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Eduardwww [97]
3 years ago
11

Write the chemical equation for the equilibrium that corresponds to Ka. Write the chemical equation for the equilibrium that cor

responds to . H+(aq)+NO−2(aq)⇌HNO2(aq) HNO2(aq)⇌H+(aq)+NO−2(aq) HNO2(aq)⇌H−(aq)+NO+2(aq) HNO2(aq)+H+(aq)⇌H2NO+2(aq) HNO2(aq)+H−(aq)⇌H2NO+2(aq)
Chemistry
1 answer:
viva [34]3 years ago
6 0

Answer:

H+AQ=POP

Explanation:

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At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal con
djyliett [7]
0.114 mol/l  
The equilibrium equation will be: 
Kc = ([Br2][Cl2])/[BrCl]^2  
The square factor for BrCl is due to the 2 coefficient on that side of the equation.  
Now solve for BrCl, substitute the known values and calculate. 
Kc = ([Br2][Cl2])/[BrCl]^2 
[BrCl]^2 * Kc = ([Br2][Cl2]) 
[BrCl]^2 = ([Br2][Cl2])/Kc 
[BrCl] = sqrt(([Br2][Cl2])/Kc)  
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) 
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) 
[BrCl] = sqrt(0.013021127 mol^2/l^2) 
[BrCl] = 0.114110152 mol/l  
Rounding to 3 significant figures gives 0.114 mol/l
4 0
3 years ago
Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c
Anna [14]

The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

d)

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

6 0
3 years ago
What is the purpose of the stairstep pattern on the periodic table of the elements?
Mamont248 [21]

Answer: The bold staircase in the periodic table allows us to classify which elements are metalloids.

Explanation: Additionally, it acts like a "divider" that allows us to properly distinguish the metals from the non-metals in the periodic table.

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                                                    (:  O_O :)

7 0
3 years ago
In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0 g of oxygen. She finds
LuckyWell [14K]

Answer:

The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.

Explanation:

2S (s) + 3O₂ (g) → 2SO₃ (g)

64g    + 96g     →  160 g

32g    + 48g     →   80 g

   x     +     y      →  100 g

   

1 mol SO₃ ___ 80g

     n _______ 100g

         n = 1.25 mol SO₃

1 mol S ___ 32 g

1,25 mol S __ 40 g

1 mol O₂ ___ 32 g

1,875 mol O₂ ___ 60 g

4 0
3 years ago
What is a Control trial
AleksandrR [38]
It is a trial aimed to reduce bias during an experiment. An example would be a sugar pill, something that has no real effect so that the results of the true trial can accurately be compared. Its like a control group.
8 0
3 years ago
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