What instruments are used to measure moleculars vibration , rotation and translation?

Answer:

atom

Explanation:

The sodium atom has a single valence electron that it can easily lose. (If the sodium atom loses its valence electron, it achieves the stable electron configuration of neon.) The chlorine atom has seven valence electrons and can easily gain one electron.

Answer:

Explanation:

First you will find the mole from the molarity and then the desired mass from the mole.

Answer:

Explanation:

Firstly, let us calculate the molar mass of the glucose. To find the molar mass we need to add the masses of individual elements which constitute one glucose molecule.

Now, we know that

Molar mass of Carbon C=12gmol−1

Molar mass of Hydrogen H=1gmol−1

Molar mass of oxygen O=16gmol−1

Therefore, Molar Mass of glucose (C6H12O6) can be calculated as shown below

⇒ 6×12+12×1+6×16

⇒72+12+96⇒180gmol−1

Given mass of glucose = 300g

Now, we can calculate the number of moles in given mass of glucose, using the below formula,

Using the Formula numberofmoles=givenmassmolarmass we get

numberofmoles=300180 = 1.7 moles or 2 moles (approx.)

Hence the number of moles present in 300 g of glucose is 1.7 moles or 2 moles approximately.

The rate of a chymotrypsin-catalyzed reaction will increase by using more

enzyme in the reaction mixture.

The rate of a chemical reaction and the enzyme concentration have a direct

relationship. As concentration increases, the rate of a chemical reaction  

also increases and vice versa.

In this scenario, we were told more enzyme was used in the reaction mixture

which signifies an increase in the concentration and a corresponding

increase in the rate of reaction.

Read more on https://brainly.com/question/12330608

Answer:

The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.

Explanation:

ok done. Thank to me :>

Answer: Mass percent of Li: total mass Li = 6.94 g/mol. % Li = mass Li mass LiClO 2 × 100 % Li = 6 . ...

Mass percent of Cl: total mass Cl = 35.45 g/mol. % Cl = mass Cl mass LiClO 2 × 100 % Cl = 35 . ...

Mass percent of O: total mass O = 32.00 g/mol.

Explanation:

Halogens combine with alkanes to yield halogenoalkanes. It is a chain reaction.

Alkanes are a group of organic molecules whose functional group is the carbon - carbon single bond. The halogenation of alkanes refers to the process by which alkanes reacts with halogen molecules to yield halogenoalkanes.

The following are true statements regarding the halogenation of alkanes;

Learn more: https://brainly.com/question/17638582

Answer:one mole of gold

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: https://brainly.com/question/22655760

Answer:  251 K

Explanation: hope this helps :)

Answer:

KNO2, KBr

Explanation:

Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.

Answer:

a.2.55e-7

b.6830

Explanation:

Iodine is much more easily polarizable than fluorine therefore temporary dipoles in the molecule are strengthened.

The halogens are members of group 17 in the periodic table. They are highly electronegative and seldom occur free in nature owing to their high level of reactivity.

We know that larger molecules are more easily polarized than smaller ones. Therefore, from Fluorine to lodine, the molecular mass increases. The  electron polarizability increases with the mass increase. In turn,  this strengthens the temporary dipole (dispersion) forces  between particles. Hence the properties of halogens change smoothly down the group.

Learn more: https://brainly.com/question/11324711

Answer:

The process by which food is broken down to release energy is called respiration

I hope it helps.

Answer:

Respiration

Explanation:

hope this helps.

Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.

If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Learn more: https://brainly.com/question/14281129

Answer:

if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.

Explanation:

As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.

The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.

Molar mass of sodium carbonate = 106 g/mol

Molar mass of potassium carbonate = 138 g/mol

Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles

Mass of HNO3 =  0.008903 moles × 63 g/mol = 0.56 g

Mass of sample added = 1.091 g

Mass of sample left over = 0.573 g

Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g

The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.

Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O

106g of Na2CO3 reacts with 126g of HNO3

x g of Na2CO3 reacts with (126 × x/106)g of HNO3

K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O

138 g of K2CO3 reacts with 126 g of HNO3

(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g

Total mass of HNO3 used;

1.19x + 0.47 + 0.91x = 0.56

2.1x + 0.47 = 0.56

2.1x = 0.56 - 0.47

2.1x = 0.09

x =  0.09/2.1

x = 0.0043 g

Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g

Mass percent of K2CO3  = 0.5137 g/ 1.091 g × 100/1 = 47.1%

Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%

Learn more: https://brainly.com/question/9743981

Answer:

Because oil have apolar molecules, while water have polar molecules

Answer:Nothing

Explanation:

The answer is nothing the tempatature isnt matched with the degrees this is false

Answer:

12 kilograms is larger.

Explanation:

Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.

Answer:

The Nucleus...

it perform many activities and stores DNA in a cell.

Using melting and boiling temperature, hardness and electric current passing testing.

Ionic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.

In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.

Finally, ionic solids are also excellent conductors of electricity.

So, their properties used to determine whether or not they are ionic solids are  melting and boiling temperature, hardness and electric current passing testing.

Learn more about ionic solids: brainly.com/question/8236583

Electrolysis is the production of a chemical reaction by passing an electric current to liquid containing ions.

Explanation:

BaF2(s) <------> Ba2+(aq) + 2F-(aq)

Ksp BaF2 =  1.0 x 10^-6.

Ksp BaF2 = [Ba2+(aq)]×[F-(aq)]^2 at equilibrium

When Qsp >Ksp, BaF2 will precipitate

Qsp = [Ba2+(aq)]×[F-(aq)]^2

[Ba2+(aq)]×[F-(aq)]^2 > 1.0 x 10^-6.

