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2 years ago

Uranium-267 undergoes alpha decay, what is the mass of the new element?

Chemistry

1 answer:

Firlakuza [10]2 years ago

3 0

Answer:

will this help ?

Explanation:

(108Hs) is a synthetic element, and thus a standard atomic weight cannot be given. Like all synthetic elements, it has no stable isotopes. The first isotope to be synthesized was 265Hs in 1984. There are 12 known isotopes from 263Hs to 277Hs and 1–4 isomers. The most stable isotope of hassium cannot be determined based on existing data due to uncertainty that arises from the low number of measurements. The confidence interval of half-life of 269Hs corresponding to one standard deviation (the interval is ~68.3% likely to contain the actual value) is 16 ± 6 seconds, whereas that of 270Hs is 9 ± 4 seconds. It is also possible that 277mHs is more stable than both of these, with its half-life likely being 110 ± 70 seconds, but only one event of decay of this isotope has been registered as of 2016.[1][2].

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Question 11 (1 point)

erastova [34]

Answer:

remove product

Explanation:

Removing the product will always shift the equilibrium to the right. This is based on the Le Chatelier's principle which states that "if any of the conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".

If a system at equilibrium is disturbed, by changing the concentration of one of the substances all the concentrations will change until a new equilibrium point is reached.
Removing the product will increase the concentration of the species on the left hand side, the equilibrium will shift to the right.

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3 years ago

What is the magnetic quantum number value for an element with n = 1?

Citrus2011 [14]

Answer:

0, l is n-1 always, ml is l to -l

6 0

3 years ago

Explain why smell is a physical property?

lawyer [7]

Answer:

because no chemical change is happening

Explanation:

5 0

3 years ago

Which of the following describes the arrangement of valence electrons in a bond between H and F?

Sloan [31]

C; The Valence electrons spend more time around the atom of F

6 0

2 years ago

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

6 0

2 years ago