Which subatomic particles are located in MOST of the space within an atom?

Classify each of the following as a homogeneous or heterogeneous substance.

Mnenie [13.5K]

Answer:

a). iron ore - is a heterogeneous substance.

b). quartz - is a homogeneous substance.

c). granite - is a heterogeneous substance.

d). energy drink - is a heterogeneous substance.

e). oil-and-vinegar salad dressing - is a heterogeneous substance.

f). salt - is a homogeneous substance.

g). rainwater - is a heterogeneous substance.

h). nitrogen - is a homogeneous substance.

Explanation:

a). iron ore - is a heterogeneous substance as iron core contains iron and other impurities

b). quartz - is a homogeneous substance as it is only made up of carbon

c). granite - is a heterogeneous substance as it is made up of multiple minerals.

d). energy drink - is a heterogeneous substance as it contains various substances mixed in it

e). oil-and-vinegar salad dressing - is a heterogeneous substance as oil and vinegar itself are different compounds

f). salt - is a homogeneous substance as it only contains one type of molecule

g). rainwater - is a heterogeneous substance as it contains water and other minerals

h). nitrogen - is a homogeneous substance as it is only made of one type of molecule.

5 0

1 year ago

Predict the formula of the ionic compound that would result from combining the monatomic ions formed by each pair of elements. M

skad [1K]

Answer:

E - Be and O

A - Mg and N

E - Li and Br

F - Ba and Cl

B - Rb and O

Explanation:

Be and O

Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).

Mg and N

Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).

Li and Br

Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).

Ba and Cl

Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).

Rb and O

Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).

4 0

3 years ago

I need help with this

Shkiper50 [21]

I highly believe it would be D
Because yes it does happen above water but it’s not just that, and the other two don’t seem correct. So sorry it’s not correct but I believe it is D

8 0

3 years ago

Plz answer question number 1,2,4,5 and 6

yulyashka [42]

Answer:

1.Handpicking,winnowing and sieving 2. distillation 3.distillation 5. winnowing 6.magnet

4 0

4 years ago

Quinine (C20H24N2O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine

dolphi86 [110]

Answer:

The pH of saturated solution of the quinine is 10.05

Explanation:

Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol

Q can behave as a weak base. Kb and pKb can be calculated for weak bases

pKb1 is to be considered when solving the question.

pKb1 = 5.1

Step 1 : Calculate the Kb of Quinine

pKb1 = - log [kb]

5.1 = - log [kb]

take Antilog of both side

[kb] = 7.94 x 10∧-6

Step 2: Calculate the concentration of saturated solution of Q in mol/dm3

From the question, 1900 ml of solution contains 1 g of Q

Therefore, 1000 ml of solution will contain........... x g of Q

x = 1000 /1900

x = 0.526 g in 1 dm3

In calculating concentration in mol/dm3,

Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass

Molar mass of Q = 324.4

Concentration in mol/dm = 0.526 /324.4

= 0.0016 mol/dm3

Step 3: Calculating the Concentration of OH-

At Equilibrium, Kb = x² / 0.0016

7.94 x 10∧-6 = x² / 0.0016

x = √ 0.0016 × 7.94 x 10∧-6

x = 1. 127 × 10∧-4 mol/dm3

The concentration of OH- = 1. 127 × 10∧-4 mol/dm

Step 4: Calculating the pH of Quinine

Recall, pOH = - log [OH-]

pOH = - log [1. 127 × 10∧-4]

pOH = 3.948

Also recall that pH + pOH = 14

pH = 14 - 3.948

pH = 10.05

8 0

4 years ago