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1 year ago

A coin collector has a bag of 50 pennies. What is the percent abundance of the post-1982 pennies? ASAP ANSWER PLS.

Chemistry

1 answer:

Novosadov [1.4K]1 year ago

5 0

The percent abundance of the post-1982 pennies is 56%.

<h3>What is percent abundance?</h3>

Percent abundance is the percentage amount of all naturally occurring isotopes of an element.

<h3> Percent abundance of the post-1982 pennies</h3>

The percent abundance of the post-1982 pennies is calculated as follows;

percent abundance = number of post-1982 pennies / total number of pennies x 100%

percent abundance = (28) / (50) x 100%

percent abundance = 56%

Thus, the percent abundance of the post-1982 pennies is 56%.

Learn more about percent abundance here: brainly.com/question/6844925

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Wind patterns choose all that apply

kumpel [21]

I think the answer is B.

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3 years ago

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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3 years ago

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ra1l [238]

Hi yes bestie just make sure you eat enough so that you can actually build the mussel

8 0

3 years ago

A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal

Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent: 0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

| masss | volume | mol

solute | | |

solvent | | |

solution | | |

Let's complete, what we have.

| masss | volume | mol

solute | 10.8g | |

solvent | | 133 mL |

solution | | 137 mL |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

| masss | volume | mol

solute | 10.8g | | 0.110 mol

solvent | 133g | 133 mL | 7.39 mol

solution | 143.8g | 137 mL | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles or solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

5 0

3 years ago

Propane is often used to heat homes. The combustion of propane follows the following reaction: C3H8(g) + 5O2(g)  3CO2 (g) + 4H

swat32

Answer:

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Explanation:

<u>Step 1</u>: Data given

C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ

This means every mole C3H8

Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)

<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat

1 mol = 2044 kJ

x mol = 7563 kJ

x = 7563/2044 = 3.70 moles

To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8

<u>Step 3: </u>Calculate mass of propane

Mass propane = moles * Molar mass

Mass propane = 3.70 moles * 44.1 g/mol

Mass propane = 163.17 grams

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

7 0

3 years ago