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Firlakuza [10]
2 years ago
9

When two ions form an ionic bond, what type of compound is formed?

Chemistry
1 answer:
Dafna1 [17]2 years ago
3 0

it is c. ionic compound just because it says ionic. can't be a crystal since there's no links to an ionic bond connected to crystals (and if it were it'd depend on the elements and the thingy majig itself right) and a covalent bond is an entirely different bond which isn't related to ionic. hope this helped, and i hope you have a good day too! :D

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An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
2 years ago
Suppose your teacher ask you to separate a mixture of sand and water.Which of the following merhods would u use:distillation fil
MrRa [10]

Answer:

Filtration

Explanation:

Filtration would be best because the sand particles would be trapped in the filter paper and the water would go through so the mixture would be separated

7 0
2 years ago
does adding boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip increase,
olga55 [171]

Addition of  boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip will have no effect on the Ksp value of ca(oh)2.

There will be no effect on the Ksp value as boiled deionised water is not able to alter the number of hydronium and hydroxide ions. As no change in the ions happen so there will be no change in Ksp value. The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

To know more about, solubility product constant, click here,

brainly.com/question/6960236

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4 0
1 year ago
Enter the chemical formula of a binary molecular compound of hydrogen and a Group 6A element that can reasonably be expected to
NARA [144]

Answer:

Any binary molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

Explanation:

Going down in a group increases the atomic radius and a greater atomic radius implyes greater ionic radius.  

When ionization takes place in these compounds they yelds protons (hidrogen ion) and an lewis base (anion). The greater the ionic radius the greater its stability, thus the periodic tendency is increaing the acidity of binary hidrogen compounds when going down a group. On the other hand going up a group decreases acidity, so any molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

3 0
3 years ago
Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
tekilochka [14]

Answer: There are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

So, the number of atoms present in 0.67 moles are as follows.

0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms

Thus, we can conclude that there are 16.14 \times 10^{23} atoms of hydrogen are present in 40g of urea, (NH_{2})_{2}CO.

7 0
2 years ago
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