Which statement describes the relative energy of the electrons in the shells of a calcium atom?(1) An electron in the first shell has more energy than an electron in the second shell.
(2) An electron in the first shell has the same amount of energy as an electron in the second shell.
(3) An electron in the third shell has more energy than an electron in the second shell.
(4) An electron in the third shell has less energy than an electron in the second shell.

Answers

Answer 1

Answer: The answer is (3) An electron in the third shell has more energy than an electron in the second shell. The energy of electron will increase when number of shell increase.

Answer 2

Answer:

Final answer:

In a calcium atom, 3) an electron in the third shell has more energy than an electron in the second shell. This is because the shell's distance from the nucleus determines the energy of an electron: the farther the shell, the higher the energy.

Explanation:

The relative energy of the electrons in the shells of a calcium atom would be best described by statement 3, which states: 'An electron in the third shell has more energy than an electron in the second shell.' This is due to the fact that electrons occupy shells (also known as energy levels) around an atom's nucleus, and the further an electron is from the nucleus, the higher its energy. This is because it is closer to the outer environment and has a higher potential energy due to its distance from the atom’s positive core.

In the case of calcium, which has an atomic number of 20, it has four energy levels. This means that an electron in the fourth shell would certainly have more energy than an electron in the first, second, or third shells, and an electron in the third shell would have more energy than one in the second or first shells.

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The cheetah has been clocked at 112 km/Hr. What speed would this be in ft/sec?

Answers

All you have to do in convert two units. You need to know that there are 3600 seconds in an hour, and one kilometer is ~3,280.8399 feet. 3,280.8399 * 112 is equal to 367454.0688 feet per hour. Now divide it by 3600 to get it into seconds. That is equal to 102.07057466666667 feet per second. That's your answer.

A) How many moles of gold atoms are present in a 295-gram sample of gold? B) How many gold atoms would there be in the sample?

Answers

The molar mass of gold, Au, as seen on the periodic table, is 196.97 grams per mole. So, do this to find the number of moles in a 295 gram sample:

$(295g\ Au)((1mol\ Au)/(196.97g\ Au)) = 1.49769mol\ Au$

So, a 295 gram sample of gold has about 1.5 moles of gold in it.

To find the number of atoms, multiply the number of moles by Avogadro's number:

$(1.49769mol\ Au)((6.022*10^(23)\ atoms)/(mol)) = 9.02 * 10^(23)\ atoms\ Au$

5 gram of 197 Au contain 1.53*10^22 atoms. so just multiply that with 59 to get 295 grams :)

Which is true of transition metals when moving from left to right on the periodic table?The d sublevels are not filled across the period.
The cation radii become larger across the period.
Atomic radii increase slightly and then start to decrease.
Atomic radii decrease slightly and then start to increase.

Answers

The correct option is D.

Transition metals are those metallic elements that are located in the central block of the periodic table. They exhibit various valency states and they generally have various colors. The atomic radius of an element refers to the measure of the size of its atom. For transition metals, their atomic radii decrease slightly and then start to increase, when one is moving from the left to the right of the periodic table.

Answer : The correct option is, Atomic radii decrease slightly and then start to increase.

Explanation :

Transition metals : It is defined as the element whose atom is ground state or ion in one of the common oxidation states, has incomplete d-subshell.

The general trend of atomic radii of transition metals is :

Generally the atomic radii of d-block elements in a series decreases with increase in atomic number but decreases in atomic size is small after midway that means almost remains same and at the end of the period there is a slight increase in the atomic radii.

Hence, the atomic radii decrease slightly and then start to increase of the transition metals when moving from left to right on the periodic table.

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 1.50 g of butane?

Answers

Answer:

6.21 × 10²² Carbon Atoms

Solution:

Data Given:

Mass of Butane (C₄H₁₀) = 1.50 g

M.Mass of Butane = 58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 1.50 g ÷ 58.1 g.mol⁻¹

Moles = 0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

Moles = Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

Number of C₄H₁₀ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of C₄H₁₀ Molecules = 0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

Number of C₄H₁₀ Molecules = 1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

1 Molecule of C₄H₁₀ contains = 4 Atoms of Carbon

So,

1.55 × 10²² C₄H₁₀ Molecules will contain = X Atoms of Carbon

Solving for X,

X = (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X = 6.21 × 10²² Atoms of Carbon

$\boxed{6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ atoms}}}$ of carbon is present in 1.50 g of butane.

Further Explanation:

Avogadro’s number indicates how many atoms or molecules a mole can have in it. In other words, it provides information about the number of units that are present in one mole of the substance. It is numerically equal to ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ . These units can be atoms or molecules.

The formula to calculate the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ is as follows:

${\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \frac{{{\text{Given mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}{{{\text{Molar mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}$ …… (1)

Substitute 1.50 g for the given mass and 58.1 g/mol for the molar mass of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ in equation (1).

$\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{1}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\n&= {\text{0}}{\text{.0258 mol}}\n\end{aligned}$

Since one mole of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ has ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . Therefore the formula to calculate the molecules of butane is as follows:

${\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \left( {{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}} \right)\left( {{\text{Avogadro's Number}}} \right)$ …… (2)

Substitute 0.0258 mol for the moles of ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ and ${\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ for Avogadro’s number in equation (2).

$\begin{aligned}{\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\mathbf{}}&=\left( {0.0258{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} * {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\n&= 1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}} \n\end{aligned}$

The chemical formula of butane is ${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}$ . This indicates one molecule of butane has four atoms of carbon. Therefore the number of carbon atoms can be calculated as follows:

$\begin{aligned}{\text{Atoms of carbon}} &= \left( {1.554 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}} \right)\n&= 6.216 * {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ C atoms}} \n\end{aligned}$

Learn more:

Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: 1.50 g, 58.1 g/mol, butane, C4H10, Avogadro’s number, $6.216*10^22$ C atoms, $1.554*10^22$ molecules, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.

He conjugate pair for a weak acid is--

Answers

The stronger acid, however, is on the left side of the equation. The general rules suggest that the stronger of a pair of acids must form the weaker of a pair ofconjugate bases. The fact that HCl is a stronger acid than the H3O+ ion implies that the Cl- ion is a weaker base than water.

Which compound contains both ionic and covalent bonds?(1) ammonia
(2) methane
(3) sodium nitrate
(4) potassium chloride

Answers

Answer is (3)- Sodium Nitrate.

Normally ionic bonds can be seen betweenmetals and non-metals while covalentbonds present betweennon-metals. Another thing that determines the bond nature is electronegativityvalue of the atoms.If the electronegativity differenceis high, then that bond tends to be an ionic bond.

Sodium nitrate consists of Na⁺ and NO₃⁻ ions. Hence, the bondbetween Na⁺ and NO₃⁻ is an ionicbond.

NO₃⁻ is made from N and O. Both are non-metallicatoms. The electronegativities of N and O are 3.0 and 3.5 respectively. Hence, there is nobig difference betweenelectronegativity values (3.5 - 3.0 = 0.5). Hence, the bondbetween N and O is a covalentbond.

Compound (3) sodium nitrate contains both ionic and covalent bonds.

Compound (3) sodium nitrate contains both ionic and covalent bonds. Sodium nitrate consists of the ions Na+ and NO3-, where the bond between Na+ and NO3- is predominantly ionic, and the bond within the NO3- ion is covalent. The sodium ion (Na+) donates an electron to the nitrate ion (NO3-), resulting in an ionic bond. However, within the nitrate ion, the nitrogen (N) and oxygen (O) atoms share electrons, forming covalent bonds.

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