1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ExtremeBDS [4]
3 years ago
6

The radioisotope sodium-24 is used to determine the levels of electrolytes in the body. A 16-μg sample of sodium-24 decays to 2.

0 μg in 45 h. What is the half-life, in hours, of sodium-24?
Chemistry
1 answer:
mario62 [17]3 years ago
8 0

Answer:

t = 15 hours

Explanation:

To get this, we need to write the general expression for the mass or concentration in life decay which is the following:

A = Ao * e^(-kt)   (1)

K is a constant decay and depends of the half life. The expression is:

k = ln2 / t₁/₂   (2)

In this question, we have the innitial mass and the final mass, and the time required to get that mass. So, with these data, we first solve for the constant k, and then, use this value to get the half life. In other words, we solve for k in (1) and then, solve for half life in (2):

2 = 16 * e(-45k)

2/16 = e(-45k)

0.125 = e(-45k)

ln(0.125) = -45k

k = -ln(0.125)/45

k = 0.0462 h⁻¹

Now that we have k, let's calculate half life:

t₁/₂ = ln2/k

t₁/₂ = ln2 / 0.0462

t₁/₂ = 15 h

This is the half life of sodium-24

You might be interested in
The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano
Lubov Fominskaja [6]

Answer:

Ka = 1.52 E-5

Explanation:

  • CH3-(CH2)2-COOH ↔ CH3(CH2)2COO-  + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

3 0
3 years ago
An experiment produced 0.10 g CO2, with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is t
Pachacha [2.7K]
The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.
3 0
3 years ago
What will most likely happen when two bromine atoms bond together?
ikadub [295]

Answer:

A<u> covalent bond</u> will hold them together.

Explanation:

The two bromine atoms will share electrons to build a stronger bond and have a full valence outer shell (which makes them stable).

Hope this helps!

5 0
3 years ago
Read 2 more answers
An ice cream sundae is a(n)
Brrunno [24]

Answer:

A frozen dessert that tastes good and makes people happy

:)

Explanation:

5 0
3 years ago
5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas
Artist 52 [7]

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)

where,

\Delta U = change in internal energy = ?

n = number of moles of gas = 5 moles

C_v = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta U=nC_v(T_2-T_1)

\Delta U=(5moles)\times (2R)\times (573-298)

\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)

\Delta U=22863.5J=22.86kJ

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)

where,

\Delta H = change in enthalpy = ?

n = number of moles of gas = 5 moles

C_p = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta H=nC_p(T_2-T_1)

\Delta H=(5moles)\times (3R)\times (573-298)

\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)

\Delta H=34295.25J=34.29kJ

The change in enthalpy of the gas is 34.29 kJ.

4 0
3 years ago
Other questions:
  • What is the number of protons that the element
    11·1 answer
  • What is the difference in concentration between a pH of 7 and 12?
    8·2 answers
  • You cannot round a number if it has a decimal point true or false
    14·1 answer
  • 150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o
    14·1 answer
  • The percent abundance of each isotope tells you how many of each kind of isotope exist in every 100 particles. What does relativ
    6·1 answer
  • What changes would occur at a molecular level if a liquid is placed in cool conditions?
    15·1 answer
  • Which need, want value, or interest was most likely involved in the
    7·2 answers
  • Comment Both propane and benzene are hydrocarbons. As a rule,
    15·1 answer
  • Help me out ? I'm not sure of my answer.<br>​
    13·1 answer
  • Tear off a small, flat sheet of waxed paper. Use the pipette to dispense a drop of water on the waxed paper. Now, dry the pipett
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!