Draw the structure of 2,4,4,5-tetramethyl-2-hexene.

Answer:

1.001

Explanation:

The Significant Figures are 1 0 0 1, This answer has 4 Significant figures, while the other three have only 2 significant figures

Answer:

See explanation

Explanation:

To write a verbal sentence to represent an equation, we must look at the equation closely in order to capture its essence.

For this equation, 3y= 4y^3, we can write it in words as:

Three multiplied by y is equal to four multiplied by y raised to power three.

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Answer:

Carbon Dioxide

Explanation:

Which is a compound:)

Answer:

2.29 m.

Explanation:

The following data were obtained from the question:

Mass of KOH = 42.3 g

Molar mass of KOH = 56.11 g/mol

Mass of water = 329 g

Molality of KOH = ?

Next, we shall determine the number of mole in 42.3 g of KOH. This can be obtained as follow:

Mass of KOH = 42.3 g

Molar mass of KOH = 56.11 g/mol

Mole of KOH =?

Mole = mass /Molar mass

Mole of KOH = 42.3/56.11

Mole of KOH = 0.754 mole

Next, we shall convert 329 g of water to kilogram (kg). This can be obtained as follow:

1000 g = 1 kg

Therefore,

329 g = 329 g /1000 g × 1 kg

329 g = 0.329 kg

Therefore, 329 g of water is equivalent to 0.329 kg

Finally, we shall determine the molality of the KOH solution ad follow:

Molality is defined as the mole of solute per unit kilogram of solvent (water) i.e

Molality = mole/ mass (kg) of water

Mole of KOH = 0.754 mole.

Mass of water = 0.329 kg.

Molality = mole/ mass (kg) of water

Molality = 0.754/0.329

Molality = 2.29 m

Therefore, the molality of the KOH solution is 2.29 m.

Answer:

2.29

Explanation:

Answer:

Matter can be broken down into two categories: pure substances and mixtures.

Pure substances are further broken down into elements and compounds. ... A chemical substance is composed of one type of atom or molecule.

A mixture is composed of different types of atoms or molecules that are not chemically bonded.

hope this helped

Matter is divided in two categories: pure substances and mixtures.Pure substances are further broken down into elements and compounds.Chemical substance is composed of one type of atom or molecule.A mixture is composed of different types of atoms or molecules .

Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.

Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.

The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.

Learn more about matter,here:

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Answer:

ΔH = 14,7kJ/mol

Explanation:

It is possible to make algebraic sum of several chemical process to obtain enthalpy of a determined reaction (Hess's law).

Sublimation of iodine is (Transition from solid to gas):

I₂(s) → I₂(g) ΔH = 60.2kJ/mol

Vaporization of iodine (From liquid to gas):

I₂(l) → I₂(g) ΔH = 45.5kJ/mol

Fusion of iodine (From solid to liquid can be obtained subtracting the sublimation process - Vaporization process:

I₂(s) → I₂(g) ΔH = 60.2kJ/mol

I₂(g) → I₂(l) ΔH = -45.5kJ/mol

I₂(s) → I₂(l) ΔH = 60.2kJ/mol - 45.5kJ/mol

Answer:

[tex]\rho =8.50g/cm^3[/tex]

Explanation:

Hello,

In this case, due to the volume difference caused by the addition of the metal, one could notice that the volume of the metal is:

[tex]V_{metal}=9.84cm^3-6.85cm^3=2.99cm^3[/tex]

In such a way, given the mathematical definition of density, it turns out:

[tex]\rho =\frac{m}{V}=\frac{25.4253g}{2.99cm^3}\\\\\rho =8.50g/cm^3[/tex]

Regards.

Answer: Carbon, because it is in group 14 and has four valence electrons

Explanation: Just did this quiz

Answer:carbon

Explanation:

Answer:

Three objects with kinetic energy

A ball rolling down the street

Moving Car

Bullet

Law of Conservation of Energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

Though technically there are limitless forms of both types of energy, officially there's five types of kinetic energy: radiant, thermal, sound, electrical and mechanical and potential energy adds gravitational, nuclear, and elastic.

