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Tasya [4]
3 years ago
14

What is the mass of oxygen in 300 grams of carbonic acid (H2CO3)

Chemistry
1 answer:
DENIUS [597]3 years ago
7 0

mass of carbonic acid = 300g

molar mass of H2CO3 = 2H + C + 3 O

= 2 x 1.008+ 12.01 + 3 x  16

= 62.03g/mol


moles of H2CO3 = mass/Molar mass

= 300/62.03

= 4.8364 moles


1 mole H2CO3 has 3 moles Oxygen


4.8364 moles H2CO3 contains  

=   3 x 4.8364  moles Oxygen  =   14.509 moles Oxygen


moles = mass/Molar mass


mass of oxygen = moles x Molar mass of Oxygen

= 14.509 x 16

= 232.15g Oxygen

mass of oxygen in 300g of carbonic acid(H2CO3) = 232.15g

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Answer: The mole allows people to calculate the number of middle schoolers entities (usually atoms or molecules.  Isabella's number is an absolute number: there are 6.022 × 1023 middle schoolers entities is in 1 mole. This can also be written as 6.022 × 1023 mol-1.

Explanation: I hope that helped !!

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Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . suppose 1
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13) Find the mass, in grams, of 5.75 liters of N2.
faltersainse [42]
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8 0
2 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

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Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

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=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

3 0
2 years ago
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