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salantis [7]
3 years 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:
jekas [21]3 years ago
4 0

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

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

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

  • pure chemical substance.
  • heterogenous mixture.
  • mixture.
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What is the pH of a 0.100 M HI solution? Is that neutral, acidic, or basic? acidic What is the pOH of a 0.100 M HI solution? Is
inessss [21]

<u>Answer:</u> The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

<u>Explanation:</u>

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

  • The pH range of acidic solution is 0 to 6.9
  • The pH range of basic solution is 7.1 to 14
  • The pH of neutral solution is 7.

We are given:

Concentration of HI = 0.100 M

1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=0.100M

Putting values in above equation, we get:

pH=-\log(0.100)\\\\pH=1

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14

pOH=14-1=13

Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.

4 0
3 years ago
15 POINTS PLEASE HELP What volume of water must be added to 35mL of 2.6m KCl to reduce its concentration to 1.2m? Please explain
BartSMP [9]
First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
M₂=the concentration we want to dilute to (in this case 1.2M)
V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute.  The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change.  That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

Solving for the above equation:
V₂=M₁V₁/M₂
V₂=(2.6M×0.035L)/1.2M
V₂=0.0758 L
That means that the solution needs to be diluted to 75.8mL to have a final concentration of 1.2M.

 Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps.  Let me know if anything is unclear.
Good luck on your quiz!
5 0
3 years ago
True or false the si in sih4 does not follow the octet rule because hydrogen is in an unusual oxidation stat
ExtremeBDS [4]
The statement is true. The octet rule refers to the general rule of thumb wherein atoms of main-group elements tend to bond with other atoms in such a way that each atom possesses eight electrons (octet) in their valence shell. They tend to form the same electronic configuration as the noble gases. However, there are some exceptions to this rule. One of which is silane, SiH₄. A hydrogen atom only has 1 valence electron and needs another electron to complete its energy level. This is unlike other atoms, for example, carbon which has 4 valence electrons and needs to form 4 covalent bonds to fill its energy levels. Thus, 4 hydrogen atoms need only 4 more electrons. This is given by the silicon atom which has 4 valence electrons. Therefore, when a silicon atom is bonded to 4 hydrogen atoms, the resulting molecule, SiH₄, is a stable one.
6 0
3 years ago
Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?
Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0
3 years ago
Consider the reaction H2(g) + I2(g) &lt;---&gt; HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 525. Which
WARRIOR [948]

Answer:

The equilibrium will shift to the left to favor the reactants.

Explanation:

Remember that the reaction quotient (Qc) is derived from initial concentrations of reactants and products. Since Qc is greater than Kc, this means that initial concentrations are heavily impacted by a high product concentration ([HI]). Therefore, the reverse reaction will occur and actually create more reactants again ([H2] and [I2]). Thus, the answer is that the equilibrium will shift to the left side to favor the reactants.

8 0
3 years ago
Read 2 more answers
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