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boyakko [2]
2 years ago
14

Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv

e particles aimed at the gold foil. Then, explain a test that might detect whether particles are emitted from thorium-234. Last, explain why it would be more difficult to detect the release of neutrons than protons.
Chemistry
1 answer:
liubo4ka [24]2 years ago
7 0

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

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7 grams of oxygen gas is reacted with excess C4H8. How many grams of CO2 gas at STP are produced?
ladessa [460]
I think 14 are produced because if you go up by that you get it
4 0
2 years ago
What is a positive ion?
BaLLatris [955]

Answer:

If the atom has more electrons than protons, it is a negative ion or ANION. If it has more protons than electrons, it is a positive ion.

Explanation:

Positive ions are typically metals or act like metals. Many common materials contain these ions. Mercury is found in thermometers, for instance, and aluminum is a metal that is found in a surprising amount of things.

8 0
3 years ago
Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Pb^{2+}/Pb]}=-0.13V

E^0_{[Cu^{2+}/Cu]}=+0.34V

E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0
2 years ago
The pKa of the α‑carboxyl group of serine is 2.21 , and the pKa of its α‑amino group is 9.15 . Calculate the average net charge
sleet_krkn [62]

Answer:

Net charge in serine at pH equal to 8.30 is "0"

Explanation:

  • At pH > pK_{a}, carboxyl group exists as -CO_{2}^{-} (charged)
  • At pH < pK_{a}, carboxyl group exists as -COOH (neutral)
  • At pH > pK_{a}, amino group exists as -NH_{2} (neutral)
  • At pH < pK_{a}, amino group exists as -NH_{3}^{+} (charged)
  • So, at pH = 8.30, both carboxyl and amino group exists in charged state.
  • Net charge in serine at pH equal to 8.30 is "0".
  • Structure of serine at pH equal to 8.30 has been shown below.

8 0
3 years ago
43 milliliters of water weighs 43 g. what is the density of the water?
Anastaziya [24]

Answer:

\rho =1g/mL

Explanation:

Hello,

In this case, since the density is defined as the ratio between the mass and the volume as shown below:

\rho =\frac{m}{V}

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

\rho =\frac{43g}{43mL}=1g/mL

Regards.

4 0
3 years ago
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