reducing agents are, H_2O_2 and Fe^{2+}.

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1 year ago
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