Making repeated separations of the various substances in the pitchblende, Marie and Pierre used the Curie electrometer to identify the most radioactive fractions. They thus discovered that two fractions, one containing mostly bismuth and the other containing mostly barium, were strongly radioactive.
<h3>What was surprising about pitchblende?</h3>
Since it was no longer appropriate to call them “uranic rays,” Marie proposed a new name: “radioactivity.”
Even more surprising, Marie next found that a uranium ore called pitchblende contained two powerfully radioactive new elements: polonium, which she named for her native Poland, and radium.
<h3>Why is radium more radioactive than uranium?</h3>
It is 2.7 million times more radioactive than the same molar amount of natural uranium (mostly uranium-238), due to its proportionally shorter half-life.
Learn more about highly radioactive elements here:
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brainly.com/question/10257016</h3><h3 /><h3>#SPJ4</h3>
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
Answer:
8.33 moles CO2 X. 25mol O2. 16mol CO2. = 13.0 moles
Explanation:
no of moles = no of atoms ÷ avogadro's number
= (9.8×10^24) ÷ (6.02×10^23)