0.0122 moldm-3 × [F-(aq)]^2 > 1.0 x 10^-6

[F-(aq)]^2 > 1×10^-6 / 0.0122 mol2dm-6

[F-(aq)]^2 > 81.96 × 10^-6 mol2dm-6

[F- (aq)] > 9.05 × 10^-3 moldm-3

So F- concentration should be more than 9.05 × 10^-3 moldm-3

Answer:

The partial pressure of water is much less than 1atm, so the free energy change for the process must be negative with ΔGrxn<0.

Explanation:

Water vapor will condense into liquid water when the water vapor is in its standard state, with a pressure of 1atm. Under normal circumstances, the partial pressure of water vapor is much less that 1atm. It is in a nonstandard state. The free energy change of this reaction has to be negative in order to make the reaction spontaneous.

The criteria of the spontaneous process are the change in the Gibbs free energy should be negative. Above the normal boiling point, the TΔS will be greater than ΔH and ΔG<0.

Gibbs free energy can be described as a state function therefore it doesn’t depend on the path. The change in free energy is equal to the change in enthalpy minus the product of entropy change and temperature of the system.

ΔG = ΔH - Δ (TS)

If the process is carried out at constant temperature, ΔT = 0:

ΔG = ΔH – TΔS

ΔG > 0 for the reaction is non-spontaneous, ΔG < 0 for the reaction is spontaneous, exergonic and ΔG = 0 for the reaction is at equilibrium.

Spontaneous can be described as a reaction that occurs by itself without any external action towards it. The non-spontaneous process needs constant external energy applied to continue and as the external action, the process will cease.

Learn more about Gibbs free energy, here:

https://brainly.com/question/9179942

#SPJ5

Answer:

Step 1: Prepare for the project.

a) Read the entire Student Guide before you begin this project.

b) If anything is unclear, be sure to ask your teacher for assistance before you begin.

c) Gather the materials you will need to complete this project.

Step 2: Review the chemical reaction.

a) The chemical reaction that you will model is shown below. Calcium oxide (CaO) is a white

solid with a crystalline structure. It is made by heating limestone, coral, sea shells, or chalk,

which are composed mainly of calcium carbonate (CaCO3). During the heating process,

carbon dioxide (CO2) is released and calcium oxide (CaO) is produced. Commercially,

calcium oxide is called lime. One of the oldest uses of lime is to make mortar, a substance

used in construction to secure bricks, stones, and blocks together.

CaCO3  CaO + CO2

Step 3: Build a model of the reactant.

a) Use gumdrops and toothpicks to build a model of CaCO3.

b) Select one color of gumdrop for calcium, a second color for carbon, and a third color for

oxygen. Be sure to include a key of your chosen color scheme with your model.

c) Use the periodic table to help you determine the number of bonds each atom will form.

Step 4: Build models of the products.

a) Use gumdrops and toothpicks to build a model of CaO and a model of CO2.

b) Be sure to use the same colors of gumdrops for calcium, carbon, and oxygen as you did in

Step 3. Include a key of your chosen color scheme with your models.

c) Use the periodic table to help you determine the number of bonds each atom will form.

Step 5: Type one to two paragraphs that describe your models and explain the conservation of

mass in the chemical reaction.

a) Create a new blank document. Type your name at the top.

b) Type one to two paragraphs that describe your models and relate them to the law of

conservation of mass. Your document should:

i. identify the names of the reactants and products in the reaction.

ii. identify the number of molecules that make up the reactants and products.

iii. identify the type and number of atoms in each molecule of the reactants and products.

iv. explain what happens during the chemical reaction.

v. explain how mass is conserved during the chemical reaction.

Step 6: Evaluate your project using this checklist.

If you can check each criterion below, you are ready to submit your project.

 Did you create an accurate model of calcium carbonate (CaCO3)? Your model should include

the correct number of gumdrops for each element in calcium carbonate, consistent use of

colors for elements in calcium carbonate, and the correct number and placement of toothpicks

(bonds).

 Did you create an accurate model of calcium oxide (CaO)? Your model should include the

correct number of gumdrops for each element in calcium oxide, consistent use of colors for

elements in calcium oxide, and the correct number and placement of toothpicks (bonds).

 Did you create an accurate model of carbon dioxide (CO2)? Your model should include the

correct number of gumdrops for each element in carbon dioxide, consistent use of colors for

elements in carbon dioxide, and the correct number and placement of toothpicks (bonds).

Did you type one to two paragraphs that describe your models and relate them to the law of

conservation of mass? Your document should include the names of the reactants and

products in the reaction, the number of molecules that make up the reactants and products,

and the type and number of atoms in each molecule of the reactants and products. It should

also explain what happens during the chemical reaction and how mass is conserved during  the reaction

Step 7: Revise and submit your project.

a) If you were unable to check off all of the requirements on the checklist, go back and make  

b) When you have completed your project, submit your models to your teacher for grading.  

c) Submit your document through the virtual classroom. Be sure that your name is on it.

Step 8: Clean up your workspace.

a) Clean up your workspace. Return any extra materials to your teacher and throw away any  trash.

Explanation:

Answer:

[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: https://brainly.com/question/12913997

This question is describing the following chemical reaction at equilibrium:

[tex]A\rightleftharpoons B[/tex]

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

[tex]K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25[/tex]

Thus, by recalling the Van't Hoff's equation, we can write:

[tex]ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

Hence, we solve for the enthalpy change as follows:

[tex]\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }[/tex]

Finally, we plug in the numbers to obtain:

[tex]\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}[/tex]

Learn more:

Answer:

It depends on how much the calories and fat there is in the meals and if you don't get enough physical activity of the same amount of food that you eat it can lead to weight gain

Explanation: