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:
<h3>
brainly.com/question/10257016</h3><h3 /><h3>#SPJ4</h3>
3! You have to ensure balance of all the different elements.
Answer:
Explanation:
final temperature of the cube
initial temperature of the cube
mass of the cube
specific heat of aluminum
1. The answer is option E, that is None of the above is correct.
As a polymer becomes more crystalline,
its melting point doesn't decreases, its density doesn't decreases, its stiffness doesn't decreases and its yield stress doesn't decreases.
2. The answer is option B, that is the molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
In the smectic A liquid-crystalline phase, molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
3. For a substitutional alloy to form, the two metals combined must have similar atomic radii and chemical bonding properties.
Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2
