State, in terms of subatomic particles, how an atom of C-13 is different from an atom of C-12

Answers

Answer 1

Answer: C-12 and C-13 are isotopes of the element carbon. Since they are both of the same element, they have the same number of protons. The number that follows the C indicates the number of neutrons. Isotopes are variations of the element with different number of neutrons. C-12 shows that that variation of carbon has 6 protons and 6 neutrons for a total mass of 12. C-13 shows that it has 6 protons but 7 neutrons for a total mass of 13

Related Questions

The saturation index for a pool measures the balance between the acid level and the amount of minerals in pool water. Balanced water has an index value of 0. Water is highly corrosive or highly scale forming if the absolute deviation of the index value from 0 is greater than 0.5. Find the index values for which pool water is highly corrosive or highly scale forming.
Which substance yields H+(aq)as the only positive ion in an aqueous solution
The two basic properties of all matter are volumelength and densitymass.
Number of CO2 molecules in 0.0180 mol CO2
The force that occurs when plates are pushed together are called____

The measurement of moles of solute per kilogram of solvent is known asa. molarity.
b. molality.
c. parts per million.

Answers

Option b: molality

That is the definition of molality.

Which formula represents a hydrocarbon?(1) CH3CH2CH2CHO
(2) CH3CH2CH2CH3
(3) CH3CH2CH2COOH
(4) CH3CH2COOCH3

Answers

2 because it is the only one that contains Carbon and Hydrogen only

$CH_(3)CH_(2)CH_(2)CH_(3)$ formula represents a hydrocarbon. Therefore, option (2) is correct.

Hydrocarbons are merely carbon and hydrogen. Aliphatic or aromatic. Aliphatic hydrocarbons are saturated or unsaturated open-chain hydrocarbons.Saturated aliphatic hydrocarbons have single bonds, while unsaturated ones have double or triple bonds.

Aromatic hydrocarbons have alternating double bonds in a ring.Only option 2— $CH_(3)CH_(2)CH_(2)CH_(3)$ —is a hydrocarbon. Butane, a saturated aliphatic hydrocarbon, with four carbon and 10 hydrogen atoms.

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Planck's constant relates the Joules of energy absorbed/released by matter to the:A. atomic number of the element

B. observed wavelength of light

C. speed of light

D. mass of the particulate matter

Answers

Answer: B. observed wavelength of light

Explanation: The relation of Planck's constant relating the Joules of energy absorbed/released by matter is usually used to determine the energy of the photon. The mathematical expression is -

E= h c / $\lambda$

Thus as we can observe that the energy is directly proportional to the speed of light and inversely proportional to the wavelength of the light.

Thus , we can say that Planck's constant relates the Joules of energy absorbed/released by matter to the observed wavelength of light.

The right answer for the question that is being asked and shown above is that: "B. observed wavelength of light" Planck's constant relates the Joules of energy absorbed/released by matter to the B. observed wavelength of light

What other forms of radiation may come from the stars?

Answers

Other forms of radiation may come from the stars. These radiations include x-ray radiation, ultraviolet radiation, electromagnetic radiation, infrared radiation, alpha ray, beta ray & gamma ray, visible light radiation, radio frequencies radiation and magnet waves.

Elements that are characterized by the filling of p orbitals are classified as _____.a. Groups 3A through 8A
b. transition metals
c. inner transition metals
d. groups 1A and 2A

Answers

Answer : The correct option is, (a) Groups 3A through 8A

Explanation :

The general electronic configurations of :

Group 1A : $ns^1$

Group 2A : $ns^2$

Group 3A : $ns^2np^1$

Group 4A : $ns^2np^2$

Group 5A : $ns^2np^3$

Group 6A : $ns^2np^4$

Group 7A : $ns^2np^5$

Group 8A : $ns^2np^6$

Transition metal : $(n-1)d^((1-10))ns^((0-2))$

Inner transition metal (Lanthanoids) : $4f^((1-14))5d^((0-1))6s^2$

Inner transition metal (Actinoids) : $5f^((0-14))6d^((0-1))7s^2$

From the general electronic configurations, we conclude that the groups 3A through groups 8A elements that are characterized by the filling of p-orbitals.

