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3 years ago

Which of the following solutes will lower the freezing point of water the most? NaCl, CaCl2!or AlBr3

1 answer:

6 0

This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.

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A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.

GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂ + I₂ ⇄ 2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

H₂ + I₂ ⇄ 2HI

In: 0.001 0.002 -

R: x x 2x

Eq: 0.001-x 0.002-x 0.00187

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is = (HI)² / (H₂) . (I₂)

0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

5 0

3 years ago

3)<br>0.09 moles of sodium sulfate in 12 mL of solution.<br>

adelina 88 [10]

Answer:

7.5 M

Explanation:

In order to find a solution's molar concentration, or molarity, you need to determine how many moles of solute, which in your case is sodium sulfate,

, you get in one liter of solution.

That is how molarity was defined -- the number of moles of solute in one liter of solution.

So, you know that you have

0.090

moles of solute in

12 mL

of solution. Your goal here will be to scale up this solution by using this information as a conversion factor to help you determine the number of moles of solute present in

6 0

2 years ago

1. How would the series of figures change in the presence of a catalyst?

statuscvo [17]

No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.

hope this explains it

If it does, please give it a brainliest :)))

8 0

2 years ago

Read 2 more answers

The molar mass of HgO is 216.59 g/mol. The molar mass of O2 is 32.00 g/mol. How many moles of HgO are needed to produce 250.0 g

Irina-Kira [14]

A reaction in which Oxygen (O₂) is produced from Mercury Oxide (HgO) would be a decomposition reaction.
2HgO → 2Hg + O₂

If 250g of O₂ is needed to be produced,
then the moles of oxygen needed to be produced = 250g ÷ 32 g/mol
= 7.8125 mol

Now, the mole ratio of Oxygen to Mercury Oxide is 1 : 2
∴ if the moles of oxygen = 7.8125 mol
then the moles of mercury oxide = 7.8125 mol × 2
= 15.625 mol

Thus the number moles of HgO needed to produce 250.0 g of O₂ is 15.625 mol

6 0

3 years ago

For centuries, sailors used tree sap (pitch) to seal the spaces between the planks on wooden boats. Explain why pitch worked wel

o-na [289]

Tree sap is a viscoelastic polymer, which means it's not really a solid but a very viscous (sticky) fluid. It's fluidity ensures that it can properly <span>seal the spaces between the planks on wooden boats. Not only that, the pitch sap is hydrophobic - it is water resistant as well. </span>

5 0

3 years ago