Value of 335k on the celsius temp scale

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

Which equation shows how to calculate how many grams (g) of Mg(OH)2 would be produced from 4mol KOH? The balanced reaction is:

Simora [160]

Answer:

4 mol of KOH would produce 116.6 g of Mg(OH)₂

Explanation:

According to the following balanced equation:

MgCl₂+ 2 KOH -----> Mg(OH)₂ + 2 KCL

One can note that 2 mol of KOH react with MgCl₂ to produce 1 mol of Mg(OH)₂.

using cross multiplication

2 mol of KOH → 1 mol of Mg(OH)₂.

4 mol of KOH → ?? mol of Mg(OH)₂.

no of moles of Mg(OH)₂ = (1 mol* 4 mol) / 2 mol =2 mol

Now we can convert moles of Mg(OH)₂ to grams using the formula

mass of Mg(OH)₂= (no. of moles * molar mass) = (2 mol * 58.3g/mol) = 116.6 g

So, 4 mol of KOH would produce 116.6 g of Mg(OH)₂.

6 0

4 years ago

1. For each of the following, convert the word equation into a formula equation, BUT do not balance! (4 pts each = 12 pts)

daser333 [38]

Answer:

1a) BaClO₃(s) → BaCl₂(g) + O₂(g)

1b) Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

1c) Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

2a) Calcium hydroxide and hydrogen gas

2b) Tin(II) silicate and Lead(IV) permanganate

2c) Magnesium oxide and water

2d) No product

2e) Mercury and iodine

2f) Calcium chloride and iodine

2g) Strontium phosphite and cesium nitride

2h) Carbon dioxide, water, and sulfur dioxide

2i) Iron oxide(III) and carbon dioxide

2j) Magnesium acetate and hydrogen gas

2k) Calcium iodide

Explanation:

1. For each of the following, convert the word equation into a formula equation, BUT do not balance!

a) Barium chlorate → Barium chloride + Oxygen

1. Chemical formulas

Barium chlorate:

It is a salt: an ionic compound.
Barium has oxidation state +2
Chlorate is the ion ClO₃⁻
Swap the oxidation numbers to write the subscripts: 2 goes to ClO₃ and 1 goes to Ba
Chemical formula Ba(ClO₃)₂
It is solid: Ba(ClO₃)₂(s)

Barium chloride:

It is a salt: an ionic compount
Barium has oxidation state +2
Chlorine is in oxidation state -1
Swap the numbers to write the subscripts: 2 goes to Cl and 1 goes to Ba
BaCl₂
It is solid BaCl₂(s)

Oxygen:

It is a diatomic gas molecule
O₂(g)

2. Write the unbalanced molecular equation:

BaClO₃(s) → BaCl₂(s) + O₂(g)

b) Chlorine + Potassium nitride → Nitrogen + Potassium chloride

1. Chemical formulas

Chlorine:

It is a diatomic gas molecule
Cl₂(g)

Potassum nitride

It is a salt

Potassium has oxidation state +1
Nitrogen is with oxidation state +3
Swap the oxidation states
K₃N
It is solid: K₃N(s)

Nitrogen:

It is a diatomic gas
N₂(g)

Potassium chloride

It is a salt (ionic compound)
Potassium has oxidation state +1
Chlorine is in oxidation state -1
Swap the oxidation numbers
KCl
It is solid: KCl(s)

2. Write the unbalanced molecular equation

Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

c) Sodium nitride + Aluminum bromate → Aluminum nitride + Sodium bromate

1. Chemical formulas

Sodium nitride

It is a salt (ionic compound)
Sodium has oxidation state +1
Nitrogen is with oxidation state -3
Swap the oxidation numbers
Na₃N
It is in aqueous solution
Na₃N (aq)

Aluminum bromate

Salt
Aluminum has oxidation state +3
Bromate is the ion BrO₃⁻
Swap the oxidation states
Al(BrO₃)₃ (aq)

Aluminum nitride

Both Al and N have oxidation state 3, which simply
AlN(s). It is not soluble in water.

Sodium bromate

Na(BrO₃)₃ (aq)

2. Write the unbalanced molecular equation

Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

<h2>This is a long answer with more than 5,000 charaters; thus, I have to add the rest of the explanations on a separate file.</h2><h2></h2><h2>The attached file contains the complete answer.</h2>

Download pdf

4 0

4 years ago

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M ac

adell [148]

Explanation:

Below is an attachment containing the solution.

7 0

4 years ago

HELP!!!!! How much does a sample of Argon weigh if it occupies 25.4 L at a pressure of 2.45 atm and a temperature of 482 K?

Artist 52 [7]

Answer:

62.82 g

Explanation:

From the question,

PV = nRT.................. Equation 1

Where P = Pressure, V = Volume, n = number of moles of argon, R = molar gas constant, T = Temperature.

But,

Number of mole (n) = mass (m)/molar mass(m')

n = m/m'................... Equation 2

Substitute equation 2 into equation 1

PV = (m/m')RT.............. Equation 3

From the question, we were asked to find m.

There make m the subject of formula in equation 3

m = PVm'/RT.............. Equation 4

Given: P = 2.45 atm, V = 25.4 L, T = 482 K

Constant: R = 0.082 atm.dm³.K⁻¹.mol⁻¹, m' = 39.9 g/mol

Substitute these values into equation 4

m = (2.45×25.4×39.9)/(0.082×482)

m = 62.82 g.

5 0

3 years ago