Classify each of the following examples as a physical change or a chemical change. Answer choices may be used more than once.

Classify Each Of The Following Examples As A Physical Change Or A Chemical Change. Answer Choices May

Answers

Answer 1

1. Physical change

Sugar dissolving in waterWater freezing into ice

In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed.

2. Chemical change

Burning a candleRust forming on a car

A chemical change is a change of materials into another, new materials with different properties and one or more than one new substances are formed. It results when a substance combines with another to form a new substance.

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Related Questions

what volume, in ml, of concentrated 16 m nitric acid would you need to use in order to prepare 500.0 ml of a 0.250 m hno3(aq) solution? (formula mass of hno3

Answers

The volume of concentrated 16 m nitric acid required is 7.81ml

Concentration of HNO3= 16M

We need to find the volume of HNO3. V=?

Concentration of required solution= 0.250M

Volume of required solution= 500M

So to solve this problem, we have to use the Molarity equation

m1v1=m2v2

So now we have to substitute the given values in the molarity equation

After doing that we get:
16M x V1 = 0.250M X 500M

V1= 0.250 X 500M/ 16M

Therefore V1= 7.81ML

Hence, the volume of concentrated 16 m nitric acid required is 7.81ml in order to prepare 500.0 ml of a 0.250 m hno3(aq) solution.

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Which material, principle, or process enables a method of numerical dating?

answer: radioactivity

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Material, principle, or process enables a method of numerical dating is the radioactivity

To establish the age of a rock or a fossil, researchers use some type of clock to determine the date it was formed geologists commonly use radiometric dating methods and using the decay of various radioactive element and you can determine a numerical age for mineral in igneous rock and the result data is the radiometric age

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what are two reasons for anodising a spoon ??? I need help ASAP pleaseeee

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Anodizing a spoon makes it 1) durable and scratch-resistant and 2) prevents the leaching of plain aluminum into food.

Anodizing is an electrochemical process that produces an oxide coating on metal surfaces. This oxide coating makes the metal surface corrosion-resistant, and durable and increases the metal's adhesive qualities (for glues and paint primers). Anodizing is usually done for aluminum and its alloys

Two reasons for anodizing a spoon:

To prevent corrosion of the spoon and make it scratch-resistant which is possible because a spoon is used extensively for eating and other purposes. It provides a smooth, hard surface that is durable.Anodizing the spoon also prevents it from reacting with acidic foods. Anodized aluminum spoons are also protected from plain aluminum leaching into the food.

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A certain red light has a wavelength of 680 nm.
a. What is the frequency of the light?
6
b. What is the energy of a quantum of light?

Answers

The frequency of light is  0.44 * 1015s-1 and the energy of a quantum of light is 0.029 * 10-17Joule

What is the quantum of light?

A tiny electromagnetic radiation energy bundle is known as a photon or light quantum.

Given:

Wavelength = 680nm

Frequency is related to the wavelength as given in the above equation:

V (frequency) = c\ (speed\ of\ light)/wavelength

C = 3 × 108 ms-1

Wavelength = 680 * 10-9m

By using the formula of frequency =  3*10/ 680 * 10-9  = 0.0044 * 1017 = 0.44 * 1015s-1

For finding the energy

Energy =  h( planck constant) * c (speed of light)/ wavelength)

h = 6.62 × 10-34 Joule seconds.