Answer: 2.7 cm

Explanation: ~~~~~ there ya go

Answer:

Define a problem, form a hypothesis, gather experimental data, form a conclusion

Answer:

Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion.

Explanation:

Answer:

The answer is D a beaker

Explanation:

Hope this helps :D

Answer:

a)- Se, S, O

b)- Na, P, Cl

c)- S, Cl, F

d)- P, N, O

Explanation:

We can solve this problem by looking at the Periodic Table. In a group, electronegativity increases from bottom to top; whereas in periods it increases from left to right.

Thus, we order the elements from lower to higher electronegativity as follows:

a. Se, O, S ⇒ order: Se, S, O

Because they are all in the same group. Se is near the bottom, followed by S and O is at the top.

b. P, Na, Cl ⇒ order: Na, P, Cl

They are in the same period. Na is at the left, followed by P and Cl is nearest the right.

c. Cl, S, F ⇒ order: S, Cl, F

P and Cl are in the same period, and P is at the left, so it has the lowest electronegativity. F is in the same group of Cl, but at the top. F has the highest electronegativity.

d. O, P, N ⇒ order: P, N, O

N and P are in the same group, but P is at the bottom so it has the lower electronegativity. N and O are in the same period, but O is at the right, so it is the most electronegative.

Answer:

The atoms of an element are represented in a chemical line drawing with its chemical formula.

Explanation:

chemical structural drawing helps to represent the pattern for which an element is formed. Chemical elements are made up of atoms that represent their single state.

The line drawing is made up of lines (representing the chemical bond between atoms) and the atoms or various atoms that make up the element.

Answer:its a law

Explanation:

I passed that quiz 100%, my answer for that question was law

so its law

Answer:

the density is 6

Explanation:

mass divided by volume equals density

Answer:

a. pH = 4.56

b. Change in pH = 0.85

c. Change in pH = 5.4

Explanation:

a. The pKa of lactic buffer is: 3.86.

Using Henderson-Hasselbalch formula for the lactic buffer:

pH = 3.86 + log [Lactate] / [Lactic acid]

Where [] is molarity of each compound but could be taken as moles

Replacing:

pH = 3.86 + log [0.050 moles] / [0.010 moles]

pH = 4.56

b. The HCl added reacts with Lactate producing lactic acid. Moles of HCl are:

5x10⁻³L * (0.5mol /L) = 0.025 moles HCl

Moles of lactate: 0.050moles - 0.025 moles = 0.025 moles

Moles lactic acid: 0.010 moles + 0.025 moles = 0.035 moles

pH = 3.86 + log [0.025 moles] / [0.035 moles]

pH = 3.71

Change in pH = 4.56 - 3.71 = 0.85

c. 1L of pure water has a pH of 7. 0.025 moles of HCl = 0.025 moles H⁺ in 1.005L:

0.025 mol / 1.005L = 0.0249M = [H⁺]

As pH = -log [H⁺]

pH = 1.6

Change in pH = 7.0 - 1.6 = 5.4

Answer:

1080 meters.

Explanation:

Answer:

A. Substance E

A. Substance C

A. Substance A  

Explanation:

Given that:

At 4 °C, Substance E has a vapor pressure of 86. torr and Substance F has a vapor pressure of 136. torr

Which has a higher boiling point?

A. Substance E

B. Substance F

C. Neither,EandF have the same boiling point

The vapor pressure varies inversely proportional to the boiling point.

[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{boiling \ point}}[/tex]

Therefore, the lower the vapor pressure, the higher the boiling point.

At 4°C, Substance E with a lower vapor pressure of 86. torr will have a higher boiling point from the given information.

2.

Recall that :

[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{enthalpy \ of \ vaporization}}[/tex]

therefore, the lower the enthalpy of vaporization, the higher the vapor pressure at any given temperature.

Given that:

Substance C has an enthalpy of vaporization smaller than that of substance D. Then, substance C has a higher vapor pressure.

3.

