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8090 [49]
3 years ago
11

Which one of the following processes is endothermic? A) Boiling a liquid. B) Freezing a solid. C) Condensation of a gas into a l

iquid. D) Condensation of a gas into a solid.
Chemistry
1 answer:
lord [1]3 years ago
8 0

Answer:

D) Condensation of a gas into a solid.

Explanation:

An endothermic process is a process that absorbs heat from the surroundings.

A) Boiling a liquid.

This gives off heat to the surroundings. An exothermic process.

B) Freezing a solid.

In this process, the water releases heat to the surroundings, so it is an exothermic process

C) Condensation of a gas into a liquid.

As water vapor condenses into liquid, it loses energy in the form of heat. Therefore, this process is exothermic.

D) Condensation of a gas into a solid.

This process absorbs heat from the surrounding, hence it is an endothermic proccess.

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Write the ionic charges (such as Ca2+) and chemical formulas and fill-in the table below.
Vikentia [17]

1) Lithium and fluorine:

Ionic charges: lithium cation Li⁺ and fluorine anion F⁻.

Chemical formula LiF.

In ionic salt lithium fluoride (LiF), fluorine has electronegativity approximately χ = 4 and lithium χ = 1 (Δχ = 4 - 1; Δχ = 3).

Fluorine attracts electron and it has negative charge and lithium has positive charge.

2) Beryllium and oxygen:

Ionic charges cation Be²⁺ and anion O²⁻.

Chemical formula is BeO.

Beryllium is metal from group 2 and oxygen is nonmetal from group 16.

Electron configuration of beryllium: ₄Be: 1s² 2s², it has two valence electrons in 2s orbital.

Beryllium lose two electrons and to gain electron configuration as noble gas helium (He).

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen gain two valence electron to form anion with stable electron configuration as noble gas neon (atomic number 10).

3) Magnesium and fluorine:  

Ionic charges cation Mg²⁺ and anion F⁻.

Chemical formula is MgF₂.

Magnesium fluoride (MgF₂) is salt, ionic compound.  

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).  

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

Fluorine (F) is nonmetal with greatest electronegativity, which means it easily gain electrons.  

Fluorine has atomic number 9, which means it has 9 protons and 9 electrons. It gain one electron to form fluorine anion (F⁻) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

4) Aluminum and chlorine:  

Ionic charges cation Al³⁺ and anion Cl⁻.

Chemical formula is AlCl₃.

The right name for AlCl₃ is aluminium chloride.

Aluminium chloride is a salt with ionic bonds.

Aluminium (metal from group 13) has oxidation number +3 and chlorine (nonmetal from group 17) has oxidation number -1, chemical compound has neutral charge (+3 + 3 · (-1) = 0).

5) Beryllium and nitrogen:  

Ionic charges cation Be²⁺ and anion N³⁻.

Chemical formula is Be₃N₂.

Atomic number of nitrogen is 7, it has 7 protons and 7 electrons.

Electron configuration of nitrogen atom: ₇N 1s² 2s² 2p³.

Nitrogen gain three electrons to form anion with stable electron configuration as noble gas neon (atomic number 10).

4 0
3 years ago
50 pts for this one If someone can solve
Arlecino [84]
This site can’t be reached
us-static.z-dn.net unexpectedly closed the connection.Try:Checking the connectionChecking the proxy and the firewallERR_CONNECTION_CLOSED

sorry bro
6 0
4 years ago
Read 2 more answers
A 64.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 32.0 ml of koh at 25 ∘c.
aleksley [76]
Hello!

The reaction between HBr and KOH is the following:

HBr+KOH→H₂O + KBr

To calculate the amount of HBr left after addition of KOH, you'll use the following equations:

HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\  \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr

That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7. 

Have a nice day!
3 0
3 years ago
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