What is the mass number of an ion with 106 electrons, 157 neutrons, and a +1 charge?

Answers

Answer 1

Answer:

264 g/mol

Explanation:

#electrons equal #protons = 106

Plus 1 charge => m protons = 106 + 1 = 107

Mass number: 107 + 157 = 264 g/mol


Related Questions

Calculate the individual percent recoveries of benzoic acid, naphthalene and 3-nitroaniline if you were able to collect 9.75 g of benzoic acid, 6.41 g of naphthalene, and 7.71 g of 3-nitroaniline from a set of extractions. The starting mass of the mixture was 26.24 g. (0.6 pt)

Answers

Answer:

Benzoic acid= 37.16%

Naphthalene = 24.43%

3-Nitroaniline= 29.38%

Explanation:

Data given:

percentage recovery of benzonic acid = 9.75/26.24 * 100 = 37.16%

Percentage recovery of napthalene = 6.41/26.24 * 100 = 24.43%

Percentage recovery of 3-nitroaniline = 7.71/26.24 * 100 = 29.38%

It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single bond could be broken b

Answers

The question is incomplete, the complete question is;

It takes 338. kJ/mol to break an carbon-chlorine single bond. Cal broken by absorbing a single photon Iculate the maximum wavelength of light for which an carbon-chiorine single bond could be Round your answer to 3 significant digits

Answer:

3.55 × 10^-7 m or 355 nm

Explanation:

Now, the energy of the photon = 338 × 10^3/6.02 × 10^23 = 5.61 × 10^-19 J

Recall that;

E= hc/λ

h= planks constant

c= speed of light

λ = wavelength

λ =hc/E

λ = 6.63 ×10^-34 × 3 × 10^8/5.61 × 10^-19

λ =3.55 × 10^-7 m or 355 nm

Pl hep help help me

Answers

It is true

I’ve done this before

PLEASE HELP!!!!

Which of the following lists describes characteristics of an acid? (3 points)
Bitter taste, high pH, and caustic
Sour taste, low pH, and dissolves metals
Sour taste, high pH, and dissolves metals
Slippery, low pH, and caustic

Answers

Acids have sour taste, low pH and dissolves metals.

The list that describes the characteristics of an acid is that it has sour taste, low pH, and dissolves metals.

Characteristics of an acid

An acid is a chemical substance that has the ability to donate hydrogen ions when involved in a chemical reaction.

The following are the characteristics or features of an acid:

They have a sour taste when tasted.

When measured using a pH scale it is less than 7(low pH).

They react with active metals to yield hydrogen gas.

Therefore, the list that describes the characteristics of an acid is that it has sour taste, low pH, and dissolves metals.

Learn more about acids here:

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Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is ______ oC and the boiling point of toluene is ______ oC. Therefore, the liquid collected first should be ______
Please put an answer in each box.

Answers

Answer: The boiling point of cyclohexane is 81oC and the boiling point of toluene is 111oC. Therefore the liquid collected first should be

CYCLOHEXANE

Explanation:

In chemistry, there are various separation techniques that can be used to separate the components of a mixture and even isolate each of these components. A typical example of such separation techniques is DISTILLATION. This is a method of separation that makes use of different boiling points of liquids in a mixture. A mixture of any number of liquids could be separated as long as they boil at different temperatures. Example include:

--> mixture of cyclohexane (boiling point is 81°C) and toluene (boiling point is 111°C)

--> mixture of alcohol (boiling point is 78°C) and water (boiling point 100°C).

The process involves heating the mixture of liquids until the more volatile liquid ( that is the one with the lower boiling point) changes to vapour. The vapour is cooled by passing it through a condenser and collected in a liquid form known as distillate.

Therefore in the mixture of cyclohexane and toluene, the liquid collected FIRST should be CYCLOHEXANE.

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Write the symbol for every chemical element that has atomic number less than 14 and atomic mass greater than 23.2 u.

Answers

Answer:

that symbol less than atomic number 14 and greater than mass number 23.2 is mg

The size of an atomic orbital is associated with:______________

a. the magnetic quantum number (ml).
b. the spin quantum number (ms).
c. the angular momentum quantum number (l).
d. the angular momentum and magnetic quantum numbers, together.
e. the principal quantum number (n).

Answers

Answer:

e. the principal quantum number (n).

Explanation:

The size of the orbital is governed and decided by the principal quantum number n, which is dependent on the overall average distance between the number of electrons as well as the nucleus. The orbital's shape is explained by the angular quantum number. The magnetic quantum number is concerned with the orbital's orientation in space. The quantum number's spin explains the spin of the electrons.

Calculate the percent dissociation of benzoic acid C6H5CO2H in a 1.3M aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits.

Answers

Answer:

the percent dissociation is  0.69 %

Explanation:

Given the data in the question;

benzoic acid C₆H₅CO₂H

C₆H₅COOH[tex]_{(aq)[/tex]  ⇔ C₆H₅COO[tex]_{-(aq)[/tex] + H[tex]_{+(aq)[/tex]

Ka =  [C₆H₅COO- ][ H+ ] / [ C₆H₅COOH ] = 6.28 × 10⁻⁵

given that it dissociated in a 1.3 M  aqueous solution.

so Initial concentration is;  

[ C₆H₅COOH ] = 1.3

[C₆H₅COO- ] = 0

[ H+ ] = 0

Change in concentration

[ C₆H₅COOH ] = -x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Concentration equilibrium

[ C₆H₅COOH ] = 1.3 - x

[C₆H₅COO- ] = +x

[ H+ ] = +x

Hence,

x² / ( 1.3 - x ) = 6.28 × 10⁻⁵

6.28 × 10⁻⁵( 1.3 - x ) = x²

8.164 × 10⁻⁵ - 6.28 × 10⁻⁵x = x²

x² + 6.28 × 10⁻⁵x - 8.164 × 10⁻⁵ = 0

solve for x

ax² + bx - c = 0

x = [ -b ± √( b² - 4ac ) ] / [ 2a ]

we substitute  

x = [ -6.28 × 10⁻⁵ ± √( (6.28 × 10⁻⁵)² - (4 × 1 × -8.164 × 10⁻⁵ ) ) ] / [ 2 × 1 ]  

x = [ -6.28 × 10⁻⁵ ± 0.01807 ] / [ 2]

x = [ -6.28 × 10⁻⁵ - 0.01807 ] / [ 2]  or [ -6.28 × 10⁻⁵ + 0.01807 ] / [ 2]

x = -0.0090664 or 0.0090036

so x = 0.0090036

hence

[ H+ ] = +x = 0.0090036 M

[C₆H₅COO- ] = +x = 0.0090036 M

Initial concentration of [ C₆H₅COOH ] = 1.3 M

concentration of C₆H₅COOH dissociated = 0.0090036 M

percent dissociation of C₆H₅COOH will be;

⇒ ( 0.0090036 M / 1.3 M ) × 100 = 0.69 %

Therefore, the percent dissociation is  0.69 %

Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.

a. HCl
b. NaCl
c. N2
d. H2O

Answers

It would be N2!!!!!!!!!!!!!!!!!!

The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprotic acid with 3 pKa values: 2.14, 6.86, and 12.4.

Answers

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

Write the balanced reaction for the methanol cannon demo that includes their Lewis structures . The reaction is the combustion of methanol (CH3OH). Include the states (s, l, g) in your balanced equation as well.

Answers

Answer:

The reaction is the combustion of methanol (CH3OH).

Write the balanced chemical equation.

Draw Lewis structures for each structure.

Explanation:

The balanced chemical equation for the combustion of methane is shown below:

[tex]2CH_3OH(g)+3O_2(g)->2CO_2(g)+ 4 H_2O(g)[/tex]

Lewis structures of the given molecules are shown below:

If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?

Answers

Answer:

0.74 M

Explanation:

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

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

Can someone please please help

Answers

Answer:

oxidizer

Explanation:

an example of an oxidizers are oxygen and hydrogen peroxide

The metal thallium becomes superconducting at temperatures below 2.39K. Calculate the temperature at which thallium becomes superconducting in degrees Celsius. Round your answer to decimal places.

Answers

Answer:

-270.76°C

Explanation:

Given that metal Thallium becomes superconducting below the temperature of 2.39 kelvin i.e. this temperature is critical temperature for Thallium and below critical temperature a metal offers no resistance to the flow of electric current. Also the metal below its critical temperature expels the magnetic field in such a way that they do not penetrate the metal and pass through its surface only.

We have the relation between kelvin scale and degree Celsius scale of temperature measurement as:

[tex]C = K - 273.15[/tex]

[tex]C=2.39-273.15\\ C=-270.76^{o}C[/tex]

Suppose a 0.042M aqueous solution of phosphoric acid (H3PO4) is prepared. Calculate the equilibrium molarity of HPO4^−2.

