Answer:
The weight of the Astroturf is 179,684.31 Newtons.
Explanation:
Length of a soccer field = 0.06214 mile = 328.0992 feet
(1 mile = 5280 feet)
Breadth of a soccer field = 253 feet
Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet
Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet
Thickness of a Astroturf with which soccer field is to be covered = h
h = ½ inch = 0.5 inch = 0.041665 feet
(1 inches = 0.08333 feet)
Volume of the Astroturf ,V= l × b × h
Mass of the Astroturf = m
Density of the Astroturf = d = 187 
1 oz = 0.0283495 kg
Weight of the Astroturf = W
W = mg
The weight of the Astroturf is 179,684.31 Newtons.
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
The answer is (2) 10.0 mol. The equation given is balanced. So the ratio of mol number of compound is the ratio of the number before the compound. The HCl and CO2 ratio is 2:1. So the answer is 10.0 mol.
You first need to write the balanced chemical reaction for what is going on.
Ca(OH)₂+2HCl→2H₂O+CaCl₂
After you make the balanced chemical reaction, First you find the moles of HCl used. To do this multiply 0.0375L by 0.124M to get 0.00465mol HCl. Then you multiply 0.00465mol HCl by (1mol Ca(OH)₂)/(2mol HCl) to get 0.002325mol Ca(OH)₂. Finally to find concentration of Ca(OH)₂ used you divide 0.002325mol by 0.020L to get 0.116M Ca(OH)₂.
Therefore the concentration of the unknown solution of Ca(OH)₂ was 0.116M.
I hope this helps. Let me know if anything is unclear.
Answer:
if the density is higher than water than the object will sink
Explanation:
