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3 years ago

Which element is rolled into a foil and used in your kitchen?

Chemistry

1 answer:

Doss [256]3 years ago

8 0

Aluminum is the correct answer

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The density of an element is 19.3 g/cm^3. What is its density in kg/m^3?

Karolina [17]

First, you need to know 1 kg = 10^3 g. And 1 m^3 = 10^6 m^3. So the 1 g/cm3 = 10^3 kg/m3. So the answer is 1.93*10^4 kg/m3.

5 0

3 years ago

Which is the correct formula for phosphorus pentachloride? PCl4 because a subscript of 4 indicates four Cl atoms PCl5 because a

bearhunter [10]

Answer:

PCl₅ because a subscript indicates 5 chlorine atoms.

Explanation:

As the name suggests, phosphorous pentachloride contains 5 chlorine atoms as penta means five and phosphorous means P atoms.

3 0

3 years ago

Read 2 more answers

1.Light waves (shown) are an example of what type of wave? (Lesson 4.03)

lesya692 [45]

Answer:

suface wave

Explanation:

hope this helps

6 0

3 years ago

If the ligand has a negative charge at a particular location, what would happen if you tried to put electrons from the metal nea

slega [8]

Answer:

The two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place

Explanation:

A complex is made up a central metal atom or ion and ligands. Ligands are lewis bases and they possess lone pairs of electrons. A complex is formed when electrons are donated from ligand species to metals.

However, if the ligand has a negative charge at a particular location and we try to put electrons from the metal near the electrons from the ligand, the two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place.

8 0

3 years ago

Calculate the percent composition of the following: A. VO3 B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago