How is it possible for compound to have both ionic and covalent bonds

Answers

Answer 1

Answer: And ionic is when a nonmetal and a metal compound bond together and a covalent is when 2 nonmetals bond together
"K2SO4 is an example of a compound that contains both ionic and covalent bonds... the sulfate ion is held together with
covalent bonds... and the potassium ions are ionically bonded to the sulfate ions. Or let's say we have a group of elements that are covalently bonded together, but still have net positive or negative charge... like PO4, OH, SO4, NH4, NO3... and these ions are both cations and anions and can have a plethora of charges.

For example... (NH4)3PO4 can be dissociated but the ions will remain as a group. So when this dissociates... we will get three ammonium and one phosphate ion. But then, the oxygen will not dissociate from the phosphorous any more than the hydrogen will dissociate from the nitrogen."

Answer 2

Answer: WHen neither atom attracts or repels the shared electon it can be both.

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If 1 bus = 12 cars, 3 cars = 1 truck, and 1000 basketballs = 1 truck, how many basketballs can be carried by 8 buses?

Answers

The factor label method (dimensional analysis) is a method used to solve problems by the given amount and value. Eightbuses can carry 32,000 basketballs.

What is dimensional analysis?

Dimensional analysis has been also called the unit factor method used to convert the various units and quantities by mathematical computation and equation formation. It can convert the fundamental and derived units.

Given,

1 bus = 12 cars

3 cars = 1 truck

1 truck = 1000 basketballs

If 3 cars have been equal to 1 truck then, 12 cars will be equal to X trucks

Solving for X as:

X = (12 cars × 1 truck) ÷ 3 cars

= 4 trucks

Hence, 4 trucks are equal to 12 cars.

Now, from above it can be said that 1 bus has been equal to 4 trucks.

So, if one bus = 4000 basketballs

8 bus = X basketballs

Solving for X:

X = 4000 × 8

= 32000 basketballs

Therefore, eight buses can contain 32000 basketballs.

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1 bus = 12 cars
1 truck = 3 cars
4 trucks= 12 cars
1 bus = 4 trucks
1 bus = 4000 basketballs
8 buses= 32,000 basketballs

Which is true about the dissolving process in water?

Answers

Water molecules that move throughout the solute.
When you put a solvent in the water, the water molecule will start to surrounds every small particles of the solute to dissolve it. solute is not actually gone, it just mixes with the water molecules.

Which bond is most polar? A) N -H
B) C -H
C) O -H
D) H -H
Can anyone explain this one?

Answers

A) N is more polar because of it electron negativity charge ex. N electron negativity charge is 3.0 and H is 2.1
B) C is more polar...electron negativity for C is 2.5 AND H is 2.1
C)O is more polar dues to it electron negativity charge being greater that hydrogen
D)H and H are both non polar because they cancel each other out due to the face the both share the electron negativity evenly

C .. ectronegativity values for each of the atoms but if you didn't know these you can just look at your periodic table and see where they are in relation to each other

How much energy is passed to the next trophic level in the food chain?

Answers

Answer:

10%

Explanation:

Out of the original layer, only ten percent is transferred to the next

I found a type of rock in the woods it's kinda rough black with some purple like a lavender purple and i can break it into pieces what could it be?

Answers

It could be a lump of coal.
Since you can break pieces off, maybe charcoal ... the real thing, not a "briquet".
Do your hands get all black when you handle it ?

An element crystallizes in a face centered cubic lattice and has a density of 1.45 g cm-3 . The edge of its unit cell is 4.52 x 10-8cm.a) How many atoms are in each unit cell?
b) What is the volume of a unit cell?
c) What is the mass of a unit cell?
d) Calculate the approximate atomic mass of the element.

Answers

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

The face-centered cubic lattice has 3 atoms from the 6 faces, and 1 atom from the eight corners. Thus, the total atoms in the face-centered lattice are four.

The face-centered lattice has been a cube.

The volume of cube = $\rm (edge)^3$

The volume of unit cell = $\rm (4.52\;*\;10^-^8\;cm)$

The volume of unit cell = $\rm 9.23\;*\;10^-^2^3\;cm^3$

The mass of a unit cell can be calculated from density. Mass can be defined as the ratio of volume to density.

Mass = $\rm (volume)/(density)$

Mass of unit cell = $\rm (9.23\;*\;10^-^2^3\;cm^3)/(1.45\;g\;cm^-^3)$

Mass of unit cell = $\rm 1.34\;*\;10^-^2^2\;g$ .

The approximate atomic mass of the element can be calculated by the mass of the carbon atom.

Mass of 1 carbon atom = $\rm (mass\;of\;1\;mole\;carbon)/(number\;of\;atoms\;in\;1\;mole\;Carbon)$

Mass of 1 carbon atom = $\rm (12)/(6.023\;*\;10^2^3)$

Mass of 1 carbon atom = 1.992 $\rm *\;10^-^2^3$ grams.

atomic mass unit per gram can be given as;

amu/gram = $\rm (12)/(1.992\;*\;10^-^2^3)$

amu/gram = $\rm 6.022\;*\;10^2^3$ amu/gram

1 gram = $\rm 6.022\;*\;10^2^3$ amu

1 amu = 1.661 $\rm *\;10^-^2^4$ gram.

The average atomic mass = mass of unit cell $*$ amu\gram

= $\rm 1.34\;*\;10^-^2^2\;g$ . $*$ 1 amu/ 1.661 $\rm *\;10^-^2^4$ gram.

= 80.7 amu.

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

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Final answer:

A face centered cubic lattice consists of 4 atoms. The volume of the unit cell is 9.22 x 10^-23 cm^3 and the mass is 1.34 x 10^-23 g. The approximate atomic mass of the element is 2.02 amu.

Explanation:

The element is said to crystallize in a face centered cubic lattice. This implies that there is one atom at each corner of the cube (8 corners for a total of 1 atom, since each corner atom is shared among 8 adjacent cubes). There is also one atom on each face of the cube (6 faces for a total of 3 atoms, since each face atom is shared among 2 adjacent cubes). Thus, a total of 4 atoms are present in each unit cell. (a)

The volume of a unit cell (edges for a cube) can be calculated by the formula 'volume = side^3', where side in this case is 4.52 x 10-8cm. Hence, the volume equals (4.52 x 10^-8cm)^3 = 9.22 x 10^-23cm^3. (b)

The density of the substance is given as 1.45g/cm^3. The formula for density is 'mass/volume' which implies that mass can be calculated as 'density x volume'. Hence, the mass of the unit cell is (1.45g/cm^3) x (9.22 x 10^-23 cm^3) = 1.34 x 10^-23 g.(c)

The atomic weight of the element can then be calculated by taking this overall mass and dividing by the number of atoms in a unit cell (4). So, the atomic weight is (1.34 x 10^-23 g) / 4 = 3.35 x 10^-24 g. But atomic weights are usually given in atomic mass units (amu), not grams, and 1 amu = 1.66 x 10^-24 g. Therefore, we have an atomic weight of (3.35 x 10^-24 g) / (1.66 x 10^-24 g/amu) = approximately 2.02 amu. (d)

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