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Arisa [49]
3 years ago
12

Sodium undergoes beta decay and becomes magnesium, write an equation. please help...

Physics
1 answer:
Tatiana [17]3 years ago
6 0

Remark

There are 2 kinds of Beta Decay. This one is an example of a neutron becomes a proton and the emitted  particle is an electron.

_{11}^{23}\text{Na}\longrightarrow_{12}^{23}\text{Mg}+_{-1}^{ 0}\text{1e-}



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The continental ...........
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Which description best models the energy transfer that occurs in the radiative zone?
Zinaida [17]

Answer : Option D) A dropped wallet is kicked around the floor of a busy train station.

Explanation : The description that best suits the model for the energy transfer that occurs in the radiative zone is -

<h3>A dropped wallet is kicked around the floor of a busy train station.</h3>

As in the radiation zone, which is also called as radiative zone or radiative region in the layer of a star's interior where energy is primarily transported toward the exterior by means of radiative diffusion and thermal conduction. This can be correlated with the example of a dropped wallet in a busy train station as the wallet would be kicked by many people who are in a hurry to travel, this represents the radiative diffusion. So, every time the wallet is being kicked by someone the energy is getting transformed.

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Explanation:

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4 years ago
A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density
Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

  • mole fraction (H2SO4)= 0.034
  • mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

  • n = m / MW
  • = ∑ AWT
  • mole fraction = moles of A component / total moles of solution
  • ρ = m /v

Where:

  • m = mass
  • n = moles
  • MW = molecular weight
  • AWT = atomic weight
  • ρ = density
  • v = volume

Information about the problem:

  • m solute (H2SO4) = 17.75 g
  • v(solution) = 100 ml
  • ρ (solution)= 1.094 g/ml
  • AWT (H)= 1 g/mol
  • AWT (S) = 32 g/mol
  • AWT (O)= 16 g/mol
  • mole fraction(H2SO4) = ?
  • mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)=  98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)=  18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092  mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092  mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092  mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

  • Solvent
  • Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0
2 years ago
Please help :)
Orlov [11]

Answer:

<em>3.5 A</em>

Explanation:

Current [A] =Electrical power [W] ÷ Potential difference [V]

KW TO W = ×1000

hence,

800W÷230V= 3.5A (1 d.p)

8 0
3 years ago
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