C = 3 × 108 ms-1

Wavelength = 680 * 10-9m

Using the above energy formula,

we got =  6.62 × 10 -34 Joule seconds * 3 × 108 ms-1/ 680 * 10-9m =  0.029 * 10-17Joule

Hence, the answer is frequency of light = 0.44 * 1015s-1

The energy of a quantum of light = 0.029 * 10-17Joule

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Copper(II) sulfate crystals, CuSO4.5H₂O, can be made by heating copper(II) oxide with dilute
sulfuric acid and then crystallising the solution formed.
a
Calculate the maximum mass of crystals that could be made from 4.00 g of
copper(II)oxide using an excess of sulfuric acid.
CuO(s) + H₂SO4(aq) → CuSO4(aq) + H₂O(1)
CuSO4(aq) + 5H₂O(l) → CuSO4.5H₂O(s)

Answers

8.05 g of CuSO₄ is the highest mass of crystals.

a) CuO(solid) + H₂SO₄(aqueous) → CuSO₄(aqueous) + H₂O(liquid)

Molecular weight,

CuO = 79 g

Molar weight of CuSO₄ = 159 g,

79 g of CuO reacted with H₂SO₄ 159 g to form CuSO₄.

∴ 4 g of CuO

= 159/79 × 4

= 8.05 g of CuSO₄

b) Percentage yield and Theoretical yield :

The Theoretical yield:

CuSO₄ of 159.6 g → CuSO₄.5H₂O of 249.5 g

8.05 g = 249.5 × 8.05 ÷ 159.6 = 12.58 g

And the percentage yield = Weight of the product ÷ Theoretical yield × 100

= 11.25 ÷ 12.58 × 100

= 89.42%

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4.47 consider the cis and trans isomers of 1,3-di-tert-butylcyclohexane (build molecular models). what unusual feature accounts for the fact that one of these isomers apparently exists in a twist boat conformation rather than a chair conformation?

Answers

Considering the cis and trans isomers of 1,3-di-tert-butylcyclohexane, trans isomer is more likely to exist in twist boat conformation.

What is twist boat ?

Conformation is the spatial arrangement of the atoms in molecule. there are a variety of conformations are possible for cyclohexane. Twist boat conformation moves the hydrogen atoms attached to carbon atoms 1 and 4 farther apart so that the steric strain is reduced.  Free rotation can be frequently observed in linear, single covalent bonds. However, rotation can also be observed in few ring structures.

t-butyl in axial position in cyclohexane is very disfavored by energy.  When you have trans 1,3 - both chairs have 1 axial and 1 equatorial, so both chairs are disfavored due to steric hindrance. Twist boat can minimize the unfavorable interactions among the t-butyls and this makes the molecule more stable.

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the addition of 0.3800 l of 1.150 m kcl to a solution containing ag and pb2 ions is just enough to precipitate all of the ions as agcl and pbcl2. the total mass of the resulting precipitate is 61.90 g. find the masses of pbcl2 and agcl in the precipitate.

Answers

The mass of PbCl2 and AgCl is respectively 23.82 grams and 38.08 grams.

Solution -

Molarity = moles of solute / liters of solution

Moles of Cl- from KCl = molarity × volume in liter = 1.150 * 0.380 = 0.437

Let mass of AgCl = x grams

Moles of AgCl = x/143.32

So moles of Cl- from AgCl = x/143.32

Moles of PbCl2 = (61.90 - x) / 278.106

So moles of Cl- from PbCl2 = 2 × (61.90-x) / 278.106 = (61.90-x) / 139.053

Now,

(x/143.32) + (61.90-x) / 139.053 = 0.437

139.053x + 8871.508 - 143.32x = 8709.0062

- 4.267x = -162.5018

x = 38.08

 

Thus,

Mass of AgCl = x = 38.08 grams

Mass of PbCl2 = 61.9 - 38.08 = 23.82 grams

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Plant reproduction. Answer as much as possible, thanks!

Answers

Answer:

Plant Reproduction

When plants reproduce asexually, they use mitosis to produce offspring that are genetically identical to the parent plant. The advantage of asexual reproduction is that it allows successful organisms to reproduce quickly. ...

When plants reproduce sexually, they use meiosis to produce haploid cells that have half the genetic information of the parent (one of every chromosome)

Explanation:

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Answers

The ΔE for the transition of an electron from n= 7 to n=4 is  9.159 ×  10⁻²⁰ J and the frequency is 1.373 × 10¹⁴ S⁻¹.

