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lorasvet [3.4K]
2 years ago
8

Medical diagnosis and ______ are common uses of radioactive elements in health care.

Chemistry
1 answer:
KengaRu [80]2 years ago
6 0
<span>The half-life of Carbon 14 and radionuclides are used to estimate the absolute (versus relative) age of pre-history items </span>
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Answer:

2nd option

Explanation:

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3 years ago
Which of the following equations has the correct products and is balanced correctly for a reaction between Na3PO4 and KOH?
ASHA 777 [7]

The answer is: A) Na3PO4 + 3KOH → 3NaOH + K3PO4, because K retains the same charge throughout the reaction.

This chemical reaction is double displacement reaction - cations (K⁺ and Na⁺) and anions (PO₄³⁻⁻ and OH⁻) of the two reactants switch places and form two new compounds.  

Na₃PO₄ is sodium phosphate.  

KOH is potassium hydroxide.  

NaOH is sodium hydroxide.  

K₃PO₄ is potassium phosphate.  

According to the mass conservation law, there are same number of atoms on both side of balanced chemical reaction.

5 0
3 years ago
2. What happens to the volume of a balloon when it is taken outside on a cold winter<br> day? Why?
forsale [732]

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4 0
2 years ago
"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
Aleksandr-060686 [28]

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

8 0
3 years ago
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