The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
Answer:
C.
Explanation:
para sakin letter C ganonn
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
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
