1. Position of Hydrogen
<span>2. Position of Isotopes</span>
The mass % of in the mixture is <u>9.93%</u>
<h3>What are hydrocarbon?</h3>
An organic substance called a hydrocarbon is made completely of hydrogen and carbon. Group 14 hydrides include, for example, 620 hydrocarbons.
The odours of gasoline and lighter fluid serve as examples of the relatively weak or insignificant odours of hydrocarbons, which are often colourless and hydrophobic.
They can be found in a wide variety of chemical forms and phases, including gases (like methane and propane), liquids (like hexane and benzene), low melting point solids (like paraffin wax and naphthalene), and polymers (such as polyethylene and polystyrene).
In the context of the fossil fuel industries, the term "hydrocarbon" refers to petroleum, natural gas, and coal that are produced naturally, as well as to their hydrocarbon derivatives and refined forms. The primary source of energy for the entire planet is the burning of hydrocarbons.
Learn more about Hydrocarbon
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Answer:
A Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.
Explanation:
Ex- when an exothermic reaction ( accompanied by or requiring absorption of heat)occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution increasing its temperature.
- When an endothermic reaction (Heat evolving reaction) occurs, the heat required is absorbed from the thermal energy of the solution, decreasing its temperature.
A burning piece of coal which is releasing heat due to its combustion and is transferred to water.
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2.48 grams.
<h3>Explanation</h3>
Start by finding the formula mass for cobalt (II) chloride and cobalt (II) chloride hexahydrate.
As a transition element in the middle d-block of the periodic table, cobalt can form ions with a plurality of charges. It is thus necessary to indicate its charge in systematic names of cobalt-containing formula.
The cation "cobalt" in the systematic name of the salt has the Roman numeral "(II)" attached to it in brackets. As a metal, cobalt forms positively-charged ion. The one here thu has charge of +2.
Chloride ions have charges -1. Charges cancel out to produce neutral compounds. Each cobalt cation in this salt would thus pair with two chloride anions. Hence the empirical formula: .
The prefix "<em>hexa-</em>" in the name cobalt (II) chloride <em>hexa</em>hydrate indicates that every formula unit of this salt contains six units of water. The hydrated salt thus has an empirical formula of .
Given the relative atomic mass for each of the elements, as seen on a modern periodic table of the elements:
- Cobalt- 58.93
- Chloride- 35.45
- Hydrogen- 1.008
- Oxygen- 16.00
Thus the formula mass of each compound
- Cobalt (II) chloride - 129.83
- Cobalt (II) chloride hexahydrate - 262.12
Cobalt (II) chloride hexahydrate decomposes under heat to produce cobalt (II) hexahydrate and water. Hence the equation:
Therefore
- Molar ratio:
- Mass ratio:
The mass ratio indicates that 262.12 grams of cobalt (II) chloride hexahydrate decomposes to produce 129.83 grams of its corresponding anhydrous salt. Accordingly, heating 5.00 grams of the hexahydrate would produce 2.48 grams of its anhydrate.
Answer:
For N₂F₂:
Molar fraction = 0.84
Partial pressure = 1.12 atm
For SF₄:
Molar fraction = 0.16
Partial pressure = 0.208 atm
Explanation:
It seems your question is missing the values required to solve the problem. However, an internet search showed me the following values for your question. If the values in your problem are different, your answer will be different as well, however the solving method will remain the same:
" A 5.00L tank at 0.7°C is filled with 16.5g of dinitrogen difluoride gas and 5.00g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. "
First we calculate the moles of each gas, using their molar mass:
- 16.5 g N₂F₂ ÷ 66 g/mol = 0.25 mol N₂F₂
- 5.00 g SF₄ ÷ 108 g/mol = 0.0463 mol SF₄
Total mol number = 0.25 + 0.0463 = 0.2963 mol
- Mole Fraction N₂F₂ = 0.25/0.2963 = 0.84
- Mole Fraction SF₄ = 0.0463/0.2963 = 0.16
Now we <u>use PV=nRT to calculate the partial pressure of each gas</u>:
P = ?
V = 5.00 L
T = 0.7 °C ⇒ 0.7 + 273.16 = 273.86 K
For N₂F₂:
- P * 5.00 L = 0.25 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 K
For SF₄:
- P * 5.00 L = 0.0463 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.86 K