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Rama09 [41]
3 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]3 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]3 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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Please help me<br> 5. Where is potential energy decreasing?<br> А B C D
Nataly [62]

Answer:

D

Explanation:

I believe it is D. your kinetic energy would be at b. A, the cart would be going at a constant rate, because there is no hill or steep slope.

7 0
2 years ago
Read 2 more answers
Select the correct answer. Given: 2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams
ivann1987 [24]

The reaction produces 2.93 g H₂.

M_r:                        133.34  2.016

        2Al + 6HCl → 2AlCl₃ + 3H₂

<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

7 0
3 years ago
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
Is this right, I feel like its wrong
hjlf

It is correct, next time re-check your answer and don't second guess yourself. ;3

7 0
3 years ago
There are 9.88x1023 molecules of O2 available.
blondinia [14]

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many  '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

7 0
2 years ago
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