Answer is: ammonia experience only dispersion intermolecular forces with BF₃ (boron trifluoride) because BF₃ is only nonpolar molecule (vectors of dipole moments cansel each other, dipole moment is zero).
The London dispersion force (intermolecular force) <span>is a temporary attractive </span>force between molecules.
Answer:
I'm thinking cooper but not sure
Answer:
This involves negatively charged particles (electrons) jumping to positively charged objects. When you rub the balloons against the fabric they become negatively charged. They take some of the electrons from the fabric and leave them positively charged.
Explanation:
Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon. When both balloons are rubbed with the wool cloth, the both receive negative charges, so they will repel each other.
Answer:
129,600kg/day
Explanation:
The river is flowing at 30.0
1 m³ is equivalent to 1000L
flowrate of river = 30*1000 =30,000L/s
Convert L/s to litre per day by multiplying by 24*60*60
flowrate of river = 30,000 * 24*60*60 L/day
= 2,592,000,000L/day
if the river contains 50mg of salt in 1L of solution
lets find how many mg of salt Y is contained in 2,592,000,000L/day
by cross multiplying we have
Y=
Y= 129,600,000,000 mg/day
convert this value to kg/day by dividing by 1 million
Y= 129,600,000,000/1000000
Y= 129,600kg/day
Answer:
![\boxed {\boxed {\sf 0.495 \ mol}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%200.495%20%5C%20mol%7D%7D)
Explanation:
We are given a number of particles and asked to convert to moles.
<h3>1. Convert Particles to Moles </h3>
1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.
So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.
![\frac { 1 \ mol }{6.022*10^{23 } \ particles}}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%201%20%5C%20mol%20%7D%7B6.022%2A10%5E%7B23%20%7D%20%5C%20particles%7D%7D)
We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.
![2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}](https://tex.z-dn.net/?f=2.98%2A10%5E%7B23%7D%20%5C%20particles%20%2A%5Cfrac%20%7B%201%20%5C%20mol%20%7D%7B6.022%2A10%5E%7B23%20%7D%20%5C%20particles%7D%7D)
The units of particles cancel.
![2.98*10^{23} *\frac { 1 \ mol }{6.022*10^{23 } }}](https://tex.z-dn.net/?f=2.98%2A10%5E%7B23%7D%20%20%2A%5Cfrac%20%7B%201%20%5C%20mol%20%7D%7B6.022%2A10%5E%7B23%20%7D%20%7D%7D)
![\frac { 2.98*10^{23}}{6.022*10^{23 } }} \ mol](https://tex.z-dn.net/?f=%5Cfrac%20%7B%202.98%2A10%5E%7B23%7D%7D%7B6.022%2A10%5E%7B23%20%7D%20%7D%7D%20%20%5C%20mol)
![0.4948522086 \ mol](https://tex.z-dn.net/?f=0.4948522086%20%5C%20mol)
<h3>2. Round</h3>
The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.
For the number we found, 3 sig figs is the thousandth place.
The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.
![0.495 \ mol](https://tex.z-dn.net/?f=0.495%20%5C%20mol)
2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>