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Step2247 [10]
3 years ago
10

For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume

there is an excess of oxygen.
2PbSs+3O2g→2PbOs+2SO2g
Chemistry
2 answers:
VMariaS [17]3 years ago
7 0

Answer:

0.356 moles of each products (PbO and SO2) are formed.

Explanation:

First step is to write out the equation and ensure it is balanced before proceeding.

2PbSs + 3O2g → 2PbOs + 2SO2g

2            3            2              2

The equation above is balanced because there are equal number of atoms of elemenents on both sides of the arrow.

By stating that oxygen is in excess, this tells that PbS is the limiting reactant (i.e the amount of products formed depends on it) and not on oxygen.

From the equation;

The products formed are PbO and SO2.

For PbO, it requires 2 moles of PbS to react completely in order to form 2 moles of PbO.

For SO2, it requres 2 moles of PbS to react completely in order to form 2 moles of SO2.

This is gotten from the equation.

PbO

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

SO2

2 moles = 2 moles

0.356 moles = x

upon cross multiplication, we are left with;

x = (0.356 * 2) / 2 = 0.356 moles

galben [10]3 years ago
3 0

Answer:

When 0.356 moles of PbS completely react 0.356 moles of PbO and 0.356 moles of SO₂ are produced.

Explanation:

2PbS (s) + 3O₂ (g) → 2PbO (s) + 2SO₂ (g)

2 mol PbS ____________ 2 mol PbO

0.356 mol PbS ________         x

          x = 0.356 mol PbO

2 mol PbS ____________ 2 mol SO₂

0.356 mol PbS ________         x

          x = 0.356 mol SO₂

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RideAnS [48]

Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve  2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})

Where Cs is the concentration in the surface of the solid

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Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055

Then we solve for x:

x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm

6 0
3 years ago
Which of the following is an irreversible physical change?
Yuliya22 [10]

Answer:

b) sharpening a pencil

Explanation:

If you melt lead, boil water, or dissolve sugar in water, you can return all of them back to their original state. If you sharpen a pencil, you can't reattach the shavings as they were originally.

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A balloon containing helium gas expands from 230
Anit [1.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

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7 0
3 years ago
Consider this equilibrium reaction at 400 K. Br2(g)+Cl2(g)↽−−⇀2BrCl(g)Kc=7.0 If the composition of the reaction mixture at 400 K
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Answer:

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Explanation:

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Now, reaction quotient, Q, is write as the same Kc but the concentrations are actual concentrations:

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