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Wittaler [7]
3 years ago
8

Can someone help me plz

Chemistry
1 answer:
Alexus [3.1K]3 years ago
4 0

Answer:

Explanation

There is a direct proportional relationship between mass and volume of water. As volume of water increases, the mass of water increases as well.

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<span>3) which is an example of a physical change

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe
Illusion [34]

Answer:

a. Kp=1.4

P_{A}=0.2215 atm

P_{B}=0.556 atm

b.Kp=2.0 * 10^-4

P_{A}=0.495atm

P_{B}=0.00995 atm

c.Kp=2.0 * 10^5

P_{A}=5*10^{-6}atm

P_{B}=0.9999 atm

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

Kp=\frac{(P_{B})^{2}}{P_{A}}

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibrium x       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Kp=\frac{(1-2x)^{2}}{x}

Working with this equation:

x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a} (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128

x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215

With x1 we get a partial pressure of:

P_{A}=1.128 atm

P_{B}=1-2*1.128 = -1.256 atm

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

P_{A}=0.2215 atm

P_{B}=1-2*0.2215 = 0.556 atm

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

P_{A}=0.505atm

P_{B}=1-2*0.505 = -0.01005 atm

Not sense.

With x2

P_{A}=0.495atm

P_{B}=1-2*0.495 = 0.00995 atm

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

X2=5*10^{-6}

With x1

P_{A}=50001atm

P_{B}=1-2*50001=-100001atm

Not sense.

With x2

P_{A}= 5*10^{-6}atm

P_{B}=1-2*5*10^{-6}= 0.9999 atm

X2 is the solution for this equation.  

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Material deposited directly by a glacier is called ______.
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C. till .....................................................................
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Read 2 more answers
Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie
aleksklad [387]
<h2>H_2O  + H_2 + O_2</h2>

Explanation:

1. Water decomposition

  • Decomposition reactions are represented by-

       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
  1. Electrolysis
  2. Photoelectrochemical water splitting
  3. Thermal decomposition of water
  4. Photocatalytic water splitting
  • Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
  • The chemical equation will be -

        H_2O  + H_2 + O_2

Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

     ∴The balanced equation is -

       2 H_2O → 2 H_2 + O_2

2. Formation of ammonia

  • The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

      N_2 + H → NH_3

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

   ∴The balanced chemical equation for the formation of ammonia gas is as  follows -

     N_2+3H→ 2NH_3.

  • When 6 moles of N_2 react with 6 moles ofH_2 4 moles of ammonia are produced.

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2 years ago
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