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3 years ago

Identify the correct charge of each atomic particle

Chemistry

2 answers:

Olegator [25]3 years ago

8 0

Answer : The correct option is, (B) Electron = negative, proton = positive, neutron = no charge

Explanation :

As we know that an atom is the smallest unit of a matter that consist of three subatomic particles which are electrons, protons and neutrons.

The protons and the neutrons are located inside the nucleus or center of the nucleus where the mass of the an atom is concentrated and the electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and the neutrons are neutral that means it has no charge.

Hence, the correct option is, (B) Electron = negative, proton = positive, neutron = no charge

MA_775_DIABLO [31]3 years ago

7 0

Answer: The correct answer is Option B.

Explanation:

There are 3 sub-atomic particles present in an atom. They are: Electrons, protons and neutrons.

Electrons: They are negatively charged particles and are present around the nucleus in the orbits.
Protons: They are positively charged particles and are present inside the nucleus of an atom.
Neutrons: They are neutral particles which means they do not carry any charge. They are present in the nucleus of an atom.

Hence, the correct answer is Option B.

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What is the mass of 975 mL of mercury? Its density is 13.5 g/mL

miss Akunina [59]

Answer:

72.22 g

Explanation:

975 mL Mercury× 13.5 g/mL = 72.22 g

7 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago

choose five elements from the periodic table and arrange them according to increasing atomic radii increasing ionization energy

gregori [183]

Choose five elements from the periodic table and arrange them according to increasing atomic radii increasing ionization energy increasing electronegativity in increasing metallic property

6 0

3 years ago

Why does salt dissolve in water?

tigry1 [53]

The water dissolves the salt because the water molecules are able to interact with the salt-forming particles, called ions. When the water Interacts with the trainers of the salt ions, the solid salt crystal structure Suffers breakdown, until the trainers of the salt ions are completely surrounded by water molecules. At this time the salt is fully dissolved by water.

hope this helps!

4 0

3 years ago

A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t

Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

moles of PbI₂
Limiting reactant
% yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

= 1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0

3 years ago