What is the percent nitrogen in NaN*O

Which statement explains the energy term in this reaction

lana66690 [7]

Answer:

Mass is lost due to the conversion of mass to energy

Explanation:

The question is not complete, the complete question is given as:

$^{235}_{72}U +^{1}_{0}n$ ⇒ $^{140}_{56}Ba+^{93}_{36}Kr+3^{1}_{0}n+energy$

total mass equals 236.053 u total mass equals 235.868 u

Which statement explains the energy term in this reaction? (1) Mass is gained due to the conversion of mass to energy. (2) Mass is gained due to the conversion of energy to mass. (3) Mass is lost due to the conversion of mass to energy. (4) Mass is lost due to the conversion of energy to mass.

Answer: From Einstein’s equation E = mc², when a radioisotope element undergoes fission or fusion in a nuclear reaction, it loses a tiny amount of mass.This mass lost is converted to energy.

The law of conservation of energy holds for this type of reaction (i.e the sum of mass and energy is remains the same in a nuclear reaction). Mass changes to energy, but the total amount of mass and energy combined remains the same before and after a nuclear reaction.

From the reaction above, the total decrease in mass = 236.053 - 235.868 = 0.185 u

4 0

2 years ago

7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.

Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0

3 years ago

The notation for a particular nucleus is r3785b. in an electrically neutral atom. How many electrons are in orbit about this nuc

Serhud [2]

$^{85}_{37}Rb$ has 37 electrons in the orbit about its nucleus.

<h3>How to write an atomic symbol with atomic mass and atomic number?</h3>

Atomic mass is the total number of neutrons and protons present in the nucleus of an element.

The atomic number is the total number of protons present in the nucleus of an element.

For writing an atomic symbol, atomic mass is written on the top left of the atomic symbol, and the atomic number is written on the downside of the left of the symbol.

$^{Atomic mass}_{Atomic number}X$

An electrically neutral atom is an atom that has an equal number of electrons as well as protons.

Thus from the above conclusion we can say that $^{85}_{37}Rb$ has 37 electrons.

Learn more about Atomic mass and atomic number here:

brainly.com/question/3187640

#SPJ4

3 0

1 year ago

What is kind of property is denisty?

Virty [35]

Answer:

INTENSIVE PROPERTY OF MATTER

Explanation:

DENSITY IS AN INTENSIVE PROPERTY OF MATTER THAT ILLUSTRATES HOW MUCH MASS A SUBSTANCE HAS IN A GIVEN AMOUNT OF VALUE,

7 0

2 years ago

[

notsponge [240]

Answer:

gime fre points

Explanation:

fre points for me!

6 0

3 years ago

What is the percent nitrogen in NaN*O_{3} ?