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Elis [28]
2 years ago
9

Calcular la presión final de un gas que inicialmente está a 21°C y 0,98 atm. Sabiendo que su temperatura aumenta a 37°C.

Chemistry
1 answer:
KatRina [158]2 years ago
7 0

Answer:

1.03 atm

Explanation:

Primero <u>convertimos 21 °C y 37 °C a K</u>:

  • 21 °C + 273.16 = 294.16 K
  • 37 °C + 273.16 = 310.16 K

Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:

  • T₁P₂ = T₂P₁

En este caso:

  • T₁ = 294.16 K
  • P₂ = ?
  • T₂ = 310.16 K
  • P₁ = 0.98 atm

Colocando los datos:

  • 294.16 K * P₂ = 310.16 K * 0.98 atm

Y <u>despejando P₂</u>:

  • P₂ = 1.03 atm
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212 pb 82 is the isotope notation for iron. what is the atomic number, mass number, and number of both protons and neutrons?
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Hello!

The mass number in isotope notation is denoted A, the atomic number is denoted as Z, and the element is denoted as X.

In the given isotope, the mass of the isotope is 212 amu, and the atomic number is 82.

We know that the number of electrons, and protons are equal to the atomic number. Therefore, there are 82 protons. Also, to find the number of neutrons, we subtract the atomic number from the atomic mass.

212 - 82 = 130 neutrons

<u>Final answers</u>:  

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You are a researcher for a golf club manufacturer. You are given two identical looking cubes of a metal alloy. You are informed
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A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
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Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

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n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

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2 years ago
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