What happens when the electron moves from the first energy level to the second energy level?
Energy is absorbed, and an emission line is produced.
Energy is released, and an emission line is produced.
Energy is absorbed by the atom.
Energy is lost from the atom.

The density given is 1.04 g/mL, or also equivalent to:

density = 1040 kg/m^3

The total mass of seawater is calculated by multiplying density and volume. conversion factor is 1 km = 1000 m:

total mass = (1040 kg/m^3) * (8.01 km^3) * (1000 m / 1 km)^3

total mass = 8.3304 x 10^12 kg

Since Mg2+ is 0.12% by weight, in fraction this is 0.0012 by weight. So:

mass Mg2+ = (8.3304 x 10^12 kg) * 0.0012

mass Mg2+ = 9.996 x 10^9 kg

Answer:

number of gold atoms = [tex]1.528X10^{22}[/tex]

Explanation:

The mass of one atom of gold is given to be = [tex]3.271X10^{-22}[/tex]

The mass of gold is given to be = [tex]5.00g[/tex]

In order to determine the number of atoms present in the 5 g of gold, we will divide the mass of the gold with the mass of each atom of the gold.

the number of atoms = [tex]\frac{massofgold}{massofatom}=\frac{5}{3.271X10^{-22}} =1.528X10^{22}[/tex]

Thus number of gold atoms = [tex]1.528X10^{22}[/tex]

To calculate the mass of oxygen per gram of sulfur for sulfur dioxide, determine the molar masses of sulfur and oxygen in each compound and use the given masses of oxygen and sulfur to calculate the ratio.

To calculate the mass of oxygen per gram of sulfur for sulfur dioxide, we need to find the molar masses of sulfur and oxygen in each compound. The molar mass of sulfur dioxide is 32.06 g/mol for sulfur and 2 x 16.00 g/mol for oxygen, while the molar mass of sulfur trioxide is 48.06 g/mol for sulfur and 3 x 16.00 g/mol for oxygen. Using the given masses of oxygen and sulfur for each compound, we can then calculate the mass of oxygen per gram of sulfur.

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Answer : The number of moles of oxygen atoms present in a sample is 0.661 moles.

Explanation :

As we know that,

1 mole of substance contains [tex]6.022\times 10^{23}[/tex] number of atoms.

As we are given that:

Number of oxygen atoms = [tex]3.98\times 10^{23}[/tex]

Now we have to determine the moles of oxygen atoms.

As, [tex]6.022\times 10^{23}[/tex] atoms of oxygen present in 1 mole of oxygen

So, [tex]3.98\times 10^{23}[/tex] atoms of oxygen present in [tex]\frac{3.98\times 10^{23}}{6.022\times 10^{23}}=0.661[/tex] mole of oxygen

Therefore, the number of moles of oxygen atoms present in a sample is 0.661 moles.

Answer is: Identical particles were produced no matter which metal he used.

J.J. Thomson placed two oppositely charged electric plates around the cathode ray. He did experiments using different metals as electrode materials and found that the properties of the cathode ray remained constant no matter what cathode material he used.

Tomson concluded that atoms are divisible and that the corpuscles are their building blocks (atoms are made up of smaller particles).

J. J. Thomson discovered the electron in 1897.

His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.

With this model, he abandoned his earlier hypothesis (the atom was composed of immaterial vortices).  

 For heavy elements the ratio of neutrons to protons that predicts a stable nucleus is 1 .5 : 1

The correct option is b. 1.5 : 1

a) 1:1

b) 1.5:1

c) 2:1

d) 10:5

Answer:

Explanation:

We can fill maximum two electrons in one orbital. The s subshell has 1 orbital that can filled upto two electrons, the p subshell has 3 orbitals that can filled upto six electrons, the d subshell has 5 orbitals that can filled upto ten electrons, the f subshell has 7 orbitals that can filled upto fourteen electrons.

According to the energy level diagram, the 2p, 3p, 4p each can hold 6 electrons because they have 3 orbitals, and 3d, 4d each can hold 10 electrons because they have 5 orbitals, the 5f, 6f, each can hold 14 electrons because they have 7 orbitals. Therefore, the maximum number of electrons would be present in f subshell or 6f orbital.

The reasoning for this is that it is more difficult to lose two electrons compared to losing just one electron.

Group 1 of the periodic table includes hydrogen and alkali metals. Because they have just one valence electron, group 1 elements are very reactive.

Alkali metals are among the most reactive metals in the periodic table and this is due to their larger atomic radii and low ionization energies. They tend to donate their electrons in reactions and have an oxidation state of +1. 4.

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Answer: Observational

Explanation:

The method in which there is a observation of characters without being interfering into their natural processes is known as observational qualitative research.

There are many different types of research method which uses different techniques or experiments to find the desired result.

