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Rus_ich [418]
3 years ago
14

Which of the following represents beta decay

Chemistry
1 answer:
Korvikt [17]3 years ago
4 0

Answer:

The reaction representing beta decay is of option C:

 _{64}^{160}\textrm{Gd}\rightarrow _{65}^{160}\textrm{Tb}+_{-1}^0\e  

Explanation:

Beta-decay: In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

 _Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta  

 _{64}^{160}\textrm{Gd}\rightarrow _{65}^{160}\textrm{Tb}+_{-1}^0\e  

Hence, the correct answer is option C.

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A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water ________
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A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

<h3>What is One Calorie ?</h3>

The amount of heat energy required to raise the temperature by 1 gram of water through 1°C is known as One Calorie.

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Learn more about the One calorie here: brainly.com/question/1061571

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Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20
KengaRu [80]

Answer:

m_{Ag}=2,265.9g

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

\Delta S=-n_TR\Sigma[x_i*ln(x_i)]

For this silver-gold mixture we write:

\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]

By knowing the moles of gold:

n_{Au}=100g*\frac{1mol}{197g} =0.508mol

It is possible to write the aforementioned formula in terms of the variable x representing the moles of silver:

20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]

Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

x=n_{Ag}=21.0molAg

So the mass is:

m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g

Best regards!

3 0
3 years ago
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