A spectral line is a darkish or vibrant line in any other case uniform and non-stop spectrum, because of an excess or deficiency of photons in a slim frequency variety, compared with the close by frequencies.

A spectral line is a darkish or shiny line in any other case uniform and continuous spectrum, as a consequence of emission or absorption of light in a slim frequency variety, in comparison with the close by frequencies. Spectral lines are regularly used to perceive atoms and molecules.

ΔE = E final - E initial = -13.6 × Z² (1/n final² -1/n initial²) eV/atom

     = +13.6 × 1² (1/7² - 1/4²)

     = 13.6 × ( 1/49 - 1/16)

     = 13.9 × ( 16 - 49) / 49 × 16

    = - 0.57244 eV/atom

    =   0.57244 × 1.6  × 10⁻¹⁹ J

   = 0.9159 × 10⁻¹⁹ J

    = 9.159 ×  10⁻²⁰ J

Frequency(ν):-

   ΔE = hν

    ν   =  ΔE/h

         =  9.159 ×  10⁻²⁰ J / 6.626 × 10⁻³⁴

         = 1.373 × 10¹⁴ S⁻¹ ( four significant figure)

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how many molecules are in 80 grams of Bromine

Answers

Answer:

[tex]6.029 \times 10^{23} molecules[/tex]

An 8. 6 mol sample of NO2 is in a 5. 1 L container. What is the pressure of this gas in atmospheres at 33 K? R= 0. 08206 L•atm / (mol•K)

Answers

8.6 mole sample of NO₂ in a 5. 1 L container has 4.6 atmospheric pressure at 33 K.

Given, number of moles of No₂(n) = 8.6 mole

Volume (V) = 5.1 liters

Temperature (T) = 33K

R= 0. 08206 L•atm / (mol•K)

Using ideal gas equation ,

PV=nRT

P (5.1L) = (8.6mol) (0.08206L•atm/(mol•K)) (33K)

P (5.1) = 8.6 × 0.08206 × 33

P (5.1) = 23.3

P = 23.3 ÷ 5.1

P = 4.6 atm

Hence, the pressure of the given gas is 4.6 atmosphere (atm).

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A tentative procedure is given below.

1. Add 50 mL of the unknown acid to a 250 mL Erlenmeyer flask.

2. ???

3. As soon as the solution turns a light pink and does not disappear after stirring, stop adding Sodium hydroxide. Record the volume in the data table below.

What should step 2 be in your procedures to perform a titration on an unknown concentration of a monoprotic acid with a known concentration of sodium hydroxide?


A) Add two drops of phenolphthalein to the buret of acid, the mixture will remain clear. Fill your buret with mixture and begin to add to the Erlenmeyer flask of acid. Use the glass stirring rod to mix the solution as you add drops of acid to the Erlenmeyer flask.

B) Add two drops of phenolphthalein to the Erlenmeyer flask of acid until it turns a pink color. Fill your buret with NaOH solution and begin to add to the Erlenmeyer flask of acid. Use the glass stirring rod to mix the solution as you add drops of NaOH to the Erlenmeyer flask.

C) Fill your buret with NaOH solution and begin to add to the Erlenmeyer flask of acid. Use the glass stirring rod to mix the solution as you add drops of NaOH to the Erlenmeyer flask.

D) Add two drops of phenolphthalein to the Erlenmeyer flask of acid, the mixture will remain clear. Fill your buret with NaOH solution and begin to add to the beaker of acid. Use the glass stirring rod to mix the solution as you add drops of NaOH to the Erlenmeyer flask.

Answers

Step 2 option B) Add two drops of phenolphthalein to the Erlenmeyer flask of acid until it turns a pink color. Fill your buret with NaOH solution and begin to add to the Erlenmeyer flask of acid. Use the glass stirring rod to mix the solution as you add drops of NaOH to the Erlenmeyer flask.

To get the maximum accurate measurement of how lots base is needed to react with the acid, we need to make sure the solutions are constantly combined collectively nicely so that it will completely react.

