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lara [203]
2 years ago
6

Phosphorus-32 (32P) is an isotope that is commonly used for medical and biological research. Phosphorus-32 has a half-life of 14

.28 days. A researcher orders a sample of 32P that has an activity of 5.0 mCi when it arrives. If delivery of the sample took 48 hours, what was the activity of the sample when it was shipped from the manufacturer
Chemistry
1 answer:
Pachacha [2.7K]2 years ago
5 0

The activity of the sample when it was shipped from the manufacturer is 4.54 mCi

<h3>How to determine the number of half-lives that has elapsed </h3>

From the question given above, the following data were obtained:

  • Time (t) = 48 hours
  • Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
  • Number of half-lives (n) =?

n = t / t½

n = 48 / 342.72

n = 0.14

<h3>How to determine the activity of the sample during shipping </h3>
  • Number of half-lives (n) = 0.14
  • Original activity (N₀) = 5.0 mCi
  • Activity remaining (N) =?

N = N₀ / 2ⁿ

N = 5 / 2^0.14

N = 4.54 mCi

Thus, the activity of the sample during shipping is 4.54 mCi

Learn more about half life:

brainly.com/question/2674699

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2 years ago
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C3H8 + 5O2 → 3CO2+ 4H2O, if 5.75L of oxygen are consumed in the above reaction, how many L of carbon dioxide are produced?
anzhelika [568]

Answer: 3.45 L carbon dioxide are produced

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{5.75L}{22.4L}=0.257moles

C_3H_8+5O_2(g)\rightarrow 3CO_2+4H_2O  

According to stoichiometry :

5 moles of O_2 produce =  3 moles of CO_2

Thus 0.257 moles of O_2 will produce=\frac{3}{5}\times 0.257=0.154moles  of CO_2  

Volume of CO_2=moles\times {\text {Molar volume}}=0.154moles\times 22.4L/mol=3.45

Thus 3.45 L carbon dioxide are produced

6 0
2 years ago
At Jim's auto shop, it takes him minutes to do an oil change and minutes to do a tire change. Let be the number of oil changes h
Juli2301 [7.4K]

Answer:

The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Explanation:

Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

5 0
3 years ago
(Digestion, Intestine, Bile)is secreted by the liver to help in the breaking down of food particles.​
Yuri [45]

Answer:

Bile

Explanation:

Bile is secreted by the liver to help in the breaking down of food particles.​

5 0
1 year ago
Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.
-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (NH_2) and a <em>carboxylic acid</em> group (COOH).  Additionally, we have an <u>alcohol </u>(CH_3CH_2OH) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (OH^-).

When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (H_2O). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base (OH^-) removes the hydrogen from the C=O bond to produce ethyl L-valinate

See figure 1

I hope it helps!

7 0
3 years ago
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