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IgorLugansk [536]
3 years ago
9

Enter the net ionic equation, including phases, for the reaction of AgNO3(aq) with K2SO4(aq). Solubility Rules are found here.

Chemistry
1 answer:
Inessa05 [86]3 years ago
5 0
Write out the equation: 
AgNO3(aq)+K2SO4(aq) --> Ag2SO4(s)+K2(NO3)... 

Balance the equation: 
2 AgNO3(aq)+K2SO4(aq) --> Ag2SO4(s)+K2(NO3)2... 

Remember to keep molecules intact when you write out the ions.
(2Ag+)+(2NO3-)+(2K+)+(SO42-) --> (Ag2SO42
Cancel out all the spectator ions. These are ions that watch the reaction that didn't change. For this problem 2K+ & 2NO3- are the spectator ions. 

You are left with your net ionic equation of
2Ag+(aq)+SO42-(aq)-->Ag2SO4(s)
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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so
Illusion [34]
Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result  in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl
4 0
3 years ago
Read 2 more answers
In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (
Natasha2012 [34]

Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>

This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>

<em />

I hope it helps!

5 0
3 years ago
Brain’s family has a compost pile at home they need to put all the weeds the pull up from the garden in it as well as any vegeta
Flauer [41]

Answer: b. It would happen faster at warmer air temperatures

Explanation:saw another site say this was the answer

5 0
3 years ago
How do compounds differ from their component elements? A. Compounds have different numbers of neutrons than their component elem
Rudiy27
The correct answer is C. Compounds have different properties than their component elements. When compounds are formed, they undergo change resulting to changes in the properties they exhibit. It will have its new set of properties different from the elements.
6 0
2 years ago
If you start with 13 mole H2O and 50 grams of O2 what is the percent yield of H202 if 100 grams of H202 is actually made?
dmitriy555 [2]

Answer:

94.1 %

Explanation:

We firstly determine the equation:

2H₂O + O₂ →  2H₂O₂

2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.

We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol

Certainly oxygen is the limiting reactant.

2 moles of water react to 1 mol of oxygen.

13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)

As we determine the limiting reactant we continue to the products:

1 mol of O₂ can produce 2 moles of H₂O₂

Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles

We convert the moles to mass: 3.125 mol . 34 g/mol= 106.25 g

That's the 100% yield or it can be called theoretical yield.

Percent yield = (Yield produced / Theoretical yield) . 100

(100g / 106.25 g) . 100 = 94.1 %

3 0
3 years ago
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