We've earlier said that:

The vapor pressure varies inversely proportional to the boiling point.

[tex]\mathbf{vapor \ pressure \ \ \alpha \ \ \dfrac{1}{boiling \ point}}[/tex]

Therefore, the lower the vapor pressure, the higher the boiling point.

As such, Substance A will have a higher boiling point.

Answer:

A property that changes if the amount of substance changes

Explanation:

An extensive property is a property that depends on the amount of matter in a sample.

Answer:

Instrumental methods

surface analysis

high performance liquid chromatography

atomic spectroscopy

potentiometry

Classical methods

precipitation titration

gravimetric analysis

Explanation:

Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.

Many instrumental methods such as HPLC rely on computer screens as readout devices.

Answer:

A. [tex]r=k[NO]^2[H_2][/tex]

B. [tex]k=0.121\frac{L^2}{mol^2*s}[/tex]

C. Second-order with respect to NO and first-order with respect to H₂

Explanation:

Hello,

In this case, for the reaction:

[tex]H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)[/tex]

The rate law is determined by writing the following hypothetical rate laws:

[tex]3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n[/tex]

Whereas we can compute m as follows:

[tex]\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2[/tex]

Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:

[tex]\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1[/tex]

A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:

[tex]r=k[NO]^2[H_2][/tex]

B) Rate constant is computed from one kinetic data:

[tex]k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}[/tex]

C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.

Best regards.

The order with respect to NO has been 2, while the order with respect to water has been 1.

The rate law has been the representation of the chemical concentration responsible for determining the rate of the reaction. The balanced chemical equation has been:

[tex]\rm H_2O\;+\;2\;NO\;\rightarrow\;N_2O\;+\;H_2O[/tex]

A. The rate equation for the reaction has been:

Rate = Rate constant [tex]\rm [H_2O]^m\;[NO]^n[/tex]

Where, m and n are the rate of the respective reactants.

The rate with respect to NO from the given data can be given as:

The ratio of Rate 1 to rate 2 with concentration of NO

[tex]\rm \dfrac{3.822\;\times\;10^-^3}{1.529\;\times\;10^-^2}[/tex] = [tex]\rm \dfrac{[0.3]^m\;[0.35]^n}{[0.6]^m\;[0.35]^n}[/tex]

m = 2

The rate of reaction with respect of NO has been 2.

The rate of reaction with water concentration has been:

[tex]\rm \dfrac{1.529\;\times\;10^-^2}{3.058\;\times\;10^-^2}[/tex] = [tex]\rm \dfrac{[0.6]^m\;[0.35]^n}{[0.6]^m\;[0.7]^n}[/tex]

n = 1

The rate of reaction with respect of water has been 1.

The rate law for reaction has been:

Rate = k [tex]\rm [H_2O]^1\;[NO]^2[/tex]

B. The rate constant can be given as:

Rate = 3.822 [tex]\rm \times\;10^-^3[/tex]

Concentration of NO = 0.6

Concentration of water = 0.35

The rate constant from the rate law can be given as:

3.822 [tex]\rm \times\;10^-^3[/tex] = k [tex]\rm [0.6]^2\;[0.35][/tex]

k = 0.121 [tex]\rm mol^-^2\;L^2\;s^-^1[/tex]

C. The order with respect to NO has been 2, while the order with respect to water has been 1.

For more information about the rate of reaction, refer to the link:

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Given :

Energy , E = 330 J .

Initial temperature , [tex]T_i=21^oC[/tex] .

Final temperature , [tex]T_f=24.6^oC[/tex] .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

[tex]n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol[/tex]

Energy required is given by :

[tex]q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}[/tex]

Hence , this is the required solution .

Answer:

14.6 %

Explanation:

Step 1: Given data

Mass of sodium chloride (solute): 6.09 g

Mass of water (solvent): 35.50 g

Step 2: Calculate the mass of solution

We will use the following expression.

m(solution) = m(solute) + m(solvent)

m(solution) = 6.09 g + 35.50 g

m(solution) = 41.59 g

Step 3: Calculate the percent by mass of the salt

We will use the following expression.

