Is earth an inertial frame of reference? justify your answer

Answers

Answer 1

Answer:

Earth is not an inertial reference frame because the Earth rotates and is accelerated with respect to the Sun.


Related Questions

A container holds 40.0 mL of nitrogen at 30° C and at a constant pressure.
Find its volume if the temperature increases to 80° C?

Answers

Answer:

The correct answer is - 46.60 mL.

Explanation:

To find the volume of the gas at its new increased temperature we need to use Charl Law that shows the direct relationship between Volume and Temperature while Pressure remains constant.

V1 = 40 ml

T1  = 30 degree C + 273 = 303 K

V2  = ?

T2  = 80 degree C + 273 = 353 K

Charl Equation is:

V 1/T 1  =  V 2/ T 2

(V1) * (T2)/ T1= V2

placing value:

40*353/303 = V2

= 14120/303

Vf = 46.60 mL

If the plant population decreased, the amount of carbon in the atmosphere would _______.



Increase


Stay the same


Decrease

Answers

Answer:

increase answer

Explanation:

i hope that is right

What is the function of a
catalyst?
A. Build enzymes
B. Speed up chemical reactions
C. Regulate the function of an enzyme

Answers

I believe it is to speed up chemical reactions

i also think it is speed up chemical reactions

what is an example of a change in genetic traits of an organism do to human affect

Answers

Answer:

A person's skin color, hair color, dimples, freckles, and blood type are all examples of genetic variations that can occur in a human population.

Explanation:

How many grams of carbon are In 25 g of CO2?

Answers

Answer: almost 7g

Explanation: 25 grams of carbon dioxide contains 6.818 grams of carbon.

Calculate the energy of an electron in the n = 2 level of a hydrogen atom.

Answers

Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

Explanation:

Given: n = 2

The relation between energy and [tex]n^{th}[/tex] orbit of an atom is as follows.

[tex]E = - \frac{13.6}{n^{2}} eV[/tex]

Substitute the values into above formula as follows.

[tex]E = - \frac{13.6}{n^{2}} eV\\= - \frac{13.6}{(2)^{2}}\\= - 3.40 eV[/tex]

The negative sign indicates that energy is being released.

Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.

At what temperature, in Celsius, will 88.0 g of Ne exert a pressure of 350. kPa in a
48.5 L container? NOTE: You must show your calculation on the attached scratch
paper, including which of the Gas Law formulas you used. *

Answers

Answer:

191 °C

Explanation:

We'll begin by calculating the number of mole in 88.0 g of Ne. This can be obtained as follow;

Mass of Ne = 88.0 g

Molar mass of Ne = 20 g/mol

Mole of Ne =?

Mole = mass /molar mass

Mole of Ne = 88 / 20

Mole of Ne = 4.4 moles

Next, we shall determine the temperature. This can be obtained as follow:

Mole of Ne = 4.4 moles

Pressure (P) = 350 KPa

Volume (V) = 48.5 L

Gas constant (R) = 8.314 KPa.L/Kmol

Temperature (T) =?

PV = nRT

350 × 48.5 = 4.4 × 8.314 × T

16975 = 36.5816 × T

Divide both side by 36.5816

T = 16975 / 36.5816

T = 464 K

Finally, we shall convert 464 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 464 K

T(°C) = 464 – 273

T(°C) = 191 °C

Thus, the temperature is 191 °C

A 0.750 M sample of phosphorus pentachloride decomposes to give phosphorus trichloride and chlorine gases.If the equilibrium concentration of PCl5 is 0.650M,what is the equilibrium constant for the reaction?
PCl5(g) --------> PCl3(g)+ Cl₂(g)
A) Keq = 0.0133
B) Keq = 0.0154
C) Keq = 0.133
D) Keq = 0.154
E) Keq = 65.0

Answers

Answer:

Explanation:

Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M

PCl₅(g) --------> PCl₃(g)+ Cl₂(g)

1 mole               1 mole      1 mole

Amount of phosphorus pentachloride reacted = 0.75 M - 0.65 M = 0.10 M

PCl₃(g) and  Cl₂(g) formed are .10 M and .10 M

Keq = [ PCl₃] x [ Cl₂ ] / [ PCl₅ ]

= .10 M x .10 M / .65 M

= .0154

B ) is the correct choice .

n-hexane boils at 68.7 oC at 1.013 bar and the heat of Vaporization is 28850
J/mol. Calculate Its Entropy ?

