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oee [108]
4 years ago
13

Write the balanced nuclear equation for β− decay of silicon−32. Include both the mass numbers and the atomic numbers with each n

uclear symbol. Use the "sup-subscript" button in the answer palette to enter these numbers correctly. Greek letters can be accessed in the drop-down menu that says "Main Keypad."
Chemistry
1 answer:
weqwewe [10]4 years ago
8 0

Answer:

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

Explanation:

Beta decay conserves the lepton number. In \beta^- decay, the atomic number of the element increases which is accompanied by the release of  an electron antineutrino, e^-+\bar{v}_e.

For example:-

^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e

The \beta^- decay of silicon-32 is shown below as:-

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

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Nitric oxide (NO) from car exhaust is a primary air pollutant. Calculate the equilibrium constant for the reaction
viva [34]

This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:

There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=-RT ln(K)

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

\Delta _rH=2*90.25 kJ/mol - (0 kJ/mol+0 kJ/mol)=180.5kJ/mol\\\\\Delta _rS=2*(0.211 kJ/mol*K)-(0.192kJ/mol*K+0.205kJ/mol*K)=0.025kJ/mol*K

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

\Delta _rG_{25\°C}=180.5-(25+298.15)*0.025=172.42kJ/mol\\\\\Delta _rG_{1496\°C}=180.5-(1496+298.15)*0.025=135.65kJ/mol

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:

K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}

Learn more:

  • (Gibbs free energy) brainly.com/question/15213613
4 0
3 years ago
Balance NO + 02 -- NO2 ​
Simora [160]

Answer:

NO + O2 -> NO2

it is balanced

3 0
3 years ago
ANSWERS PLEASE HELP 20 MINUTES LEFT????!!!!?!!???
sasho [114]

Answer:

Explanation:

b

3 0
3 years ago
Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2 is 7.
murzikaleks [220]

Answer: The rate of the loss of O_3 is 0.52M/s

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

2O_3(g)\rightleftharpoons 3O_2(g)

Rate of disappearance of O_3 =-\frac{1d[O_3]}{2dt}

Rate of formation of O_2 =+\frac{1d[O_2]}{3dt}

-\frac{1d[O_3]}{2dt}=+\frac{1d[O_2]}{3dt}

Rate of formation of O_2 = 7.78\times 10^{-1}M/s

Thus Rate of disappearance of O_3 =\frac{2d[O_2]}{3dt}=\frac{2}{3}\times 7.78\times 10^{-1}M/s=0.52M/s

8 0
4 years ago
A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
Juliette [100K]

V = maximum capacity of human lung = 3 liter = 3 x 0.001 m³ = 0.003 m³      (Since 1 liter = 0.001 m³)

P = pressure of oxygen = 21.1 kilo pascal = 21.1 x 1000 = 21100 Pa      (since 1 kilo = 1000)

T = temperature of air = 295 K

n = number of moles of oxygen

Using the ideal gas equation

PV = n RT

inserting the above values in the equation

(21100) (0.003) = n (8.314) (295)

n = 0.026 moles



4 0
3 years ago
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