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MatroZZZ
2 months ago
14

!!15 points!!Which of the following phrases describes how the position of an electron

Chemistry
2 answers:
Alekssandra [3K]2 months ago
8 0
A. The energy of an electron increases as its distance from the nucleus increases.
Tems11 [2.7K]2 months ago
4 0
A. The energy of an electron increases as its distance from the nucleus increases.
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Which list of radioisotopes contains an alpha emitter, a beta emitter, and a positron emitter?
alisha [2963]

Answer: The correct option is 3.

Explanation: Radioisotopes that emit alpha-particles are termed alpha-emitters. These isotopes undergo alpha-decay.

Those radioisotopes that emit beta-particles (_{-1}^0\beta ) are called beta-emitters. They undergo beta-minus decay, in which a neutron converts to a proton and an electron.

Isotopes that emit positrons (_{+1}^0\beta ) are known as positron-emitters, undergoing beta-plus decay where a proton becomes a neutron.

From the options given,

Option 1: All three isotopes undergo beta-minus decay.

Option 2: Cs-137 and Tc-99 undergo beta-minus decay.

Fr-220 undergoes alpha-decay.

Option 3: Kr-85 undergoes beta-minus decay.

_{36}^{85}\textrm{Kr}\rightarrow _{37}^{85}\textrm{Rb}+_{-1}^0\beta

Ne-19 undergoes positron decay.

_{10}^{19}\textrm{Ne}\rightarrow _{9}^{19}\textrm{F}+_{+1}^0\beta

Rn-222 undergoes alpha decay.

_{86}^{222}\textrm{Rn}\rightarrow _{84}^{218}\textrm{Po}+_{2}^4\alpha

Option 4: All three isotopes undergo beta-minus decay processes.

Therefore, the correct choice is 3.

6 0
1 month ago
Read 2 more answers
A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane
Tems11 [2777]

Explanation:

The rate at which gases effuse is inversely related to the square root of their molar masses.

In this case, half of the helium (1.5 L) passed through the membrane in 24 hours. Therefore, we can calculate the effusion rate of He gas as follows.

          \frac{1.5 L}{24 hr}

            = 0.0625 L/hr

Given that the molar mass of He is 4 g/mol and for O_{2} it is 32 g/mol.

Now,

   \frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}

                               = 2.83

Thus, the effusion rate of O_{2} = \frac{\text{Rate of He}}{2.83}

                       = \frac{0.0625 L/hr}{2.83}

          Rate of O_{2} = 0.022 L/hr.

This implies that 0.022 L of O_{2} gas will effuse in one hour.

Consequently, to find the duration needed for 1.5 L of O_{2} gas to effuse, we calculate as follows.

         \frac{1.5 L}{0.022 L/hr}

                = 68.18 hours

Thus, we can conclude that it will require 68.18 hours for half of the oxygen to effuse through the membrane.

3 0
2 months ago
Which procedure cannot be performed on a hot plate, requiring a Bunsen burner instead
Tems11 [2777]

Answer: The process of heating a crucible to eliminate moisture from a hydrate.

Explanation:

The available choices are:

a. Heating a solvent to aid in the dissolution of a solute.

b. Heating a solid in isolation to remove moisture.

c. Bringing water to a boil for use in a water bath.

d. Heating a crucible to eliminate moisture from a hydrate.

Possible actions that can be done on a hot plate include:

a. Heating a solvent to assist a solute in dissolving.

b. Heating a solid in isolation to dry it.

c. Heating water to boiling for a water bath.

However, it's important to note that using a hot plate for heating a crucible to remove water from a hydrate is not advisable. Silica or ceramic materials are not meant to be heated on a hot plate.

Consequently, the correct procedure is heating a crucible to remove water from a hydrate.

4 0
3 months ago
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