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lbvjy [14]
3 years ago
12

How many moles are there in 814.504 g of H2O?

Chemistry
1 answer:
earnstyle [38]3 years ago
6 0
First you need to calculate the molar mass of H2O. To do that, look at the periodic table and add up the AMUs (atomic mass units) 

H = 1.01
O = 16.0 

1.01 • 2 + 16.0 = 18.02 

Next, use stoichiometry to convert grams into moles. To do that, divide 814.504g by the number of grams in one mole of H2O.

814.504 ÷ 18.02 = 45.2 moles 

There are 45.2 moles in 814.504 grams of H2O.
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B is a precipitation reaction.

Explanation:

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A very loud sound has a high _____.
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Resting Point

Explanation:

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For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G
Phantasy [73]

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the wavelength of radiation emitted  is \mathbf{\lambda= 2169.62 \ nm}

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}

For n = 7:

\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}

\mathtt{E_7 =-0.27755 \ eV}

For n = 4

\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}

\mathtt{E_4 =- 0.85\ eV}

The  electron goes from the n = 7 to the n = 4, then :

\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}

\mathtt{= 0.57245\ eV}

Wavelength of the radiation emitted:

\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}

where;

hc  = 1242 eV.nm

\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}

\mathbf{\lambda= 2169.62 \ nm}

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