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poizon [28]
3 years ago
9

Given the equation representing a system at equilibrium:

Chemistry
1 answer:
polet [3.4K]3 years ago
3 0
I think the one that cause the equilibrium to shift would be :
3. adding a noble gas
Adding the noble gas will add more concentration to the KNO3, which will create different amount of equilibrium

hope this helps
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A sample of nitrogen gas expands in volume from 16 L to 5.4 L at constant
xxTIMURxx [149]

a. W = 0 J

b. W = - 308.028 J

<h3>Further explanation</h3>

Given

Nitrogen gas expands in volume from 1.6 L to 5.4 L

Required

The work done

Solution

Isothermal :

W = -P . ΔV

Input the value :

a. At a vacuum, P = 0

So W = 0

b. At pressure = 0.8 atm

W = - 0.8 x ( 5.4 - 1.6)

W = -3.04 L.atm ( 1 L.atm = 101.325 J)

W = - 3.04 x 101.325

W = - 308.028 J

6 0
3 years ago
What Is a combination of substances that are combined physically but not chemically
adoni [48]
It is a heterogenous mixture
7 0
3 years ago
Read 2 more answers
Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.
Sever21 [200]

Answer:

The spectator ions is: K^+   and  SO^{2-}_4

Explanation:

The equation of reaction between  H₂ SO₄ and KOH is:

H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}

Rewriting this equation as ionic;

[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+  SO_4^{2-} + 2H_2O ]

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

5 0
3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
3 years ago
A double replacement reaction is a reaction in which one element replaces a similar element within a compound.
ladessa [460]
False. What you’re describing is a displacement reaction.
6 0
3 years ago
Read 2 more answers
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