%w/w = mass of NaCl / mass of solution × 100%

%w/w = 6.09 g / 41.59 g × 100%

%w/w = 14.6 %

Answer:

After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M.

Explanation:

Based on the reaction of the problem, you have as general kinetic law for a first-order reaction:

ln[HI] = -kt + ln [HI]₀

Where [HI] is actual concentration after time t,

k is rate constant

and [HI]₀ is initial concentration of the reactant.

Initial concentration of HI is 0.310M,

K is 0.0660s⁻¹,

And the actual concentration is 0.0558M:

ln[HI] = -kt + ln [HI]₀

ln[0.0558M] = -0.0660s⁻¹*t + ln [ 0.310M]

-1.7148 = -0.0660s⁻¹*t

26.0s = t

Answer:

a. 7.528x10²² atoms Ti

b. 7.528x10²² atoms Ti

c. 1.5055x10²³ atoms Ti

d. 2.258x10²³ atoms Ti

Explanation:

Using the molecular formula of the structures, we can determine the moles of titanium in 0.125 moles of each compound. With moles of titanium we can convert these to atoms using Avogadro's number as follows:

a. FeTiO₃: 1 mole of Ti per mole of FeTiO₃.

Moles of Ti are 0.125 moles. Atoms are:

0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =

b. Titanium (IV) chloride = TiCl₄: 1 mole of Ti per mole of TiCl₄.

Moles of Ti are 0.125 moles. Atoms are:

0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =

c. Ti₂O₃: 2 moles of Ti per mole of Ti₂O₃.

Moles of Ti are 2*0.125 moles = 0.25 moles. Atoms are:

0.25 moles Ti * (6.022x10²³ atoms / 1 mole) =

a. Ti₃O₅: 3 mole of Ti per mole of Ti₃O₅.

Moles of Ti are 3*0.125 moles = 0.375 moles of Ti. Atoms are:

0.375 moles Ti * (6.022x10²³ atoms / 1 mole) =

From the formula for calculating number of atoms, the number of atoms of Ti in 0.125 moles of the compounds FeTiO₃, TiCl₄, Ti₂O₃, Ti₃O₅ are 7.528 * 10²² atoms Ti, 7.528 x 10²² atoms Ti, 1.5055 * 10²³ atoms Ti and 2.258 * 10²³ atoms Ti respectively.

The number of atoms present in a substance is calculated using the formula:

Moles of Titanium in the compounds:

a. FeTiO₃ contains 1 mole of Ti per mole of FeTiO₃.

Moles of Ti are 0.125 moles.

number of atoms = 0.125 * 6.022x10²³ atoms

number of atoms = 7.528 * 10²² atoms Ti

b. 1 mole of Titanium (IV) chloride TiCl₄ contains 1 mole of Ti per mole of TiCl₄.

Moles of Ti are 0.125 moles.

Atoms of titanium = 0.125 * 6.022x10²³ atoms / 1 mole)

number of atoms = 7.528 x 10²² atoms Ti

c. 1 mole of Ti₂O₃ contains 2 moles of Ti per mole of Ti₂O₃.

Moles of Ti are 2 * 0.125 moles = 0.25 moles

number of atoms = 0.25 * 6.022x10²³ atoms

number of atoms = 1.5055 * 10²³ atoms Ti

d.1 mole of Ti₃O₅ contains 3 mole of Ti per mole of Ti₃O₅.

Moles of Ti are 3 * 0.125 moles = 0.375 moles of Ti.

number of atoms = 0.375  * 6.022x10²³ atoms

number of atoms = 2.258 * 10²³ atoms Ti

Therefore, the number of atoms of Ti in 0.125 moles of the compounds FeTiO₃, TiCl₄, Ti₂O₃, Ti₃O₅ are 7.528 * 10²² atoms Ti, 7.528 x 10²² atoms Ti, 1.5055 * 10²³ atoms Ti and 2.258 * 10²³ atoms Ti respectively.

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