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1 year ago

Choose all that apply for Grasslands

Chemistry

1 answer:

Fynjy0 [20]1 year ago

4 0

Answer:

no tree shrubs rainfall is low contains grasses and woody plants correct me if I'm wrong

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Sprinter set in starting blocks at the beginning of a race rely on what kind of energy conversion

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They convert chemical potential energy (stored in muscle cells) to kinetic energy (running)

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2 years ago

All of the following are true about the Earth’s inner core EXCEPT

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that it is cooler than the lithosphere.

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3 years ago

What is the percent by mass of oxygen in mg(oh)2

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54.868% because well la la la (sorry had to be 20 characters)

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3 years ago

Selah is making a mobile of an atom of beryllium to hang from the ceiling in her science class. She uses a clear plastic bag for

mel-nik [20]

Answer:

See explanation

Explanation:

The question is incomplete because the images of the models are absent. However, i will try to give you a general description of what the correct answer should be.

Beryllium is a member of group 2 in the periodic table. Beryllium has an atomic number of 4. This implies that it has four protons in its nucleus and four electrons in its shells. In a neutral atom, the number of electrons on the shells is equal to the number of protons in the nucleus.

The electronic configuration of Beryllium is 1s2 2s2. This implies that it should have two shells each containing only two electrons each.

Since we are using white foam balls for protons and black foam balls for neutrons, the clear plastic will contain four white foam balls and five black foam balls since the mass number of beryllium is 9 and number of neutrons = mass number - number of protons.

Four blue foam balls hanging from strings will represent the electrons around the nucleus.

Any model that corresponds to the description above is the correct answer.

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2 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