1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.
2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.
The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.
This are some of the transuranium elements:
Neptunio - 93
Plutonium - 94
Americium - 95
Curium - 96
Berkelium - 97
Californium - 98
Einstenium - 99
And so all the known elements (the last one is the 118).
3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.
4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:
0
β, which means 0 mass and charge - 1.
-1
5) This is, then, an example of a β decay equation for one transuranium element:
239 239 0
Np → Pu + β
93 94 -1
As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.
The ignition occurs when a mixture of fuel & oxygen encounter an external heat source with sufficient heat<u> Piloted ignition.</u>
<h3>
What is Piloted ignition?</h3>
When a volatile fuel is close to a nearby local energy source (pilot) and reaches its lower limit of flammability in air, piloted ignition may be possible. The flame that originates in the premixed system spreads outward from the pilot. The first law of thermodynamics for systems with fixed mass only describes the energetics of this process.
<h3>Definition of thermodynamics </h3>
The science of thermodynamics examines how heat, work, temperature, and energy are related. The general topic of thermodynamics is the transfer of energy from one location or form to another. The fundamental idea is that heat is a type of energy that is equivalent to a specific quantity of mechanical labor.
To learn more about ignition
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Answer:
It's IF5. ................
<span>Answer:
1170 ml is the final volume of the sample of gas, if the pressure and moles in the container is kept constant.</span>
Answer:
396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.
Explanation:
To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;
C6H12O6 + 6O2 -------> 6CO2 + 6H2O
1 mole of C6H12O6 reacts to form 6 moles of CO2
Then, calculate the molar mass of the two variables;
Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol
Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol
Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles
1* 180 g of glucose yields 6 * 44 g of CO2
180 g of glucose = 264 g of CO2
If 270 g of glucose were to be used, how many grams of CO2 will be produced;
so therefore,
180 g of glucose = 264 g of CO2
270 g of glucose = x grams of CO2
x = 264 * 270 / 180
x = 71 280 / 180
x = 396 g of CO2.
In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.