Answers

He one 28850




C. C c c.

Which of the following statement is true?

a. The resonance effect of the hydroxyl group stabiizes the anionic intermediate.
b. The resonance effect of the hydroxyl group stabilizes the cationic
c. The inductive effect of the hydroxyl group stabiizes the cationic intermediate
d. The inductive effect of the hydroxyl group stabiizes the anionic intermediate

Answers

Answer:

a. The resonance effect of the hydroxyl group stabilizes the anionic intermediate

Explanation:

The resonance effect stabilizes the the charge through the delocalization of the pi bonds. The resonance stabilization mainly occurs in the conjugated pi systems.

For example, phenol forms a strong hydrogen bonds than the nonaromatic alcohols as the [tex]$O_2-H_2$[/tex] dipole present in the hydroxyl group is being stabilized by the presence of the aromatic ring of phenol.

Thus the resonance effect of the hydroxyl group stabilizes the anionic intermediate.

A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pressure of the gas after it is heated to 278 K?

Answers

Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

Explanation:

Given: [tex]T_{1}[/tex] = 178 K,      [tex]P_{1}[/tex] = 0.3 atm

[tex]T_{2}[/tex] = 278 K,           [tex]P_{2}[/tex] = ?

According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.

Formula used to calculate the pressure is as follows.

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\[/tex]

Substitute the values into above formula is as follows.

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm[/tex]

Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

A 16.0 L sample contains 0.80 moles of a gas. If the gas is at the
same pressure and temperature, what is the number of moles when
the volume increases to 24.0 L? NOTE: You must show your
calculation on the attached scratch paper, including which of the
Gas Law formulas you used. *
A. 0.83 mol
B. 1.2 mol
C. 0.53 mol
D. 1.9 mol
(Please show your work)

Answers

[tex]\huge\bf\red{\underline{\underline{Answer:-}}}[/tex]

Here, 16 L = 0.80 Mole

Or, 1 L = ( 0.80 / 16 ) Mole

Or, 1 L = 0.05 Mole

_____________________________

Now, 24 L = 0.05 × 24

= [tex] \frac{5}{100} \times 24[/tex]

= 120/100

= 12/10

= 1.2 Mole ( Ans )
B 1.2 mole should be your answer
hope this helps

Enthalpy change is the
a. amount of energy absorbed or lost by a system as heat during a process at constant pressure.
b. entropy change of a system at constant pressure.
c. pressure change of a system at constant temperature.
d. temperature change of a system at constant pressure.

Answers

Answer:

I believe is A

Explanation:

Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure.

how to solve x² in differential​

Answers

Answer:

x² = mutiphy by them self

Explanation:

What is the pH of a 6.2 x 10-5 M NaOH solution?

Answers

Answer:

4.21

Explanation:

The pH of a solution can be found by using the formula

[tex]pH = - log [ {H}^{+} ][/tex]

From the question we have

[tex]ph = - log(6.2 \times {10}^{ - 5} ) \\ = 4.207608[/tex]

We have the final answer as

4.21

Hope this helps you

Estimate the volume of a solution of 5M NaOH that must be added to adjust the pH from 4 to 9 in 100 mL of a 100 mM solution of a phosphoric acid?

