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Sergio039 [100]
3 years ago
8

One way in which elements differ from each other is the structure of the electron cloud in each elements Atoms in electron cloud

an electron that is farther away from the nucleus has
Chemistry
2 answers:
Bond [772]3 years ago
5 0

The question is incomplete, the complete question is:

One way in which elements differ from each other is the structure of the electron cloud in each element’s atoms. In an electron cloud, an electron that is farther away from the nucleus has

a greater charge than an electron near the nucleus.

a smaller charge than an electron near the nucleus

a higher energy than an electron near the nucleus.

a lower energy than an electron near the nucleus.

Answer:

a higher energy than an electron near the nucleus.

Explanation:

Energy levels increase outwards from the nucleus. An electron that is nearest to the nucleus is said to be in its ground state, this is the lowest energy state of that electron. As we move outwards from the nucleus, the energy levels increase progressively. Hence, the farther electrons are from the nucleus, the greater their energy level and the greater the energy of the individual electrons.

BartSMP [9]3 years ago
4 0

Not sure what you are asking. I have two possible answers though...

It could either be more negatively charged, or valence electrons.

The more away from the nucleus a electron is, the more negatively charged it is.

The electrons on the outermost electron shell is valence electrons.

Again, I don't know what you were asking, but one of these answers may be correct.

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How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0
3 years ago
A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr
sergey [27]

Answer:

184.62 ml

Explanation:

Let p_1, v_1, and T_1 be the initial and p_2, v_2, and T_2 be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

p_1=p_2 ...(i)

v_1 = 200 ml

T_1= 26 ^{\circ}C = 273+26 =299 K

T_2= 3 ^{\circ}C = 273+3 =276 K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)

For the final condition,

p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)

Equating equation (i), and (ii)

\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

\frac{v_1}{T_1}=\frac{v_2}{T_2}  [from equation (i)]

v_2=\frac{T_2}{T_1} \times v_1

Putting all the given values, we have

v_2=\frac{276}{299} \times 200 = 184.62 \; ml

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0
3 years ago
What makes physical change a physical change and a chemical change a chemical change
Snowcat [4.5K]
A physical change is any change in a substances form that does not change its chemical makeup. Examples of physical changes are breaking a stick or melting ice. A chemical change occurs when atoms of a substance are rearranged, and the bonds between the atoms are broken or formed. HOPE THIS HELPS!!
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2 years ago
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olga2289 [7]

Answer:

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3 0
3 years ago
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Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)

Explanation:

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V

In an electrochemical cell at 25°C  the potentials of the  semi-reactions are usually measured  in the sense of reduction  and generally the standard potential between both electrochemical cells will be:

E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

<em>E⁰=1.40 V</em>

<em><u>The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)</u></em>

7 0
3 years ago
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