Determine the mass percent of metal and its atomic mass.

2. What group is the atom in? 2 or 2A 6 14 1

Juliette [100K]

Answer:

$\Large \boxed{\mathrm{Group \ 4}}$

Explanation:

The number of valence electrons tells us the group number of the neutral atom.

The atom has 4 valence electrons.

The atom is in group 4.

5 0

3 years ago

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A substance has a atomic number of 80.How many protons and electrons do atoms of the substance have ?what is the substance

Anarel [89]

Answer: Substance is Mercury (Hg)

number of protons = 80

number of electrons = 80

Explanation:

Atomic number which is symbolized by "Z" is the number of protons in nucleus of an atom.

Hence,

number of protons = atomic number = 80

By knowing atomic number, we can find the element by using periodic table. Since the atomic number 80, the element is Mercury (Hg).

If the substance is in its neutral state, which means there is no charge on it, then the number of protons equals to the number of electrons of that element.

Hence,

number of electrons = number of protons = 80

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3 years ago

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What’s are the different Types of chemical weathering

Ira Lisetskai [31]

Answer:

Types of chemical weathering

1. Hydrolysis

2. Oxidation

3. Carbonation

4. Acid rain

5. Acids produced by lichens

Explanation:

Chemical weathering occurs when rocks undergo chemical reactions to form new minerals.

Hydrolysis : is when water dissolves minerals present in rocks forming new compounds.

Oxidation : is when oxygen reacts with rocks eg. rust formation

Carbonation : it uses an acid known as carbonic acid, it is important in the formation of many caves and sinkholes.

8 0

2 years ago

Why does every plant cell have a nucleus

Lena [83]

Every plant cell has a nucleus because a nucleus controls the cell's main activities.

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2 years ago

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A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

2 years ago