<h2><u>Answer:</u></h2>
It wasn't an adjustment in the condition of issue on the grounds that the vitality in the can did not change. Additionally, since this was a physical change, the atoms in the can are as yet similar particles. No synthetic bonds were made or broken. You added enough vitality to make a stage change from strong to fluid.
The main changes recorded which don't include framing or breaking substance bonds would bubble and liquefying. Bubbling and liquefying are physical changes as opposed to synthetic changes, so no new items are shaped.
Answer:
c
Explanation:
igneous, sedimentary, and metamorphic. igneous forms from molten rock cools and solidifides
Answer:
heated test tube in cold water
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer:
Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)
If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C
Using dc = (1/4πεo)qQ/Eα we have
dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm
Note: 1meter = 10^15fentometer
Explanation:
This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.