Answers

Answer:

3mL of 5M NaOH must be added to adjust the pH to 7.20

Explanation:

When NaOH is added to phosphoric acid, H₃PO₄, the reaction that occurs are:

NaOH + H₃PO₄ ⇄ NaH₂PO₄ + H₂O pKa1 = 2.15

NaOH + NaH₂PO₄ ⇄ Na₂HPO₄ + H₂O pKa2 = 7.20

NaOH + Na₂HPO₄ ⇄ Na₃PO₄ + H₂O pKa3 = 12.38

We can adjust the pH at 7.20 = pKa2 if NaH₂PO₄ = Na₂HPO₄. To make that, we must convert, as first, all H₃PO₄ to NaH₂PO₄ and the half of NaH₂PO₄ to Na₂HPO₄. To solve this question we need to find the moles of phophoric acid in the initial solution. 1.5 times these moles are the moles of NaOH that must be added to fix the pH to 7.20:

Moles H₃PO₄:

100mL = 0.100L * (0.100mol / L) = 0.0100 moles H₃PO₄

Moles NaOH:

0.0100 moles H₃PO₄ * 1.5 = 0.0150 moles NaOH

Volume NaOH:

0.0150 moles NaOH * (1L / 5moles) = 3x10⁻³L 5M NaOH are required =

3mL of 5M NaOH must be added to adjust the pH to 7.20

3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.

What is pH?

It is the negative log of the concentration of Hydrogen ions in the solution.

To calculate the volume of NaOH first, calculate the moles of NaOH and H₃PO₄.

Moles of H₃PO₄.

[tex]\rm moles \ of \ H_3PO_4 = 100\rm \ mL = 0.100\rm \ L \times (0.100 \ mol / L)\\\\\rm moles \ of \ H_3PO_4 = 0.01[/tex]

The moles of NaOH:

[tex]\rm Moles \ of \ NaOH =0.01 \ moles \ H_3PO_4\times 1.5 \\\\\rm Moles \ of \ NaOH= 0.0150[/tex]

The volume of NaOH:

[tex]\rm Volume\ of \ NaOH = \rm 0.0150\ moles\ NaOH \times (1 \ L / 5 \ moles) \\\\\rm Volume\ of \ NaOH = 3\times 10^{-3} L[/tex]

Therefore, 3 mL of 5 Molar NaOH is required to adjust the pH of phosphoric acid.

Learn more about pH:

https://brainly.com/question/3262317

What is the Ksp expression for the dissociation of calcium oxalate?Immersive Reader
(4 Points)

Ksp=[Ca⁺²] x [C₂O₄⁻²]

Ksp=[Ca⁺²]² x [C₂O₄⁻²]

Ksp=[Ca⁺²]⁴ x [C₂O₄⁻²]

Ksp=[Ca⁺²] x [C₂O₄⁻²]²

Answers

Answer:

Ksp = [Ca⁺²] × [C₂O₄⁻²]

Explanation:

Step 1: Write the balanced reaction for the dissociation of calcium oxalate

CaC₂O₄(s) ⇄ Ca⁺²(aq) + C₂O₄⁻²(aq)

Step 2: Write the expression for the solubility product constant (Ksp) of calcium oxalate

The solubility product constant is the equilibrium constant for the dissociation reaction, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It doesn't include solids nor pure liquids because their activities are 1.

Ksp = [Ca⁺²] × [C₂O₄⁻²]

Which of the following
describes the zone of the
ocean where no light reaches?
A. up to 200 meter depth and includes
photosynthetic plants, sea anemones,
sponges, crabs, and clams
B. the "twilight zone" between 200-1000
meters deep and includes whales and octopi
and little life
C. permanent darkness below 1000 meters
with bioluminescent bacteria, bottom
feeders, and angler fish

Answers

Answer:

Bathypelagic

54% of the ocean lies in the Bathypelagic (aphotic) zone into which no light penetrates. This is also called the midnight zone and the deep ocean. Due to the complete lack of sunlight, photosynthesis cannot occur and the only light source is bioluminescence.

Explanation:

The small surface zone that has light is the photic zone. The entire rest of the ocean does not have light and is the aphotic zone.

Permanent darkness below 1000 meters with bioluminescent bacteria, bottom feeders, and angler fish is where no light reaches.

What is Darkness?

This is referred to the state of being dark as a result of absence of light in the area.

