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Aleks04 [339]
2 years ago
11

Assume that 8.5 L of iodine gas are produced at STP according to the following balanced equation:

Chemistry
2 answers:
katovenus [111]2 years ago
3 0
Balanced chemical equation is as;

<span>                                  2 Kl  +  Cl</span>₂     →     <span>2 KCl  +  I</span>₂

1)  Moles of I₂<span> produced;

             22.4 L volume is occupied by</span>   =  1 mole of I₂ at STP
So,
             8.5 L of I₂ will be occupied by  =  X mole of I₂ at STP

Solving for X,
                       X  =  (8.5 L × 1 mol) ÷ 22.4 L

                       X  =  0.379 Moles of I₂ 

2)  Moles of Cl₂ used;

                         22.4 L I₂ utilized  =  1 mole of Cl₂  gas
So,
                    8.5 L of I₂ will utilize =  X mole of Cl₂  gas

Solving for X,
                       X  =  (8.5 L × 1 mol) ÷ 22.4 L

                       X  =  0.379 Moles of Cl₂ 

3)  <span>Grams of Cl</span>₂<span> used;

As,
                       Moles  =  Mass / M.mass
Or,
                       Mass  =  Moles </span>× M.mass

                       Mass  =  0.379 mol × 70.90 g.mol⁻¹

                       Mass  =  26.87 grams of Cl₂
Nat2105 [25]2 years ago
3 0

Answer:

  1. 0.38
  2. 0.38
  3. 27

             

Explanation:

right on edge

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Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)
gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}, where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr   7.96 torr  0       0            <em> initial partial pressure</em>

-x                  -x            +x     +x          <em> react</em>

105.1-x       7.96-x      x        x            <em>equilibrium</em>

Then:

Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}

2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0
2 years ago
Read 2 more answers
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0
2 years ago
An electrically neutral atom of gallium has 31 electrons and 39 neutrons. What is the mass number for an atom of gallium?. A. 31
hammer [34]
If the atom is neutral (meaning, not charged) the number of electron is equal to the number of protons. The mass number of an atom is the sum of the number of proton and the number of neutrons. From the given above, the mass number of gallium is 31 + 39. The answer is letter D. 70.
4 0
2 years ago
Read 2 more answers
Use the specific heat capacity that you calculated for granite to determine how many grams of granite at the initial temperature
Valentin [98]

Answer:

Explanation:

specific heat of granite s = .79 J / g / k

let the mass of granite = m

heat lost by granite = heat gained by water

heat lost = mass x specific heat x drop in temperature

= m x .79 x (80 - 20.45)

heat gained by water

= 3000 x 4.186 x (20.45- 20)

heat lost by granite = heat gained by water

m x .79 x 59.55  =  3000 x 4.186 x .45

m = 120.12 g .

4 0
2 years ago
A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a
Nikolay [14]

This is an incomplete question, the given sketch is shown below.

Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)

Explanation :

Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.

There are three types of unit cell.

  • SCC (simple-centered cubic unit cell)
  • BCC (body-centered cubic unit cell)
  • FCC (face-centered cubic unit cell)

In SCC, the atoms are arranged at the corners.

Z=\frac{1}{8}\times 8=1

The number of atoms of unit cell = Z = 1

In BCC, the atoms are arranged at the corners and the body center.

Z=\frac{1}{8}\times 8+1=2

The number of atoms of unit cell = Z = 2

The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

Z=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4

The number of atoms of unit cell = Z = 4

Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)

7 0
2 years ago
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