Elements that are characterized by the filling of p orbitals are classified as $\boxed{{\text{a}}{\text{. Groups 3A through 8A}}}$ .

Further Explanation:

In order to make the study of numerous elements easier, these elements are arranged in a tabular form in increasing order of their atomic numbers. Such a tabular representation of elements is called a periodic table. Horizontal rows are called periods and vertical columns are called groups. A periodic table has 18 groups and 7 periods.

a. Groups 3A through 8A

The elements from group 3A to 8A has the general outermost electronic configuration of $n{s^2}n{p^(1 - 6)}$ . So the added electrons are to be filled in p orbitals.

b. Transition metals

These metals have the general valence configuration of $\left( {n - 1} \right){d^(1 - 10)}n{s^(0 - 2)}$ . This indicates that the added electrons enter either s or d orbitals.

c. Inner transition metals

These are classified as lanthanoids and actinoids. The general outermost configuration of lanthanoids is $4{f^(1 - 14)}5{d^(0 - 1)}6{s^2}$ while that of actinoids is $5{f^(0 - 14)}6{d^(0 - 1)}7{s^2}$ . In both cases, the added electron enters either d or f orbitals.

d. Groups 1A and 2A

The elements of group 1A have the general valence electronic configuration of $n{s^1}$ . It implies the last or valence electron enters in the s orbital. The group 2A elements have a general configuration of $n{s^2}$ . Here also the last electron enters the s orbital.

So elements from groups 3A to 8A are classified by the filling of p orbitals and therefore option a is correct.

Learn more:

Which ion was formed by providing the second ionization energy? brainly.com/question/1398705
Write a chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: periodic table, configuration, ns1, ns2, d, p, f, 3A, 8A, transition metals, inner transition metals, lanthanoids, actinoids, orbitals, 1A, 2A.

Balance the following oxidation-reduction reaction: Fe(s)+Na+(aq)→Fe2+(aq)+Na(s) Express the coefficients as integers separated by commas (e.g., 4,1,3,2, where 1 indicates the absence of a coefficient).

Answers

Answer:

The answer to your question is: 1, 2, 1, 2

Explanation:

1 Fe(s) + 2 Na⁺(aq) → 1 Fe²⁺(aq) + 2 Na(s)

Fe⁰ - 2e⁻ ⇒ Fe⁺² Oxidases

Na⁺ + 1 e⁻ ⇒ Na⁰ Reduces

1 x ( 1 Fe⁰ ⇒ 1 Fe⁺²) Interchange number of

2 x ( 2Na⁺ ⇒ 2 Na⁰ ) electrons

Final answer:

To balance the oxidation-reduction reaction Fe(s) + Na+(aq) → Fe2+(aq) + Na(s), follow these steps: balance atoms other than hydrogen and oxygen, balance hydrogen atoms, balance oxygen atoms, verify charges

Explanation:

To balance the oxidation-reduction reaction Fe(s) + Na+(aq) → Fe2+(aq) + Na(s), we need to balance the number of atoms for each element and the total charge on both sides of the reaction. Here's the step-by-step process:

First, balance the atoms other than hydrogen and oxygen. There is only one atom of Fe on each side, so this is already balanced.
Next, balance the hydrogen atoms. We have no hydrogen on the left side and one hydrogen on the right side, so we need to add a coefficient of 2 in front of Na(s): 2Na(s).
Now, balance the oxygen atoms. We have no oxygen on the left side and one oxygen in Fe2+(aq), so we need to add a coefficient of 2 in front of Fe2+(aq): 2Fe2+(aq).
Finally, verify that the charges are balanced. The charge on the left side is 1+ (from Na+) and the charge on the right side is 0 (from Na). To balance the charges, we need to add a coefficient of 2 in front of Na+: 2Na+(aq).

So, the balanced oxidation-reduction reaction is: Fe(s) + 2Na+(aq) → 2Fe2+(aq) + Na(s).

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