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lyudmila [28]
2 years ago
10

Xenon-135 is produced in a nuclear reactor by two primary methods. one is directly from fission, the other is from the decay of:

_______
Chemistry
1 answer:
LekaFEV [45]2 years ago
6 0

Xenon-135 is produced from the  the decay of iodine-135.

<h3>What is a nuclear reactor?</h3>

The term nuclear reactor refers to an instrument where nuclear reaction takes place. Now we know that nuclear reactions can be used for several purposes and one of those purposes is the generation of electricity.

Now, Xenon-135 is produced in a nuclear reactor by two primary methods. one is directly from fission, the other is from the decay of iodine-135.

Learn more about nuclear reactor:brainly.com/question/13869748

#SPJ1

Missing parts

Xenon-135 is produced in a reactor by two primary methods. One is directly from fission; the other

is from the decay of...

A. cesium-135.

B. iodine-135.

C. xenon-136.

D. iodine-136.

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What is the total number of oxygen atoms on the right-hand side of this chemical equation?
vivado [14]

The chemical equation is

Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)

Answer:

12

Explanation:

In the right hand side of the equation, there are three compound which contains O2, which are;

Cu(NO3)2 , number of oxygen atoms =3*2 =6

2NO2, number of oxygen atoms = 2*2=4

2H2O, number of oxygen atoms =2*1=2

Total number of oxygen atoms on the right side of equation = 6+4+2 =12

8 0
3 years ago
Bubba falls out of a plane flying over his peanut field. What will bubba speed be after falling for 8 seconds?
chubhunter [2.5K]

Variables we know:

t = 8 seconds

Vi = 0 m/s

g = -9.81

Δy = ?

Vf = ?

Equation we will be using to solve for Vf: Vf = Vi + gt

Steps to solve:

Vf = (0) + (-9.81)(8)

Vf = -78.48 m/s

Hope this helps!! :)

5 0
3 years ago
Read 2 more answers
What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?
MArishka [77]
Combined gas law is 
       PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
     P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm                                     P₂ = standard pressure = 1 atm
V₁ = 720.0 mL                                    T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
  
By substituting,
 3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
                                          V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL
5 0
3 years ago
Given the balanced chemical equation, SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = −184 kJ Determine the mass (in grams) of H
Ostrovityanka [42]

Answer:

mass HF = 150.05 g

Explanation:

  • SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )

∴ MwHF = 20.0063 g/mol

∴ mol HF = 4 mol

∴ ΔH°rxn = - 184 KJ

∴ Q = 345 KJ

mass HF ( mHF ):

⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn

⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ

⇒ mHF = 150.05 g HF

3 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

#SPJ1

7 0
2 years ago
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