What types of radiation cause the parent isotope to change into a different element?

Chemistry

1 answer:

kvv77 [185]2 years ago

3 0

Answer:

Beta decay is most common in elements with a high neutron to proton ratio. Gamma decay follows the form: In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state.

Explanation:

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Why does warm air raise and cold air sink

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Suppose that you are performing a titration on a monoprotic acid. Titration of this monoprotic acid required a volume of 0.1L of

Alex73 [517]

Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions

Explanation:

The monoprotic acid (HA) has a valency of 1 and diprotic acid $(H_2A)$ has a valency of 2.

As the concentration and volume of the diprotic acid and the monoprotic acids are equal.

The neutralization reaction for monoprotic acid is:

$HA+BOH\rightarrow BA+H_2O$

The neutralization reaction for diprotic acid is:

$H_2A+2BOH\rightarrow B_2A+2H_2O$

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$