The light ray penetration decreases with increase in depth thereby making areas below 1000 meters dark with bioluminescent bacteria, bottom feeders, and angler fish which is why option C was chosen.

Read more about Darkness here https://brainly.com/question/24581271

Thermal energy naturally flows from _________ matter to _______ matter.

Answers

Answer:

Warmer

Cooler

Explanation:

Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen,
and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50​

Answers

Answer:

C₃H₈O

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1.031 g

Mass of CO₂ = 2.265 g

Mass of H₂O = 1.236 g

Empirical formula =?

Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:

For carbon, C:

Mass of CO₂ = 2.265 g

Molar mass of CO₂ = 44.01 g/mol

Molar mass of C = 12.01 g/mol

Mass of C =?

Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂

Mass of C = 12.01/44.01 × 2.265

Mass of C = 0.618 g

For hydrogen, H:

Mass of H₂O = 1.236 g

Molar mass of H₂O = 18.02 g/mol

Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

Mass of H =?

Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O

Mass of H = 2.02/18.02 × 1.236

Mass of H = 0.139 g

For oxygen, O:

Mass of compound = 1.031 g

Mass of C = 0.618 g

Mass of H = 0.139 g

Mass of O =?

Mass of O = Mass of compound – (mass of C + mass of H)

Mass of O = 1.031 – ( 0.618 + 0.139)

Mass of O = 1.031 – 0.757

Mass of O = 0.274 g

Finally, we shall determine the empirical formula. This can be obtained as follow:

C = 0.618 g

H = 0.139 g

O = 0.274 g

Divide by their molar mass

C = 0.618 / 12.01 = 0.051

H = 0.139 / 1.01 = 0.138

O = 0.274 / 16 = 0.017

Divide by the smallest

C = 0.051 / 0.017 = 3

H = 0.138 / 0.017 = 8

O = 0.017 / 0.017 = 1

Therefore, the empirical formula of the compound is C₃H₈O

Based on the data provided, the empirical formula of the compound is C₃H₈O

What is empirical formula?

The empirical formula of compound is its simplest formula showing the mole ratio of the elements in the compound.

First, the mass of carbon, hydrogen and oxygen in the compound is determined first as follows:

For carbon, C:

Mass of CO₂ = 2.265 g

Molar mass of CO₂ = 44.01 g/mol

Molar mass of C = 12.01 g/mol

Mass of C = 12.01/44.01 × 2.265

Mass of C = 0.618 g

For hydrogen, H:

Mass of H₂O = 1.236 g

Molar mass of H₂O = 18.02 g/mol

Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

Mass of H = 2.02/18.02 × 1.236

Mass of H = 0.139 g

For oxygen, O:

Mass of compound = 1.031 g

Mass of C = 0.618 g

Mass of H = 0.139 g

Mass of O = Mass of compound – (mass of C + mass of H)

Mass of O = 1.031 – ( 0.618 + 0.139)

Mass of O = 0.274 g

To determine the empirical formula of a compound, the mole ratio of the elements are determined as follows:

Mole ratio = reacting mass/molar mass

Mole ratio of the elements:

Carbon                Hydrogen         Oxygen

0.618 g/12.01    0.139 / 1.01         0.274 / 16

0.051                   0.138              0.017

Divide by the smallest ratio to convert to whole numbers

0.051 / 0.017     0.138 / 0.017  0.017 / 0.017

3           :            8           :               1

Therefore, the empirical formula of the compound is C₃H₈O

Learn more about empirical formula at: https://brainly.com/question/24297883

0=4
Balance this equation
H₂sicl2+ H₂O → H8Si4O4 + HCl

Answers

Balanced Equation is

4H2SiCl2+4H2O → H8Si4O4 + 8HCl

1. Calculate and interpret the equilibrium constant. Using the reaction below.

The equilibrium concentrations 0.60 M for E, 0.80 M for F, and 1.30 M for G. (Note: E, F, and G are all gases.) Do not include your solution.

