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Iteru [2.4K]
4 years ago
6

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Chemistry
1 answer:
Rufina [12.5K]4 years ago
6 0
Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t =  ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co  = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr
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The molecular weight of potassium ferricyanide is 329.24 grams/mol. How would you make 1 liter of a 500mM potassium ferricyanide
True [87]

Answer:

w = 164.62 g

Explanation:

molarity of a solution is given as -

Molarity (M)  =  ( w / m ) / V ( in L)

where ,

m = molecular mass ,

w = given mass ,

V = volume of solution ,

From the question ,

M = 500 mM = 0.5 M

( since , 1 mM =  1 / 100 M)

As we know , the molecular mass of potassium ferricyanide = 329.24 g/ mol

V = vol.of solution = 1 L

w = ?

<u>To find the value of w , using the above formula , and putting the respected values , </u>

Molarity (M)  =  ( w / m ) / V ( in L)

0.5  =  ( w / 329.24 ) / 1 L

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6 0
3 years ago
the specific heat capacity of water is 1 cal/gc how much heat in calories is released when 25.7 of water is cooled from 85 to 49
liraira [26]

Answer:

The amount of heat that is released is -925.2 cal

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body can receive or release without affecting its molecular structure, that is, it does not change the state (solid, liquid, gaseous).  In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state.

The equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • c= 1 \frac{cal}{g*C}
  • m= 25.7 g
  • ΔT= Tfinal - Tinitial= 49 °C - 85 °C= -36 °C

Replacing:

Q= 1 \frac{cal}{g*C} *25.7 g* (-36 C)

Solving:

Q= -925.2 cal

<u><em>The amount of heat that is released is -925.2 cal</em></u>

8 0
3 years ago
What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3
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Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})

We are given:

K_a = Dissociation constant of propanoic acid = 1.3\times 10^{-5}

[CH_3CH_2COONa]=0.254 M

[CH_3CH_2COOH]=0.329 M

pH = ?

Putting values in above equation, we get:

pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})

pH = 4.77

4.77 is the pH of the given buffer .

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3 years ago
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