In this case the moneys are observed in their habitat without being interfering into their habitat. How they walk, how they react, how they eat and many more circumstances are observed only.

Substances are classified based on their chemical composition into elements (e.g., gold), compounds (e.g., salt, pure water, carbon dioxide), homogeneous mixtures (e.g., salt water, pure air), and heterogeneous mixtures (e.g., soil, bronze).

Substances can be classified into four main categories based on their chemical composition: elements, compounds, homogeneous mixtures, and heterogeneous mixtures. The classification for each of the given substances is as follows:

An element is a pure substance that cannot be broken down into simpler substances. A compound is a pure substance that consists of two or more elements chemically combined in a fixed ratio. A homogeneous mixture, or solution, has uniform composition throughout, whereas a heterogeneous mixture has a composition that varies from point to point.

Compounds can be exemplified in two main ways. These are Chemical Formula and Molecular Model.

Chemical formula includes:

Structural formula – this is a formula that shows the arrangement of atoms in the molecule of a compound.

Molecular formula – this is a formula giving the number of atoms of each elements to present in one molecule of a specific compound.

Empirical formula – this is a formula giving the proportions of elements present in a compound.

Molecular model:

Space-filling model - represents atoms fill the space between each other.

Ball and stick model - represents atoms as balls and chemical bonds as sticks.

Remember that a compound is a substance that is formed when two or more elements are bonded together.

Compounds can be represented in several ways including Molecular Formulas, Empirical Formulas, Structural Formulas, Ball-and-Stick Models, and Space-Filling Models. Each method provides different levels of detail and helps us understand the composition and structure of chemical compounds.

Compounds can be represented in several ways: Molecular Formulas, Empirical Formulas, Structural Formulas, Ball-and-Stick Models, and Space-Filling Models.

Molecular formulas use chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. For instance, the molecular formula for water is H2O, indicating there are two hydrogen atoms and one oxygen atom.

Empirical formulas give the simplest whole-number ratio of atoms in a compound. For water, the empirical formula would still be H2O.

Structural formulas indicate the bonding arrangement of the atoms in the molecule, giving more detailed information about the molecule's structure.

Ball-and-stick models show the geometric arrangement of the atoms in a molecule with atomic sizes not to scale. These models often use balls to represent atoms and sticks to represent chemical bonds.

Lastly, Space-filling models also show the geometric arrangement of atoms, but in these models, the relative sizes of the atoms are represented accurately and the model gives a more realistic view of the molecule's shape.

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Density is the amount of matter within a given amount of space. 

Water has a density of 1.0 gram per centimeter squared.

Mercury has a density of 13.6 grams per centimeter squared. 

Thus, within the same volume, mercury has more mass than water. Thus, mercury has a higher density. 

Density is an intensive property as it does not depend on the quantity of the substances Whereas mass and volume are extensive property. Mercury is more denser than water.

Density tells about the compactness of the substances, how much dense is the substances in other words. Object that is more denser than water they just sink in the water.

Mathematically,

Density = Mass of the substance ÷volume of substance

From above formula we can see that density is directly proportional to mass of substances at constant volume.

The  volume of  water  and mercury is same that is 1 liter. Since the mass of  mercury that is 13.6 kilograms is more than mass of water that is 1 kilogram. So, the density of mercury is more than the density of water which means mercury is denser than water.

Therefore mercury is more denser than water.

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The mass of mercury is divided by density.

100 g / 13.54 g/ml = 7.3855 ml

Since, we only have 3 significant digits in 100 and round the result to 3 significant digits. So, 7.3855 ml = 7.39 ml.

Density is defined as the mass per unit volume it means that mass present in 1 meter cube is called density. The S.I unit of density is kg/m^3 and in C.G.S it is gram/cm^3.

So In above question can understand that density, mass, and volume all are convert to each other it means that if we know any two variable then third one will be calculated easily.  The S.I unit of density is kg/m^3 and in C.G.S it is gram/cm^3.

Mathematically,

Formula for density will be as mentioned below:-

Density = Mass/Volume.

From above formula we can calculate mass which is as follows:-

Mass = Volume × Density.

S.I unti of mass is kilograms and C.G.S unit is gram.

Therefore, The mass of mercury is divided by density.

100 g / 13.54 g/ml = 7.3855 ml

Since, we only have 3 significant digits in 100 and round the result to 3 significant digits. So, 7.3855 ml = 7.39 ml.

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We can solve this problem using Raoults law. The formula for Raoults law is:

P_a = X_a * P

where,

P_a = the partial pressure of substance A

X_a = the mole fraction of substance A

P = total pressure

Therefore calculating for P:

P = P_a / X_a

P = 0.580 atm / 0.35

P = 1.657 atm

0.52 on edg 2020

Explanation: