Answer:
45 mL
Explanation:
Tenemos los siguientes datos:
V = 1 L
C = 4,5% v/v
El porcentaje en volumen (%v/v) expresa el volumen de soluto (alcohol en este caso) que hay cada 100 mL de solución. Si la solución tiene una concentración del 4,5% v/v eso quiere decir que hay 4,5 ml de alcohol cada 100 ml de solución, de acuerdo a lo siguiente:
4,5% v/v alcohol = volumen alcohol/ volumen solución x 100 = 4,5 mL alcohol/100 mL solución= 4,5 mL alcohol/0,1 L alcohol
Por lo tanto, al multiplicar por el volumen total de la solución (1 L), obtenemos la cantidad total de alcohol:
4,5 mL alcohol/0,1 L alcohol x 1 L = 45 mL
Answer:
This is because, within a period or family of elements, all electrons are added to the same shell.
The following is the introduction to a special e-publication called Determining the Age of the Earth (click the link to see a table of contents). Published earlier this year, the collection draws articles from the archives of Scientific American. In the collection, this introduction appears with the title, “Stumbling Toward an Understanding of Geologic Timescales.”
Well when a particle of air is becomes heated it rises, right? So you could write some like you started off close to the earth (aka the troposphere) until you became heated then you started to rise and as you reached higher elevations you cooled down and you were recycled into cool air and you moved back down and became new fresh cool air until the next time you'll become heated and rise again to be recycled into fresh cool new air.
The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980
