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snow_tiger [21]
3 years ago
10

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

days. If you begin with 34.7 mg of this isotope, what mass remains after 8.52 days have passed
Chemistry
1 answer:
evablogger [386]3 years ago
8 0

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

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How many moles of sucrose<br> are in 5.25x1029 sucrose<br> molecules? (in scientific notation)
Lorico [155]

Answer:

8.72 × 10^5 moles

Explanation:

To find the number of moles in 5.25 x 10^29 molecules of sucrose, we divide the number of molecules by Avagadro constant (6.02 × 10²³ molecules). That is;

no. of moles = no. of molecules ÷ 6.02 × 10²³ molecules

In this case of sucrose, no of moles contained is as follows;

5.25 × 10^29 ÷ 6.02 × 10²³

5.25/6.02 × 10^ (29-23)

0.872 × 10^6

= 8.72 × 10^5 moles

5 0
3 years ago
Describe four stresses that upset the equilibrium of a chemical reaction
julia-pushkina [17]

Answer:

Concentration, pressure, and temperature are the three factors that affect chemical equilibrium

5 0
3 years ago
3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.
tiny-mole [99]

Answer:

The Avogadro's  number is N_A     =  6.02289 *10^{23}

Explanation:

From the question we are told that

   The edge length is  L   = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100}  =  5.02 * 10^{-10}

    The density of the metal is \rho =  5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3}  * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3

     The molar mass of  Ba is  Z  =  137.3 \ g/mol = \frac{137.3}{1000} =  0.1373 \  kg / mol

     

Generally the volume of a unit cell is  

       V =  L^3

substituting value

        V =  [5.02 *10^{-10}]^3

         V = 1.265*10^{-28}\ m^3  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        V_a  =  \frac{V}{2}  * 0.68

substituting value

        V_a  =  \frac{ 1.265*10^{-28}}{2}  * 0.68

        V_a  = 4.301 *10^{-29} \ m^3

The Molar mass of barium is mathematically represented as

      Z  =  N_A V_a *  \rho

Where N_A is the Avogadro's number

 So  

      N_A     =   \frac{ Z}{ V_a *  \rho}

substituting value

     N_A     =   \frac{ 0.1373}{ 4.301*10^{-29} *  5.3*10^{3}}

     N_A     =  6.02289 *10^{23}

4 0
3 years ago
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
How many molecules are there in 79g of Fe2O3?
weqwewe [10]
There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms
6 0
3 years ago
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