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vlabodo [156]
3 years ago
10

How many days have passed if 4.80 mg of calcium-47 decayed to 1.2 mg of calcium-47?

Chemistry
1 answer:
LenaWriter [7]3 years ago
6 0
(1) 36 days is 8 half-life periods,
so 24mg becomes 24mg times (1/2)^8 = 24 mg/256 = 0.094 mg

(2) 9.6/1.2 = 8, which means three half-life periods went by.
At the end of 4.5 days, there were 4.8 mg
At the end of 9.0 days, there were 2.4 mg
At the end of 13.5 days, there were 1.2 mg

(3) The activity is proportional to the amount of radioactive material remaining,
so after two half-life periods, the activity should be 1/4 of what it originally was.
So the initial activity was probably 12.0 Ci
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Answer:

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Explanation:

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3 years ago
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In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed
finlep [7]

The ratio of H⁺ ions to OH⁻ ions at a pH = 2 is 10¹⁰

<h3>Further explanation</h3>

Given

ph = 2

Required

The concentration of H⁺ and OH⁻ ions

Solution

  • The concentration of H⁺ ions

pH=-log[H⁺]

2=-log[H⁺]

[H⁺]=10⁻²

  • The concentration of OH⁻ ions

pH+pOH=14

pOH=14-2

pOH=12

pOH=-log[OH⁻]

12=-log[OH⁻]

[OH⁻]=10⁻¹²

  • The ratio of H⁺ ions to OH⁻ ions at a pH = 2

\tt \dfrac{10^{-2}}{10^{-12}}=10^{10}

6 0
2 years ago
Not sure....HELP!!!!!!!
Murrr4er [49]
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4 0
3 years ago
Which of these is a chemical property of matter?
Hitman42 [59]

Answer:

A. Ability to burn

Explanation:

8 0
2 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.  

8 0
3 years ago
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