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Lyrx [107]
3 years ago
12

A balloon is filled with 2.00 L of helium gas at sea level, 1.00 atm and 12.0ºC. The balloon is released and it rises to an alti

tude of 30,000 ft. If the pressure at this altitude is 0.300 atm and the temperature is -55.0ºC, what is the volume of the balloon?
Chemistry
1 answer:
pshichka [43]3 years ago
3 0

Answer:

THE VOLUME OF THE BALLOON IS 1.45 L

Explanation:

At sea level:

Volume = 2 L

Pressure = 1 atm

Temperature = 12 °C

At 30000 ft altitude:

Pressure = 0.30 atm

Temperature = -55°C

Volume = unknown

Using the general gas formula:

P1 V1 / T1 = P2 V2 / T2

Re-arranging the formula by making V2 the subject of the equation, we have;

V2 = P1 V1 T2 / T1 P2

V2 = 1 * 2 * 12 / 0.30 * 55

V2 = 24 / 16.5

V2 = 1.45 L

The volume of the balloon at the temperature of -55 C and 0.30 atm is 1.45 L

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Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
Oxygen. A glowing wooden splint relights in a test tube of oxygen.
5 0
3 years ago
When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb
aliya0001 [1]

The heat released in the combustion of lactic acid is absorbed by the

calorimeter and in the decomposition of the lactic acid.

ΔH°f of lactic acid is approximately <u>-716.2 kJ</u>

Reasons:

Known parameters are;

Mass of the lactic acid = 3.93 grams

Heat  capacity of the bomb calorimeter = 10.80 kJ·K⁻¹

Change in temperature of the calorimeter, ΔT = 5.34 K

ΔHrxn = ΔErxn

ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹

ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹

The chemical equation for the reaction is presented as follows;

  • C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O

The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ

Molar mass of C₃H₆O₃ = 90.07 g/mol

Number of moles of C₃H₆O₃ = \dfrac{3.93 \, g}{90.07 \, g/mol} = 0.043633 moles

Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles

Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ

Moles of H₂O produced = 0.130899 moles

Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ

Therefore, we have;

Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter

Which gives;

Heat absorbed by lactic acid  = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ

The heat absorbed by the lactic acid ≈ -31.249 kJ

  • \Delta H^{\circ}f \ of \ C_3H_6O_3 = \dfrac{-31.249}{0.043633} \approx  -716.2

ΔH°f of C₃H₆O₃ ≈ -716.2 kJ

Heat of formation of lactic acid ≈ <u>-716.2 kJ</u>.

Learn more here:

brainly.com/question/13185938

5 0
2 years ago
Describe the modem system of cassification,
yaroslaw [1]

Answer: The system classifies organisms into eight levels: domain, kingdom, phylum, class, order, family, genus, and species.

Explanation:

The scientific name given to an organism is based on binomial nomenclature.

5 0
3 years ago
What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?
djverab [1.8K]

The empirical formula of the compound is CHCl₃.

<h3>Calculation:</h3>

Given,

Mass of carbon = 5.03 g

Mass of hydrogen = 0.42 g

Mass of chlorine = 44.5 g

Molecular weight of carbon = 12 g

Molecular weight of hydrogen = 1 g

Molecular weight of chlorine = 35.4 g

First, calculate the moles of each element,

                      Moles = given mass/ molecular weight

Moles of carbon = 5.03/12 = 0.42

Moles of hydrogen = 0.42/1 = 0.42

Moles of chlorine = 44.5/ 35.4 = 1.26

Divide the moles of each element by the smallest number of moles,

0.42 mol of C/ 0.42 = 1 C

0.42 mol of H/ 0.42 = 1 H

1.26 mol of Cl/0.42 = 3 Cl

The ratio of elements is 1:1:3

Therefore the empirical formula of the compound will be CHCl₃.

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5 0
2 years ago
3. Oblicz stężenie procentowe 423 g roztworu zawierającego 17 g rozpuszczonego siarczanu (VI) miedzi (II). Zapisz obliczenia i o
goldfiish [28.3K]

Answer:

17 / 423 = 0.04018912529

~ 4%

proszę bardzo

6 0
3 years ago
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