Answers

Answer:

2.89x10⁻⁵M = [HPO₄²⁻]

Explanation:

The equilibrium of H3PO4 in water occurs H2PO4-:

H3PO4(aq) + H2O(l) ⇄ H3O⁺(aq) + H2PO4⁻(aq)

pKa = 2.16. And as pKa = -log Ka; Ka = 10^-2.16

Ka = 6.9183x10⁻³ = [H3O⁺] [H2PO4⁻] / [H3PO4]

As both [H3O⁺] and [H2PO4⁻] comes from the same equilibrium,

[H3O⁺]=[H2PO4⁻] :

[H3O⁺] = X

[H2PO4⁻] = X

[H3PO4] = 0.042 - X

Where X is reaction coordinate

Replacing:

6.9183x10⁻³ = [X] [X] / [0.042 - X]

6.9183x10⁻³ = X² / 0.042 - X

2.905686x10⁻⁴ - 6.9183x10⁻³X - X² = 0

Solving for X:

X = -0.02M. False solution. There is no negative concentration.

X = 0.014M. Right solution

[H2PO4⁻] = 0.014M

In the second equilibrium:

H2PO4⁻(aq) + H2O(l) ⇄ HPO4-(aq) + H3O+(aq)

Based on the same principles of the last equilibrium:

pKa2 = 7.21

Ka2 = 6.166x10⁻⁸ = [HPO4-] [H3O+] / [H2PO4⁻]

[HPO4-] = X

[H3O+] = X

[H2PO4⁻] = 0.014M - X

6.166x10⁻⁸ = X² / [0.014M - X]

8.3623x10⁻¹⁰ - 6.166x10⁻⁸X - X² = 0

Solving for X:

X = -0.0000289485. False solution.

X =

2.89x10⁻⁵M = [HPO₄²⁻]

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to lowest. CoCl3, NH4Cl, Li2SO4

Answers

Answer:

NH4Cl > Li2SO4 > CoCl3

Explanation:

Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.

Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.

Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Answers

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

What is determined by calculating the slope of the position versus time graph distance

Answers

Answer:

Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. ... If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position versus time graph tells it all.

Explanation:

#carryonlearning

Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s.

Which statement is true about molarity and percent by mass? (3 points)
They have the same unit.
They are inversely related.
They are different units of dilution.
They are different units of concentration.

Answers

Answer:

The guy above used photoshop here is the actual answer.

Explanation:

The true statements about molarity and percent by mass is that they are the different units of concentration

What is concentration?

Concentration of any substance present in any solution guves idea about their relative amount in that and it can de described in terms of molarity and percent by mass.

Molarity is define as the moles of solute present in per unit volume of the solution and has a unit of mol/L.Percent by mass is define as the mass of solute present in total mass of solution and it is a unit less quantity.They both gives idea about the concentration, not of the dillutions.They are not inversely related else they are directly related to each other as they are depends on the moles or mass of solute.

Hence, molarity and percent by mass are the different units of concentration and directly related to each other.

To know more about molarity & percent mass, visit the below link:

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The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the temperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.

Answers

Answer:

A. P₂ / P₁ = 2

B. P₂ / P₁ = 1.1

Explanation:

A. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 46 K

Final temperature (T₂) = 92 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/46 = P₂/92

Cross multiply

46 × P₂ = P₁ × 92

Divide both side by P₁

46 × P₂ / P₁ = 92

Divide both side by 46

P₂ / P₁ = 92 / 46

P₂ / P₁ = 2

B. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K

Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/308.4 = P₂/342

Cross multiply

308.4 × P₂ = P₁ × 342

Divide both side by P₁

308.4 × P₂ / P₁ = 342

Divide both side by 308.4

P₂ / P₁ = 342 / 308.4

P₂ / P₁ = 1.1

Describe the electron configuration of an atom using principal energy level, sublevels, orbitals, and periodic table. Give one example others may not think about and why you made this selection.

Silicon is not allowed.

Answers

Explanation:

The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.

The sample atom is Carbon with electron configuration; 1s² 2s² 2p².

The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2

The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;

s and p energy sublevels.

The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.

Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.

Read more:

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Suppose that you move from a Suppose that you move from a town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust? town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust?

Answers

Answer:

all I can say is town near the ocean atmospheric changes will be cooler, warm, sea breeze, and fresh healthy air. Then when it comes to the mountain lot of change firstly there's a dry air

PLEASE HELP FAST!!

