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Taya2010
2 months ago
6

An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

ases. All these gases are evenly distributed throughout the sample of air. Which term or terms could be used to describe this sample of air? Check all that apply. View Available Hint(s)
- solution,
- pure chemical substance,
- heterogenous mixture,
- mixture,
- compound,
- homogeneous mixture,
- element.
Chemistry
1 answer:
Anarel [2.9K]2 months ago
4 0

Explanation:

A mixture refers to a substance that contains two or more different kinds of substances that are combined physically.

For instance, air is a mixture that contains oxygen, nitrogen, and other gases.

A heterogeneous mixture is characterized by uneven distribution of solute particles in the solvent.

For example, sand suspended in water is a heterogeneous mixture.

Conversely, a homogeneous mixture is one where the solute particles are uniformly distributed in a solvent.

A homogeneous mixture appears as a clear solution.

For instance, when salt dissolves in water, it forms a homogeneous mixture.

A solution is defined as a mixture of two or more substances combined together.

A compound consists of two or more different elements chemically bonded together in a specific mass ratio.

An element is the simplest form of a substance that comprises only one type of atom.

For example, a piece of sodium is composed solely of sodium atoms.

Conversely, a pure substance refers to a material consisting of just one type of molecule or atom.

For example, O_{2}, N_{2} etc are considered pure substances.

Thus, it can be concluded that the air sample can be described using the terms:

  • pure chemical substance.
  • heterogeneous mixture.
  • mixture.
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Tems11 [2777]

84.34 grams of iron (III) chloride is the maximum produced since iron is the limiting reagent, and chlorine gas is in excess.

Explanation:

Balanced equation:

2 Fe + 3 Cl2 → 2 FeCl3​

DATA PROVIDED:

iron =  atoms

mass of chlorine = 67.2 liters

mass of FeCl3 =?

The number of moles of iron will be calculated as

number of moles = \frac{total number of atoms}{Avagaro's number}

number of moles = \frac{3.18 x 10^23}{6.022x 10^23}

number of moles = 0.52 mol of iron

moles of chlorine gas

number of moles = \frac{mass}{molar mass of 1 mole}

Substituting the values into the equation:

n = \frac{67200}{70.96}               (molar mass of chlorine gas = 70.96 g/mol)

   = 947.01 moles

As iron is the limiting reagent therefore

2 moles of Fe lead to 2 moles of FeCl3

0.52 moles of Fe will yield

\frac{2}{2} = \frac{x}{0.52}

0.52 moles of FeCl3 is produced.

To express this in grams:

mass = n x molar mass

         = 0.52 x 162.2                   (molar mass of FeCl3 is 162.2g/mol)  

          = 84.34 grams        

3 0
2 months ago
How many moles of ammonium ions are in 6.985 g of ammonium carbonate?
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 <span>(NH4)2CO3 -> 96.09 g/mol

(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate

In this calculation, the unit 'grams' cancels out as it's present in both the numerator and the denominator, leading to 'mol' being the remaining unit.

Examining the formula (NH4)2CO3, it can be interpreted as:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3

This means every mole of ammonium carbonate yields one mole of carbonate ions and two moles of ammonium ions.

(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion) + (0.363981 mol ammonium ion) </span>
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2 months ago
How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
Alekssandra [3086]
First, we need to identify the half-reaction for magnesium. It can be represented as:

Mg2+ + 2e- = Mg

Next, we will determine the overall charge generated during the electrolysis using the information derived from the half-reaction. The calculation follows:

4.50 kg Mg (1000 g / 1 kg) (1 mol / 24.305 g) (2 mol e- / 1 mol Mg) (96500 C / 1 mol e-) = 35733388.2 C

The provided EMF is given in voltage. Since 1 V equals J/C, 5 V translates to 5 J/C.

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eduard [2782]
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