In adding the measurements 11.074 m, 18.2m, and 16.943 m, what should be the number of scientific figures in the result

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

6 0

3 years ago

2. A cell with few energy needs would most

sattari [20]

A Cell with few energy needs would most likely contain a small number of Mitochondria.

All cells require energy to function, but cells typically have significant energy needs that can only be met by the mitochondria, the cell's powerhouse.
They transform glucose into ATP, a chemical with a huge energy storage capacity.
Muscles have a large number of mitochondria, allowing them to react rapidly and powerfully to the body's ongoing need for energy.
Macromolecules, defunct cell components, and microbes are all digested by lysosomes.
Vacuoles are typically tiny and aid in the sequestration of waste.
The ribosome, an intercellular structure consisting of both RNA and protein, is where a cell produces new proteins.

Therefore out of all these cell organelles, the cell has fewer mitochondria for less energy need.

Learn more about cell organelles here:

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5 0

2 years ago

Write down the formula for B<br> example:<br> Hydrogen + Fluorine = Hydrogen Fluorine <br><br> help

Tom [10]

Answer:

P³⁻ + Cl⁻ --> PCl₃

Explanation:

PCl₃: phosphorus trichloride. prefix in front of chloride is "tri"–meaning three.

5 0

3 years ago

Read 2 more answers

What is the formula for the ionic compound formed between sodium and oxygen?

CaHeK987 [17]

Answer:The formula for the compound formed between sodium and oxygen will be Na2O; a ratio of 2 Na atoms for each O atom.

Explanation:

8 0

3 years ago

Read 2 more answers

When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo

Marta_Voda [28]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

7 0

3 years ago