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3 years ago

explain using diagrams how potassium forms the compound potassium flouride when it reacts with flourine

Chemistry

1 answer:

Katarina [22]3 years ago

7 0

Answer:

The answer to your question is given below.

Explanation:

Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.

Fluorine (F) has 9 electrons with electronic configuration of 2, 7.

Fluorine needs 1 electron to complete it's octet configuration.

Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.

**** Please see attached photo for further details.

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Which of the following 0.300 M solutions would contain the highest concentration of potassium ions?

mojhsa [17]

Here is a list of 0.300 M solutions that you can consider to select which has the highest concetration of potassium ions:

Select one:

a. potassium oxide

b. potassium phosphate

c. potassium hydrogen carbonate

d. potassium hypochlorite

e. potassium iodide

Answer:

Option b. potassium phosphate

Explanation:

First, you must write the chemical formula of each one of the potasssium compounds. Then, assume all the compounds ionize 100%.

Since all the solutions have the same concentrations, by assuming 100% ionization, the ionization equations will show which one produces the highest number of potassium ions, K⁺¹, and that is the solution with the highest concentration of such ions.

1. Chemical formulae

Compound Formula

a. potassium oxide K₂O

b. potassium phosphate K₃PO₄

c. potassium hydrogen carbonate KHPO₃

d. potassium hypochlorite KClO₄

e. potassium iodide KI

2. Dissociation (ionization):

a. potassium oxide, K₂O

K₂O does not dissociate but react with water to product KOH, then KOH dissociates, but there is not such thing as a 0.300 M solution of K₂O.

b. potassium phosphate:

K₃PO₄ → 3K³⁺ + PO₄¹⁻ 3 K⁺¹ ions per unit formula

c. potassium hydrogen carbonate

KHPO₃ → K⁺¹ + HPO₃⁻¹ 1 K⁺¹ ion per unit formula

d. potassium hypochlorite

KClO₄ → K⁺¹ + HClO₄⁻¹ 1 K⁺¹ ion per unit formula

e. potassium iodide

KI → K⁺¹ + I⁻¹ 1 K⁺¹ ion per unit formula.

Hence, the 0.300 M solution of potasssium phosphate produces the highest concentration of potassium ions.

6 0

4 years ago

The pressure and temperature inside a bike tires 10 ATM and 10 K receptively. What will the pressure be come in the tire when th

pishuonlain [190]

Answer:

P = 20atm

Explanation:

P1 = 10atm

T1 = 10K

P2 = ?

T2 = 20K

This question requires the use of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided its volume remains constant

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3=........=Pn / Tn

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (10 × 20) / 10

P2 = 20atm

The final pressure of the gas is 20atm

4 0

3 years ago

How do you know which element is the most reactive in the periodic table?

Snezhnost [94]

The most reactive metal in the periodic table (Cesium) with the most reactive non-metal (Fluorine).

3 0

3 years ago

Which option correctly describes the influence of electronegativity difference on bond character?

Bezzdna [24]

Electronegativity is a property of a certain element to attract an electron from other elements towards itself during a reaction. The greater the difference of the electronegativities of two elements in a compound, the stronger is their binding force. The strongest bond is the ionic bond, then following by covalent bonds.

7 0

3 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago