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Ierofanga [76]
3 years ago
7

Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?

Chemistry
1 answer:
kipiarov [429]3 years ago
5 0
The answer is evaporation<span>.Evaporation is the vaporization of the liquid from its surface into gaseous phase, without boiling the liquid. When all the liquid has passed to gaseous phase the salt dissolved in the salt water will remain as solid crystals.</span><span />
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Use the changes in oxidation numbers to identify which atom is oxidized, reduced, the oxidizing agent, and the reducing agent. 5
Vinil7 [7]

Answer:

Reaction A:

  • Hydrogen atoms in H₂ are oxidized.
  • Oxygen atoms in O₂ are reduced.
  • Hydrogen gas H₂ is the reducing agent.
  • Oxygen gas O₂ is the oxidizing agent.

Reaction B:

  • Oxygen atoms in KNO₃ are oxidized.
  • Nitrogen atoms in KNO₃ are reduced.
  • Potassium nitrate (V) KNO₃ is both the oxidizing agent and the reducing agent.

Explanation:

  • When an atom is oxidized, its oxidation number increases.
  • When an atom is reduced, its oxidation number decreases.
  • The oxidizing agent contains atoms that are reduced.
  • The reducing agent contains atoms that are oxidized.

Here are some common rules for assigning oxidation states.

  • Oxidation states on all atoms in a neutral compound shall add up to 0.
  • The average oxidation state on an atom is zero if the compound contains only atoms of that element. (E.g., the oxidation state on O in O₂ is zero.)
  • The oxidation state on oxygen atoms in compounds is typically -2. (Exceptions: oxygen bonded to fluorine, and peroxides.)
  • The oxidation state on group one metals (Li, Na, K) in compounds is typically +1.
  • The oxidation state on group two metals (Mg, Ca, Ba) in compounds is typically +2.
  • The oxidation state on H in compounds is typically +1. (Exceptions: metal hydrides where the oxidation state on H can be -1.)

For this question, only the rule about neutral compounds, oxygen, and group one metals (K in this case) are needed.

<h3>Reaction B</h3>

Oxidation states in KNO₃:

  • K is a group one metal. The oxidation state on K in the compound KNO₃ shall be +1.
  • The oxidation state on N tend to vary a lot, from -3 all the way to +5. Leave that as x for now.
  • There's no fluorine in KNO₃. The ion NO₃⁻ stands for nitrate. There's no peroxide in that ion. The oxidation state on O in this compound shall be -2.
  • Let the oxidation state on N be x. The oxidation state of all five atoms in the formula KNO₃ shall add up to zero. 1\times (+1) + 1 \times (x) + {\bf 3} \times (-2) = 0\\x = +5. As a result, the oxidation state on N in KNO₃ will be +5.

Similarly, for KNO₂:

  • The oxidation state on the group one metal K in KNO₂ will still be +1.
  • Let the oxidation state on N be y.
  • There's no peroxide in the nitrite ion, NO₂⁻, either. The oxidation state on O in KNO₂ will still be -2.
  • The oxidation state on all atoms in this formula shall add up to 0. Solve for the oxidation state on N: 1\times (+1) + 1 \times (y) + {\bf 2}\times (-2) = 0\\y = +3. The oxidation state on N in KNO₂ will be +3.

Oxygen is the only element in O₂. As a result,

  • The oxidation state on O in O₂ will be 0.

\rm\stackrel{+1}{K}\stackrel{\bf +5}{N}\stackrel{\bf -2}{O}_3 \to \stackrel{+1}{K}\stackrel{\bf+3}{N}\stackrel{\bf -2}{O}_2 + \stackrel{\bf 0}{O}_2.

The oxidation state on two oxygen atoms in KNO₃ increases from -2 to 0. These oxygen atoms are oxidized. KNO₃ is also the reducing agent.

The oxidation state on the nitrogen atom in KNO₃ decreases from +5 to +3. That nitrogen atom is reduced. As a result, KNO₃ is also the oxidizing agent.

<h3>Reaction A</h3>

Apply these steps to reaction A.

H₂:

  • Oxidation state on H: 0.

O₂:

  • Oxidation state on O: 0.

H₂O:

  • Oxidation state on H: +1.
  • Oxidation state on O: -2.
  • Double check: {\bf 2} \times (+1) + (-2) = 0.

\rm \stackrel{}{2}\; \stackrel{\bf 0}{H}_2 + \stackrel{\bf 0}{O}_2\stackrel{}{\to} \stackrel{}{2}\;\stackrel{\bf +1}{H}_2\stackrel{\bf -2}{O}.

The oxidation state on oxygen atoms decreases from 0 to -2. Those oxygen atoms are reduced. O₂ is thus the oxidizing agent.

The oxidation state on hydrogen atoms increases from 0 to +1. Those hydrogen atoms are oxidized. H₂ is thus the reducing agent.

4 0
3 years ago
The ground-state configuration of a negative ion of a halogen
mafiozo [28]

Answer:

The ground state configuration for the negative ion of a halogen. ... A possible excited state electronic configuration. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1 6s^1. Its symbol is the double solid lin, // or ||, in a cell diagram.

5 0
3 years ago
Which type of energy in the bungee cord prevents the bungee jumper from hitting the ground
Effectus [21]
Elastic potential energy
7 0
3 years ago
Water is capable of dissolving a variety of different substances, which is why it is such a good solvent. it is often called the
Tema [17]

Water is often referred as a <span>universal solvent </span>because it is capable dissolving much more solutes as compared to any other solvent.  This is because, water is a high polar molecule. In water, H has partial positive charge while O has partial negative charge. 


Due to this, water favors dissociation of molecules into positively and negatively charged ions. Positively charge ions gets attracted  towards oxygen i.e. negatively charges, while negatively charged ions get attracted towards positive end of water molecule. 


However, it is worth nothing that, despite water being referred as universal solvent, many compounds are insoluble or partially soluble in water. For instance, most of the hydroxide displays poor solubility in water.

6 0
3 years ago
Read 2 more answers
How many grams of K2Cr207 are in 250. mL of 0.500M of K2Cr207:
goldenfox [79]

<u>Analysing the Question:</u>

We are given a 250 mL solution of 0.5M K₂Cr₂O₇

Which means that we have:

0.5 Mole in 1L of the solution

0.125 moles in 250 mL of the solution      <em>[dividing both the numbers by 4]</em>

<em />

<u>Mass of K₂Cr₂O₇ in the given solution:</u>

Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol

<em>we know that we have 0.125 moles in the 250 mL solution provided</em>

Mass = Number of moles * Molar mass

Mass = 0.125 * 194

Mass = 36.75 grams

7 0
3 years ago
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