Answers

Answer:

kc = [G]² / [E] [F], kc = [1.30M]² / [0.60M] [0.80M]

Explanation:

The reaction is:

E + F ⇄ 2G

The equilibrium constant, kc, must be written as the ratio of the molar concentrations of products over reactants. Each concentration powered to its coefficient.

For the reaction of the problem, kc is:

kc = [G]² / [E] [F]

Replacing the given concentrations:

kc = [1.30M]² / [0.60M] [0.80M]

when 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?

Answers

Answer:

si.mple fe

Explanation:

ghhsshzhzbbzhh

3.05 moles of H2O is equivalent to (blank amount) molecules of water.

Show your work.

Answers

Answer:

[tex]1 \: mole \: = 6.02 \times {10}^{23} \: molecules \\ 3.05 \: moles \: = (3.05 \times 6.02 \times {10}^{23} ) \: molecules \\ = 1.8361 \times {10}^{24} \: molecules[/tex]

chemical properties of citric acid​

Answers

Answer:

Citric Acid is a weak acid with a chemical formula C6H8O7.

...

Properties of Citric Acid – C6H8O7.

C6H8O7 Citric Acid

Molecular Weight/ Molar Mass 192.124 g/mol

Density 1.66 g/cm³

Boiling Point 310 °C

Melting Point 153 °C

Hope it helps❤️

Chemical formula: C6H8O7
Properties: No color, no smell, Crystal like appearance, Boiling point of 175 C, Melting point of 153 C, Density of 1.66 g/ml, Weak acid

The blending together of some genes is called:

Answers

Answer:

its called molding

Explanation:

Which molecule will have double bonds?
(4 Points)

Fluorine

Nitrogen

Hydrogen

Oxygen

Answers

I believe its oxygen
hope this helps

At the start of a reaction, there are 0.0249 mol N2,
3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a
3.50 L reaction vessel at 375°C. If the equilibrium constant, K, for the reaction:
N2(g) + 3H2(g)= 2NH3(g)
is 1.2 at this temperature, decide whether the system is at equilibrium or not. If it is not, predict in which direction, the net reaction will proceed.​

Answers

Answer:

Explanation:

The reaction is given as:

[tex]N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}[/tex]

The reaction quotient is:

[tex]Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}[/tex]

From the given information:

TO find each entity in the reaction quotient, we have:

[tex][NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}[/tex]

[tex][N_2] = \dfrac{0.024 }{3.5}[/tex]

[tex][N_2] = 0.006857[/tex]

[tex][H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}[/tex]

[tex][H_2] = 9.17 \times 10^{-3}[/tex]

[tex]Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135[/tex]

However; given that:

[tex]K_c = 1.2[/tex]

By relating [tex]Q_c \ \ and \ \ K_c[/tex], we will realize that [tex]Q_c \ \ < \ \ K_c[/tex]

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

How many unpaired electrons are in the following complex ions?
a. Ru(NH3)62+ (low-spin case)
b. Ni(H2O)62+
c. V(en)33+

Answers

Solution :

A complex ion may be defined as a metal ion that is located at the center with other molecules or ions surrounding the center ion.

In the context, the number of the unpaired electrons for the following complex ions are :

1. [tex]$Ru(NH_3)62+$[/tex]  (low spin)

  [tex]$Ru^{2+} = [Kr] \ 4d^6 \ 5s^0$[/tex]

   Here all the d-electrons are paired. Therefore, there are no unpaired electrons.

2. [tex]$Ni(H_2O)62+$[/tex]  (high spin)

   [tex]$Ni^{2+} = [Ar] \ 4s^0 \ 3d^8$[/tex]

   Therefore, it has two unpaired electrons.

3. [tex]$V(en)33+$[/tex]  (high spin)

    [tex]$V=[Ar] \ 3d^2 \ 4s^0$[/tex]

   It has 2 unpaired electrons

What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?

Answers

Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: [tex]V_{1}[/tex] = 61 L,      [tex]T_{1}[/tex] = 183 K,      [tex]P_{1}[/tex] = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L[/tex]

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

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