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Rama09 [41]
4 years ago
13

Chemist and physicist marie curie became famous for her pioneering research on radioactivity. What are the two chemical elements

that she discovered?
Chemistry
2 answers:
Kryger [21]4 years ago
5 0

Answer: She discovered polonium and radium.

Explanation: In 1896, intrigued by the physicist Henri Becquerel’s accidental discovery of radioactivity, Curie began studying uranium rays; Pierre soon joined her in her research. Two years later, the Curies discovered polonium—named after Marie’s homeland—and radium. In 1903 they shared the Nobel Prize in physics with Becquerel for their groundbreaking work on radioactivity.

wolverine [178]4 years ago
4 0

Answer:

The two elements are POLONIUM and RADIUM.

Explanation:

Maria Curie is a French physicist and chemist, though she was of a Polish naturals. She was the first woman to receive a Noble Price which she earned for conducting leading and head way research on radioactivity. She discovered the theory of radioactivity; also the techniques isolating radioactive isotopes. These helped her and her husband discover Polium and Radium.

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3 years ago
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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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