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Andrej [43]
3 years ago
6

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The

decays proceed through a series of alpha-particle and beta-particle emissions.A.How many of alpha-particle emissions are involved in this series?B.How many of beta-particle emissions are involved in this series?
Chemistry
1 answer:
dsp733 years ago
3 0

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

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nekit [7.7K]

Answer:

14700J

Explanation:

From the question given, the following were obtained:

M = 100g

ΔT = 35° C

C = 4.2J/g °C

Q=?

The heat transferred can calculated for by using the following equation

Q = MCΔT

Q = 100 x 4.2 x 35

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3 years ago
NEED HELP ASAP!!! TYSM!!!! What are the coefficients when the following equations are balanced?
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Answer:

Picking up marbles and placing them in a sack

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3 years ago
The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to rea
mihalych1998 [28]

<u>Answer:</u> The mass of tetraphosphorus decaoxide formed is 64.81g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

  • <u>For potassium chlorate:</u>

Given mass of potassium chlorate = 93.3 g

Molar mass of potassium chlorate = 122.55 g/mol

Putting values in above equation, we get:

\text{Moles of potassium chlorate}=\frac{93.3g}{122.55g/mol}=0.761mol

For the given chemical reaction:

10KClO_3+12P\rightarrow 3P_4O_{10}+10KCl

Red phosphorus is given in excess . So, it is considered as an excess reagent and potassium chlorate is considered as a limiting reagent.

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 3 moles of tetraphosphorus decaoxide

So, 0.761 moles of potassium chlorate will react with = \frac{3}{10}\times 0.761=0.2283moles of tetraphosphorus decaoxide

Calculating the mass of tetraphosphorus decaoxide by using equation 1, we get:

Molar mass of tetraphosphorus decaoxide = 283.886 g/mol

Moles of tetraphosphorus decaoxide = 0.2283 moles

Putting values in equation 1, we get:

0.2283mol=\frac{\text{Mass of tetraphosphorus decaoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decaoxide}=64.81g

Hence, the mass of tetraphosphorus decaoxide formed is 64.81g

5 0
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3 years ago
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