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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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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

If 6.4 miles of Fe react with excess O2, how many miles of Fe2O3 can be formed? 4 Fe + 3 O2 -> 2 Fe2O3

Yuri [45]

I'm assuming that by "miles" you mean moles.

If O2 is the excess reactant, that means Fe is the limiting reactant. That means that the amount of product being formed depends on the amount of Fe reactant present. To calculate the moles of Fe2O3 formed, start with the given 6.4 moles of Fe and use the mole to mole ratio given by the reaction as shown below:

6.4 mol Fe x $\frac{2 mol Fe2O3}{4 mol Fe}$ = 3.2 mol Fe2O3

6 0

3 years ago

Could the Periodic Table be arranged differently?

coldgirl [10]

Answer:

No because it is stayed that way and you can't define them differently.

7 0

3 years ago

Read 2 more answers

Determine the rate of a reaction that follows the rate law:

natima [27]

Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

6 0

2 years ago

A physical change _____ change the composition or identity of the substance. will will not 2. An example of a physical change is

Sav [38]

Answer:

not sure

Explanation:

a physical change (can)change.,....

6 0

3 years ago