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Mkey [24]
2 years ago
6

Use the periodic table to identify the number of core electrons and the number of valence electrons in each case below.Magnesium

(Mg): 1s22s22p63s2
Chemistry
2 answers:
amm18122 years ago
3 0

Answer:

1s1

2s2/2p6

3s2⇨⇨ period 3 - group 2

laila [671]2 years ago
3 0
There’s 2 valence electrons in Mg
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In 2011, a computer named Watson competed against two humans on a popular television quiz show and won. Watson listened to each
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Answer: The Computer: Similarities and Differences. Both use electrical signals to send messages. The brain uses chemicals to transmit information; the computer uses electricity. Even though electrical signals travel at high speeds in the nervous system, they travel even faster through the wires in a computer

Explanation: I hoped that helped. -wear,mask

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PLEASE HELP!!! PLEASE.
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Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄:  The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

<u><em>Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl? </em></u>

  • As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.
  • The normality of the NaOH and HCl = Their molarity.

<em>∵ (NV)NaOH = (NV)HCl</em>

<em>∴ N of HCl = (NV)NaOH / (V)HC</em>l = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

<u><em>Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?</em></u>

  • As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.
  • <em>The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.</em>

<em>∵  (NV)H₂SO₄ = (NV)KOH</em>

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

<em>∴ The concentration of KOH = 7.675 mN = 7.675 mM.</em>

<em></em>

<u><em>Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?</em></u>

  • As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

<em>The normality of NaOH = Molarity of NaOH = 0.5 N.</em>

<em>∵ (NV)H₂SO₄ = (NV)NaOH</em>

<em>∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄</em> = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

<em>∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.</em>

<em></em>

<u><em>Q₄: Explain the difference between an endpoint and equivalence point in a titration.</em></u>

  • The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.
  • The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.
  • End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.
  • The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.
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In Philip’s French class, the students are learning how to pronounce closed vowels and open vowels. The students are most likely
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For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re
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Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

Ba2+(aq) + SO42-(aq) → BaSO4(s)

Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

Hide answer

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

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