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muminat
2 years ago
9

The empirical formula for a compound is C2H4NO. If its molar mass is 232.2 g/mol, what is the molecular formula of the compound?

Chemistry
1 answer:
Irina18 [472]2 years ago
4 0

Empirical formula mass

  • C2H4NO
  • 2(12)+4(1)+14+16
  • 30+24+4
  • 58g/mol

Molar mass=232.2g/mol

Find n

  • Molar mass/Empirical formula mass
  • 232.2/58
  • 4

Molecular formula

  • n×Empirical formula
  • 4(C2H4NO)
  • C8H16(NO)_4
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Answer:

Beta decay is most common in elements with a high neutron to proton ratio. Gamma decay follows the form: In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state.

Explanation:

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3 years ago
When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.
dsp73

We have that the work function of the metal

\phi=1.227*10^{-18}J

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy K.E =3.54*10-19 J.

Generally the equation for Kinetic energy    is mathematically given as

KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}

\phi=1.227*10^{-18}J

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What is the energy of an electron in the hydrogen atom determined by?
mezya [45]

Answer:

The energy of a hydrogen atom's electron is determined by which principal quantum number n value corresponds to the energy state the electron occupies. where n=1,2,3,... is the quantum number that quantizes the energy levels. That is, they are discrete energy values proportional to 1n2 .

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If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?
IceJOKER [234]

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

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