What is the most likely decay for the fe-53 nucleus?

1 answer:

7 0

The element number of Fe is 26, which means that the number of neutrons of the following isotopes is:

53-Fe: 27 
54-Fe: 28 
56-Fe: 30 
57-Fe: 31 
58-Fe: 32 
59-Fe: 33 

Because 53-Fe has too few neutrons when compared to the other isotopes we can rule out any decay that involves losing a neutron: (alpha decay, beta decay) So, this isotope will prefer to decay by electron capture or positron emission (assuming the system has about 1 MeV to spare), but in each case the product will be the same: 53-Mn (long-lived radioisotope).

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

Answer: The volume of concentrated solution required is 9.95 mL

Explanation:

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated nitric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted nitric acid solution

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

6 0

2 years ago

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres

Marizza181 [45]

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$