If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? <span>
A. one-fourth</span>
When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.
<u>Explanation</u>:
- Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
- The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.
- The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.
Answer: 1.27 bar
Explanation:
1 atm = 1.01325 bar
1.25 atm = Z (let Z be the unknown value)
To get the value of Z, cross multiply
Z x 1 atm = 1.25 atm x 1.01325 bar
1 atm•Z = 1.2665625 atm•bar
To get the value of Z, divide both sides by 1 atm
1 atm•Z/1 atm = 1.2665625 atm•bar/1atm
Z = 1.2665625 bar
(Round up Z to the nearest hundredth as 1.27 bar)
Thus, 1.25 atm when coverted gives 1.27 bar
The answer is: the pressure inside a can of deodorant is 1.28 atm.
Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
p₁/T₁ = p₂/T₂.
p₁ = 1.0 atm.; initial pressure
T₁ = 15°C = 288.15 K; initial temperature.
T₂ = 95°C = 368.15 K, final temperature
p₂ = ?; final presure.
1.0 atm/288.15 K = p₂/368.15 K.
1.0 atm · 368.15 K = 288.15 K · p₂.
p₂ = 368.15 atm·K ÷ 288.15 K.
p₂ = 1.28 atm.
As the temperature goes up, the pressure also goes up and vice-versa.
Answer: sorry I’m late but it is 11 electrons
Explanation:
