What is the empirical formula of C9H18

Answer:

C. 3

Explanation:

I hope my answer will be useful

Answer:

Estoy deprimido y odio la vida. Así que sí

Answer: [tex]CsF[/tex] > [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

Explanation:

The percent ionic character is used to determine if a bond is more ionic or covalent. More is the percent ionic character, the more ionic the molecule is. If the electronegativity difference between two elements is higher than 1.7 it will be considered as ionic.

The electronegativity difference for CsF is (3.98-0.79) = 3.19

The electronegativity difference for [tex]OF_2[/tex] is = (3.98-3.44) = 0.54

The electronegativity difference for [tex]SrF_2[/tex] is (3.98-0.95) = 3.0

The electronegativity difference for [tex]F_2[/tex] is  0.

Thus decreasing order of  ionic character of the bonds is CsF >  [tex]SrF_2[/tex] >  [tex]OF_2[/tex]  > [tex]F_2[/tex]

Answer:

the moral amount of the gas.

Answer: a) [tex]1.7\times 10^{-4}[/tex]

b) [tex]3.4\times 10^{-4}[/tex]

Explanation:

The reaction is :

[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]

Rate = Rate of disappearance of [tex]N_2O_5[/tex] = Rate of appearance of [tex]NO_2[/tex]

Rate =  [tex]-\frac{d[N_2O_5]}{2dt}[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

Rate of disappearance of [tex]N_2O_5[/tex] = [tex]\frac{\text {change in concentration}}{time}[/tex] = [tex]\frac{0.100-0.066}{200-0}=1.7\times 10^{-4}[/tex]

a) Rate of disappearance of [tex]N_2O_5[/tex] = [tex]-\frac{d[N_2O_5]}{2dt}[/tex]

Rate of appearance of [tex]NO_2[/tex] = [tex]\frac{d[NO_2]}{4dt}[/tex]

b) Rate of appearance of [tex]NO_2[/tex] =  [tex]\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}[/tex]

A) Find the rate of disappearance of [tex]N_2O_5[/tex] from t = 0 s to t = 200s

[tex]Rate = \frac{1}{2}(\frac{-\delta N_2O_5}{\delta t})\\\\Rate = -\frac{1}{2}(\frac{0.066 - 0.100}{200 - 0})\\\\Rate = 8.5*10^{-5}[/tex]

B) Find the rate of appearance of [tex]NO_2[/tex] from t = 0 s to t = 200s

According to rate law,

[tex]\frac{1}{2}(\frac{-\delta N_2O_5}{\delta t}) = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\8.5*10^{-5} = \frac{1}{4}(\frac{\delta NO_2}{\delta t})\\\\\frac{\delta NO_2}{\delta t} = 4 * 8.5*10^{-5}\\\\Rate = 3.4*10^{-4}[/tex]

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Answer:

One could be What kind of insulator do you use in the winter?

Explanation:

Answer:

Because of reaction between solid and liquid

Answer:

shouldn't take that long i would say from 30 to 40 minutes bc its melting in warm water its turning from a solid to a liquid.

Explanation:

Answer:

132.14 g/mol

Explanation:

Answer:

Liquids take the shape of their containers because they are stuck together enough to not go flying off like gases but not so tightly that they stay in a stiff shape like solids do.

Explanation:

Hope this helps you! ^^

Answer:

Cu    +    2AgNO₃  →   2Ag    +    Cu(NO₃)₂

Explanation:

The reactants are:

         Copper metal  = Cu

         Silver nitrate  = AgNO₃

The products are:

          Silver metal  = Ag

           Copper (II) nitrate  = Cu(NO₃)₂

So, the reaction equation is given as;

      Cu    +    2AgNO₃  →   2Ag    +    Cu(NO₃)₂

This is a single displacement reaction

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

Answer:

When converting from liters to moles or moles to liters, you will use the equality 1 mole=22.4 L for a gas at STP.

If you have moles and want to convert to (Volume) liters, 1 mole = 22.4 L is the conversion factor that can be used.

The area that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere. Various forms have various volumes. We have studied the several solids and forms that are specified in three dimensions, such as cubes, cuboids, cylinders, cones, etc., in 3D geometry. We will discover how we can determine the volume for each of these shapes. One mole of any gas have 22.4 L at STP.

1 mole = 22.4 L

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Answer:

She overcame her disabilities to compete in the 1956 Summer Olympic Games, and in 1960, she became the first American woman to win three gold medals in track and field at a single Olympics. Later in life, she formed the Wilma Rudolph Foundation to promote amateur athletics.

Explanation:

Answer:

9 . this was not able the first one. it has to go on, I will not sure about

Answer:

Here:

Explanation:

Determine the number of electrons in the atom from its atomic number. (See Below.)

Add electrons to the sublevels in the correct order of filling.

Add two electrons to each s sublevel, 6 to each p sublevel, 10 to each d sublevel, and 14 to each f sublevel.

To check your complete electron configuration, look to see whether the location of the last electron added corresponds to the element’s position on the periodic table.

Predicting the Order of Filling of the Orbitals

There are three ways to predict the order of filling of the orbitals. Probably the least reliable method is to memorize the following list (even though it shows the order of filling of all the orbitals necessary for describing the ground state electron configurations of all of the known elements).

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Instead of relying on memorization, you can use the memory aid shown below to remind you of the correct order of filling of the sublevels. The following steps explain how to write it and use it yourself.

Write the possible sublevels for each energy level in organized rows and columns like the image below. To do this, you need to remember that there is one sublevel on the first principal energy level, two on the second, three on the third, etc. Every principal energy level has an s orbital. The second principal energy and all higher energy levels have a p sublevel. The d sublevels start on the third principal energy level, the f sublevels start on the fourth principal energy level, etc.

Draw arrows like those you see below .

Starting with the top arrow, follow the arrows one by one in the direction they point, listing the sublevels as you pass through them.

The sublevels that are not needed for describing the known elements are enclosed in parentheses below .

Image of the memory aid that allows you to determine the order of filling for the sublevels

We can also use the block organization of the periodic table, as shown below, to remind us of the order in which sublevels are filled. To do this, we move through the elements in the order of increasing atomic number, listing new sublevels as we come to them. The type of sublevel (s, p, d, or f ) is determined from the block in which the atomic number is found. The number for the principal energy level (for example, the 3 in 3p) is determined from the row in which the element is found and the knowledge that the s sublevels start on the first principal energy level, the p sublevels start on the second principal energy level, the d sublevels start on the third principal energy level, and the f sublevels start on the fourth principal energy level.

We know that the first two electrons added to an atom go to the 1s sublevel.

Atomic numbers 3 and 4 are in the second row of the s block (look for them in the bottom half of in image below), signifying that the 3rd and 4th electrons are in the 2s sublevel.

Atomic numbers 5 through 10 are in the first row of the p block, and the p sublevels start on the second energy level. Therefore, the 5th through 10th electrons go into the 2p sublevel.

Atomic numbers 11 and 12 are in the third row of the s block, so the 11th and 12th electrons go into the 3s sublevel.

Because atomic numbers 13 through 18 are in the p block, we know they go into a p sublevel. Because the p sublevels begin on the second principal energy level and atomic numbers 13 through l8 are in the second row of the p block, the 13th through 18th electrons must go into the 3p sublevel.

The position of atomic numbers 19 and 20 in the fourth row of the s block and the position of atomic numbers 21 through 30 in the first row of the d block show that the 4s sublevel fills before the 3d sublevel.

Moving through the periodic table in this fashion produces the following order of sublevels up through 6s:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s

Notice that atomic numbers 57 through 70 on the periodic table below are in the 4f portion of the table. It is a common mistake to forget that the 4f sublevel is filled after the 6s sublevel and before the 5d sublevel. In order to make the overall shape of the table more compact and convenient to display, scientists have adopted the convention of removing the elements with atomic number 57 through 70 and 89 through 102 (the latter being the 5f portion of the table) from their natural position between the s and d blocks and placing them at the bottom of the table. Electrons go into the 5f sublevel after the 7s sublevel and before the 6d sublevel. The second periodic table below shows how the blocks on the periodic table would fit together if the inner transition metals­—the f block—were left in their natural position.

Answer:

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

T

1

P

1

V

1

=

T

2

P

2

V

2

where,

P_1,V_1\text{ and }T_1P

1

,V

1

and T

1

are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2P

2

,V

2

and T

2

are the final pressure, volume and temperature of the gas

We are given:

\begin{gathered}P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K\end{gathered}

P

1

=760mmHg

V

1

=175L

T

1

=15

o

C=[15+273]K=288K

P

2

=640mmHg

V

2

=198L

T

2

=?K

Putting values in above equation, we get:

\begin{gathered}\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K\end{gathered}

288K

760mmHg×175L

=

T

2

640mmHg×198L

T

2

=274K

Hence, the temperature when the volume and pressure has changed is 274 K

Taking into account the definition of density, the mass of the sample of oil with a volume of 10 cm³ and a density of 0.8 [tex]\frac{g}{cm^{3} }[/tex]  is 8 grams.

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

[tex]density=\frac{mass}{volume}[/tex]

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.

In this case, you know that:

Replacing in the definition of density:

[tex]0.8 \frac{g}{cm^{3} } =\frac{mass}{10 cm^{3} }[/tex]

Solving:

mass=  0.8 [tex]\frac{g}{cm^{3} }[/tex] × 10 cm³

mass= 8 g

In summary, the mass of the sample of oil with a volume of 10 cm³ and a density of 0.8 [tex]\frac{g}{cm^{3} }[/tex]  is 8 grams.

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Answer: C- one should have high electron affinity and the other low ionization energy.

Explanation:

Ionization energy is the energy required to remove the electron from an isolated gaseous atom.

Electron affinity is the tendency of an isolated atom to attrcat the electrons towards itsellf.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and it can donate electron easily if it has low ionization energy.

The element which accepts the electrons is known as electronegative element or non metal and it can easily accept the electron if it has high electron affinity.

Answer:

f = 4.6 × 10¹⁴ Hz

Explanation:

Given data:

Wavelength  of light = 6.50x10^-7m

Speed of light = 3.00 x 10^8 m/s

Frequency of light = ?

Solution:

Formula:

Speed of radiation = frequency × wavelength

by putting values,

3×10⁸ m/s  =  f × 6.50x10^-7m

f  =  3×10⁸ m/s   /  6.50x10^-7m

f  =  0.46 × 10¹⁵ s⁻¹

f = 0.46 × 10¹⁵ Hz

f = 4.6 × 10¹⁴ Hz

How would you calculate the change in entropy when the pressure of a perfect gas is changed isothermally from P_i to P_f

Answer:

ΔS = nRIn(P_i/P_f)

Explanation:

Formula for entropy change is;

ΔS = nC_p•In(T_f/T_i) + nRIn(P_i/P_f)

Since the pressure of the perfect gas is changed isothermally, it means that temperature remains the same and as such T_f = T_i

Thus, we now have;

ΔS = nC_p•In(1) + nRIn(P_i/P_f)

In(1) = 0

Thus;

ΔS = nRIn(P_i/P_f)

Answer: 1.  C. polar covalent: electrons shared between silicon and sulfur but attracted more to the sulfur

2. B)  [tex]NH_3[/tex]

3. B) Fluorine

Explanation:

1. A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of sulphur- electronegativity of silicon = 2.5 -1.8 = 0.7

Thus as electronegativity difference is less than 1.7 , the cond is polar covalent and as electronegativity of sulphur is more , the electrons will be more towards sulphur.

2. A molecular compound is usually composed of two or more nonmetal elements. Example: [tex]NH_3[/tex]

Ionic compound is formed by the transfer of electrons from metals to non metals. Example: [tex]Mg_3N_2[/tex] , [tex]AlCl_3[/tex] and [tex]LiBr[/tex]

3.  For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here K is having an oxidation state of +1 and as the compound formed is KZ, the oxidation state of non metallic element Z should be -1. Thus the element Z is flourine which exists as diatomic gas [tex]F_2[/tex]

Heat needed to raise the temperature : Q = 3432 J

Given

c = The specific heat capacity = 2.20 J/g-K

m = mass = 5 g

Δt = Temperature difference = 75-36 =39 °C=312 K

Required

Heat needed

Solution

Heat can be formulated

Q = m.c.Δt

Input the values :

Q = 5 g x 2.2 J/g-K x 312 K

Q = 3432 J

Answer:

cup...................

Answer:

Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are classified as either physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes.

Answer:

I think its C

Explanation:

but don't trust me 2 much

The environment provides a vast array of materials and services that people use to live. Often these resources have competing uses and values.  In order to conserve environment change our resource use habits. The correct option is C.

An environmental resource is any material, service, or information from the environment that is valuable to society. This can refer to anything that people find useful in their environs, or surroundings. Food from plants and animals, wood for cooking, heating, and building, metals, coal , and oil are all environmental resources.

Clean land, air, and water are environmental resources, as are the abilities of land, air, and water to absorb society's waste products. Heat from the sun, transportation and recreation in lakes, rivers, and oceans, a beautiful view, or the discovery of a new species are all environmental resources.

Thus the correct option is C.

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Answer:

an atom is the smallest part of an element that has the properties of that element

it can't be broken further into smaller substances

an element is the simplest chemical

it consists of one type of atoms

so you can consider atoms as the building block of an element

Answer:

0.64 in trial 2

Explanation:

Answer:

See explanation and image attached

Explanation:

Now, we have chosen the alkene 1-propene in our example.

In the first reaction of  1-propene  with HBr, the reaction proceeds by ionic mechanism leading to the formation of 2-bromo propane.

In the second reaction of  1-propene  with the bromine molecule, the first step is the formation of the brominium cation which is a cyclic intermidiate followed by the addition of Br^- yielding the 1,2- dibromopropane product