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gladu [14]
2 years ago
13

Charlie is frying an egg in a pan located over a gas burner. He develops a model to determine the energy produced by the flame i

n the gas burner by calculating the energy absorbed by the egg. Which assumption will Charlie need to make in order for his model to be considered a closed system?
Chemistry
1 answer:
frosja888 [35]2 years ago
8 0

Answer: B.)

Explanation:  Heat flows from pan to sounroundings

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For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

6 0
2 years ago
Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
2 years ago
Read 2 more answers
Marie discovers boxes of elements in a storage room. The boxes do not say which elements they contain, but they do have informat
Mazyrski [523]
<span>The elements in a Periodic Table are grouped according to their classifications. The major classifications are Metals, Non-metals, and Metalloids. Their level of reactivity can be gauged by simply looking at their position in the table. For Metals, their reactivity increases as you move to the left then going down. Non-metal reactivity increases as you move to the right then going up, starting at the bottom of the table.</span>
5 0
2 years ago
Read 2 more answers
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
2 years ago
Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 10
vlada-n [284]

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

<em>Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.</em>

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

<h3>a. both temperature changes will be the same</h3>

5 0
2 years ago
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