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aliina [53]
3 years ago
7

1. According to the equation, what mass of hydrogen fluoride is necessary to produce 2.3 g of sodium fluoride?

Chemistry
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

1.096g

Explanation:

You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:

Hydrogen: 1.008g/mol

Fluorine: 18.99g/mol

Sodium: 22.98g/mol

Next, find the composition percentage of NaF

22.98 + 18.99 = 41.97

Fluorine is 18.99/41.97 =45.25%

Sodium is 100-45.25 = 54.75%

Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g

Find comp. percentage of HF

18.99+1.008 = 19.998; H/total F/total

Hydrogen 5.041%

Fluorine 94.959%

Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g

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umka21 [38]

Hi!


The correct options would be:

1. Cathode - <em>reduction</em>

The cathode is the negatively charged electrode, and so has an excess of electrons. Cations (positively charged ions) are attracted to the cathode, and gain electrons to acquire a neutral charge. The process in which a gain of electron occurs is called reduction.


2. Anode - <em>oxidation</em>

The opposite occurs at the anode which is positively charged and attracts negatively charged ions, anions. These anions lose their electrons at the anode to acquire a neutral charge, and the process involving loss of electrons is known as oxidation.


3. Salt Bridge - <em>ion transport </em>

Salt bridge is a physical connection between the the anodic and cathodic half cells in an electrochemical cell and is a pathway that facilitates the flow of ions back and forth these half cells. Salt bridge is involved in maintaining a neutral condition in the electrochemical cells, and its absence would result in the accumulation of positive charge in the anodic cell, and negative charge in the cathodic cell.


4. Wire - <em>electron transport </em>

Wires have a universal role of being a pathway for the transport of electrons in circuit. This role is also the same in the wires involved in an electrochemical cells where they are used to transport electrons from the anodic half cell, and this electron transport results in the generation of electricity in the internal circuit of the electrochemical cell.


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4 0
3 years ago
Which of the analogies below is a location-based analogy?
nydimaria [60]
D is the correct answrt
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3 years ago
Pls help
Irina-Kira [14]

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

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5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

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3 0
2 years ago
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STatiana [176]

Answer: 3.79*10^24 atoms

Explanation:

1 mole = 6.02214076*10^23 atoms

8 0
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