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Degger [83]
2 years ago
14

How many kilograms of gasoline fill a 12.0-gal gas tank

Chemistry
1 answer:
Sergeu [11.5K]2 years ago
8 0

Answer:

29.98kg

Explanation:

12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg

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The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
BARSIC [14]
The answer is: D) centi
5 0
2 years ago
What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and
Fudgin [204]

is this for a test or are you genuinely interested? molality = mols sugar/kg solvent

Solve for molality

delta T = Kf*m

Solve for delta T and subtract from zero C to find the new freezing point.

or

-5.58

4 0
2 years ago
Read 2 more answers
6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an
xz_007 [3.2K]

Answer:

ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K=1.53x10^{14}

Explanation:

1) Reduction potentials

First of all one should look up the reduction potentials for the species envolved:

Ce^{4+} + e→Ce^{3+}         E°red=1.61V

Fe^{3+} + e→Fe^{2+}         E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>

1)  Ce^{4+} reduces while  Fe^{2+} oxidates

  (oxidation)               Fe^{2+}→Fe^{3+} + e          E°oxi=-0.771V

  (reduction)               Ce^{4+} + e→Ce^{3+}         E°red=1.61V

  (overall equation)    Fe^{2+}+Ce^{4+}→Ce^{3+}+Fe^{3+} E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°-\frac{RT}{nF} Ln Q

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°=\frac{RT}{nF} Ln K

Isolating for 'K' gives:

K=e^{\frac{nFE^{o} }{RT} }

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K=1.53x10^{14}

6 0
2 years ago
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Paul [167]

Answer:

1.3 mL

Explanation:

First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:

1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.

It should have 2 significant figures, because 1.2kg has 2 and we are dividing.

7 0
2 years ago
consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q
Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.  

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado!  :)

7 0
1 year ago
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