Which of the following ions is formed when an acid is dissolved in a solution?
H+
O−
OH−
SO42+

Answers

Answer:

H+

Explanation:

acid produce H+ when it is dissolved in a solution

8.7 Two products are formed in the following reaction in a 50:50 mixture. Would the resulting solution be optically active

Answers

Answer:

Yes. The solution would be optically active.

Explanation:

Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.

In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.

So the answer is Yes.

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 13 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time (in K/min) at that instant if n = 10 mol.

Answers

Answer:

The rate of change of T with respect to time is 0.40 K/min

Explanation:

The gas law equation is:

[tex] PV = nRT [/tex]

We can find the rate of change of T with respect to time by solving the above equation for T and derivating with respect to time:

[tex] \frac{dT}{dt} = \frac{d}{dt}(\frac{PV}{nR}) [/tex]

[tex] \frac{dT}{dt} = \frac{1}{nR}(V\frac{dP}{dt} + P\frac{dV}{dt}) [/tex]

Where:

n: is the number of moles = 10 mol

R: is the gas constant = 0.0821

V: is the volume = 13 L

P: is the pressure = 8.0 atm

dP/dt: is the variation of the pressure with respect to time = 0.13 atm/min

dV/dt: is the variation of the volume with respect to time = -0.17 L/min

Hence, the rate of change of T is:

[tex] \frac{dT}{dt} = \frac{1}{10*0.0821}(13*0.13 - 8.0*0.17) = 0.40 K/min [/tex]    

Therefore, the rate of change of T with respect to time is 0.40 K/min

I hope it helps you!    

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Which statement is true about the experiments? (5 points)
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
The same amount of heat is absorbed in both the experiments because the heat absorbed depends only on the final temperature.
The heat absorbed in Trial 2 is about 3,674 J greater than the heat absorbed in Trial 1.
The heat absorbed in Trial 2 is about 5,021 J greater than the heat absorbed in Trial 1.

Answers

Answer:

Explanation:

Using the formula below to calculate the heat absorbed in each trial:

Q = m × c × ∆T

Where;

Q = amount of heat absorbed (J)

m = mass of substance (g)

c = specific heat of water (4.184J/g°C)

∆T = change in temperature (°C)

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Q = 30 × 4.184 × (40 - 0)

Q = 30 × 4.184 × 40

Q = 5,020.8J

Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Answer:

The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.

Explanation:

Explanation:

Using the formula below to calculate the heat absorbed in each trial:

Q = m × c × ∆T

Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.

Q = 30 × 4.184 × (40 - 0)

Q = 30 × 4.184 × 40

Q = 5,020.8J

Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.

Q=40*4.184*30

Q=5020.8J

A compound has a formula mass of 228.0 and an empirical formula of C2H4O3. What is the molecular formula

Answers

Answer:

C₆H₁₈O₉

Explanation:

First we calculate the molar mass of the compound represented by the empirical formula:

Molar Mass = (Molar mass of C) * 2 + (Molar Mass of H) * 4 + (Molar Mass of O) * 3Molar Mass = 12 * 2 + 1 * 4 + 16 * 3 = 76 g/mol

Then we divide the given formula mass by the calculated molar mass:

228 / 76 = 3

Thus we multiply by 3 the subscripts in the empirical formula:

The molecular formula is C₆H₁₈O₉

1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear

A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian

Answers

Answer:

1.Precipitation

2.Transpiration

3.Condensation

4.Evaporation

5.Evaporation

3.Condensation

Explanation:

Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.

Given 200ul of a 0.5mg/ml stock solution of BSA, how much do you pipet into a test tube so that you are adding 5ug of BSA to the test tube

Answers

Answer: [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.

Explanation:

We are given:

Mass of BSA to be formed = [tex]5\mu g=0.005mg[/tex]      (Conversion factor: [tex]1mg=1000\mu g[/tex]

Volume of stock solution = [tex]200\mu L=0.2mL[/tex]    (Conversion factor: [tex]1mL=1000\mu L[/tex]

It is also given that for the mass of BSA is 0.5 g, the volume used up is 1 mL

In order to have, 0.005 g, the volume of stock solution needed will be = [tex]\frac{1mL}{0.5g}\times 0.005g=0.01mL=10\mu L[/tex]

Hence, [tex]10\mu L[/tex] of volume needs to be pipetted out in the test tube.

What is the phase change from solid to gas?
O A. Condensation
O B. Sublimation
O C. Freezing
O D. Vaporization

Answers

Answer:

The answer is B, sublimation.

Answer:

The correct answer

B . Sublimation

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