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stealth61 [152]
3 years ago
8

What’s is element classification why K2O Is an ionic component

Chemistry
1 answer:
Darya [45]3 years ago
7 0
It has bonds between a metal and a nonmetal, so it is ionic
You might be interested in
6. The pOH of a solution of NaOH is 11.30. What is the [H+<br><br> ] for this solution?
sashaice [31]

Answer:

The [H⁺] for this soluton is 2*10⁻³ M

Explanation:

pH, short for Hydrogen Potential and pOH, or OH potential, are parameters used to measure the degree of acidity or alkalinity of substances.

The values ​​that compose them vary from 0 to 14 and the pH value can be directly related to that of pOH by means of:

pH + pOH= 14

In this case, pOH=11.30, so

pH + 11.30= 14

Solving:

pH= 14 - 11.30

pH= 2.7

Mathematically the pH is the negative logarithm of the molar concentration of the hydrogen or proton ions (H⁺) or hydronium ions (H₃O):

´pH= - log [H⁺] = -log [H₃O]

Being pH=2.7:

2.7= - log [H⁺]

[H⁺]= 10⁻² ⁷

[H⁺]=1.995*10⁻³ M≅ 2*10⁻³ M

<u><em>The [H⁺] for this soluton is 2*10⁻³ M</em></u>

8 0
4 years ago
If 110g of copper sulphate is present in 550g of solution calculate the concentration of solution
Bingel [31]

Answer:

20%

Explanation:

mass by mass percentage of a solution =(mass of solute)/(mass of solution)

mass of solute=550g

therefore 110×100/550=20%

hope u will understand .:") credit to the owner

7 0
4 years ago
What percentage of strontium-90 remains in a sample after three half-lives have passed? 12.5% 15% 25% 60%?
Lyrx [107]

12.5% of strontium-90 would remain in a sample after three half-lives have passed. Half-life automatically means 50% of the original amount would remain.

4 0
3 years ago
Read 2 more answers
Describe how you would prepare approximately 2 l of 0.050 0 m boric acid, b(oh)3.
Elenna [48]

The given concentration of boric acid = 0.0500 M

Required volume of the solution = 2 L

Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.

Calculating the moles of 0.0500 M boric acid present in 2 L solution:

2 L * \frac{0.0500 mol B(OH)_{3} }{1 L} = 0.100 mol B(OH)_{3}

Converting moles of boric acid to mass:

0.100 mol B(OH)_{3} * \frac{61.83 g}{mol B(OH)_{3}}   = 6.183 g

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.


5 0
3 years ago
Name the product formed by the oxidation of d-gulose.
faust18 [17]
Oxidation of D-glucose are usually found on the photosynthesis which has a by product of carbon dioxide (CO2) and water  (H2O) chemically written as,

C6H12O6 +6O2----> 6CO2 +6H2O

Moreover, D-glucose can also be chemically react with oxygen to form D-gluconic acid and D-glucoronic acid.
8 0
3 years ago
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