Answer:
no of atoms = 13.9 x 10^25
Explanation:
No. of moles = mass of compound / molar mass of compound
As ; mass of sodium = 5.3 x 10^3 g
Molar mass of sodium = 22.9 g/ mol
putting values
n = 5.3 x 10^3 / 22.9
n = 231.4 mol
Also; no of mol (n) = no of particles / Avagadros number
so no of particles = n x Avagadros no.
put n = 231.4 and Avagadros no = 6.023 x 10^23
no of particles = 231.4 x 6.023 x 10^23
= 13.9 x 10^25
Answer:
589.038 m/s
Explanation:
i dont know if did this right tho
The electrostatic potential is the work done to remove the charge. The molecule showing the potential map shows that it is HI.
<h3>What is electronegativity?</h3>
Electronegativity is the elemental property to attract the electron toward the atom and is affected by atomic number and the atomic radius. The map shows that the molecule is made of two different atoms and has linear geometry.
Due to the linear geometry, the molecule cannot be Br₂ and SO₃ as they are nonpolar and nonlinear respectively. Similarly, BrF and ICl can be eliminated as they are interhalogen compounds. In the HI molecule, the hydrogen atom is a cation and the iodine atom is an anion that has high electronegativity differences.
Therefore, HI is the molecule depicted in the potential map.
Learn more about the electrostatic potential here:
brainly.com/question/14889552
#SPJ4
Answer:
7628 y
Explanation:
Carbon-14 is radioactive and it follows the first-order kinetics for a radioactive decay. The first-order kinetics may be described by the following integrated rate law:
![ln(\frac{[A]_t}{[A]_o})=-kt](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%3D-kt)
Here:
is the mass, moles, molarity or percentage of the material left at some time of interest t;
is the mass, moles, molarity or percentage of the material initially, we know that initially we expect to have 100 % of carbon-14 before it starts to decay;
is the rate constant;
is time.
The equation becomes:
![ln(\frac{[A]_t}{[A]_o})=-\frac{ln(2)}{T_{\frac{1}{2}}}t](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%3D-%5Cfrac%7Bln%282%29%7D%7BT_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7Dt)
Given:
![\frac{[A]_t}{[A]_o} = \frac{40.0 %}{100.0 %}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%20%3D%20%5Cfrac%7B40.0%20%25%7D%7B100.0%20%25%7D)

Solve for time:
![t = -\frac{ln(\frac{[A]_t}{[A]_o})\cdot T_{\frac{1}{2}}}{ln(2)}](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7Bln%28%5Cfrac%7B%5BA%5D_t%7D%7B%5BA%5D_o%7D%29%5Ccdot%20T_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bln%282%29%7D)
In this case:
