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prisoha [69]
3 years ago
11

Un gas ocupa un volumen de 358L a 152°C y 470 mmHg ¿Qué volumen ocupara el gas, si la temperatura aumente a 500 K y 6 atm?

Chemistry
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

42 L

Explanation:

de los parámetros en la pregunta;

V1 = 358L

T1 = 152 ° C + 273 = 425 K

P1 = 470 mmHg × 1 atm / 760 mmHg = 0.6atm

V2 =?

P2 = 6 atmósferas

T2 = 500 K

P1V1 / T1 = P2V2 / T2

P1V1T2 = P2V2T1

V2 = P1V1T2 / P2T1

V2 = 0,6 × 358 × 500/6 × 425

V2 = 107400/2550

V2 = 42 L

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3 moles

Explanation:

SrCO3

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Number of moles = mass / molar mass

Number of moles = 442.8 / 147.6

Number of moles = 3

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Water is a great solvent. For example, it causes ionic compounds like sodium chloride (table salt) to dissociate. One of these r
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A polar molecule has both a partial positive and a partial negative. Since NaCl (sodium chloride) is made out of Na+ and Cl- ions, the positive and negative parts of the water will pull on these ions individually since positive attracts negative and negative attracts positive. This magnetic force causes the Na+ and the Cl- ions to be pulled apart.

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Answer:

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Explanation:

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Let's have a further discussion of the evidences.

1. Change of Color- Color change is caused by the combination of two or more substance with different molecular structures. A popular example of this is the Statue of the Liberty, which is made of copper plates. Due to the exposure of copper to elements like water, it changed color.

2. Change of Odor- This can be best presented with rotting food. During the rotting process, the food undergoes a chemical reaction. The result is a rotten smell.

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3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
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Answer:

Explanation:

From the information given:

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Pressure = 1 bar

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The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

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The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

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