Titrations are used to decide the quantity of 1 substance present by way of reacting it with a known amount of some other substance. as instance, you could discover the molar mass of acid with the aid of titrating the acid with a solution of the base of regarded concentration.

The answers will start with turning blue, but after a few moments, it will become colorless. To carry out the demo, stopper the flask and shake the solution vigorously, ensuring to maintain the stopper in place with one hand. the answer will flip blue after enough shaking.

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what is the molarity of a solution of hno2 that contains 0.20 moles of hn03 in 1.5 l of solution

Answers

Answer:

0.13M

Explanation:

The formula of molarity is moles of solute÷litres of solution hence:

M= 0.20÷1.5

M= 0.13

help pls

A 36.4-l volume of methane gas is heated from 25°c to 88°c at constant pressure. what is the final volume of the gas?

Answers

Answer:

44.1 liters

Explanation:

We can use the combined gas law:

P1V1/T1 = P2V2/T2,

where P, V, and T are the  pressures(P), volumes(V), and temperatures(T),  for the initial (P1,V1,T1) and final states (P2,V2,T2).  Note that the temperatures must be in Kelvin (add 273.15 to C to make it K).

We are given all the conditions except V2, the final volume.  Let's rearrange the gas law to solve for V2:

 V2 = V1(T2/T1)(P1/P2)

Note the way I organized the temperature and pressures into ratios of their starting and final conditions.  This makes it easier to visualize how changes will impact the final volume.  If the temperatue goes up, (T2/T1) will increase and V2 will increse.  But it we increase pressure, (P1/P2) will drop, casuing a reduction in volume.

Enter the data.

V1 =36.4L

T1 = 298.2K

T2 = 361.2K

Pressures are not given, but are said to remain the same:  "at constant pressure."  We need a pressure, so we can assume for the sake of simplicity, that the pressure is 1 atm (both P1 and P2).  We can see from the ratio of the two, that the absolute value of the pressure makes no difference (P1/P2) since P1=P2 and the ratio is simply 1.

V2 = V1(T2/T1)(P1/P2)

V2 = (36.4L)(361.2K/298.2K)(1atm/1atm)

The ratio of the temperatures clearly tells us that the final volume should increase, by around 20% (about 60K higher than 300K).  Now do the calculation and see if the volume change is indeed around 20% higher.

V2 = 44.1L

This is around a 20% increase and is higher than the initial volume, so let's claim our work is done.

Give the year and opponent for each of Muhammed Ali's heavyweight titles. He won the
championship in 1964 he was fighting floyd patterson.

Answers

Muhammad Ali is a professional boxer and American activist. He is known as "The Greatest" and is regarded as one of the greatest sports stars of all time.

Explain about Muhammed Ali's heavy weight titles?

Ali famously said, "I am the greatest!" after defeating the much favoured Sonny Liston in six rounds on February 25, 1964, to win the heavyweight title.

He became a Muslim in 1961. He won the heavyweight world championship by defeating Sonny Liston on February 25, 1964, when he was only 22 years old. He considered to his old name as a "slave name" and changed it to Muhammad Ali that year.

The first heavyweight championship battle, for the WBC/Ring/Lineal title, was on November 22, 1965, and the second, for the North American Boxing Federation (NABF) heavyweight title, happened on September 20, 1972.

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the products and reactants of a(n) reaction will form at the same rate when equilibrium is attained. decomposition exchange synthesis reversible

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The products and reactants of reversible reaction will form at the same rate when equilibrium is attained.

The chemical reaction in which the reactants react to form products and at the same time products reacts to form product is known as reversible reaction.

Equilibrium is attained in reversible reactions when the rate at which the reaction is going in the forward direction is equal to the rate of reverse reaction.

The double arrow  ⇌ are used to indicate reversible reactions.

The general form of reversible reaction is :

                                          A + B  ⇌ C + D

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What do these two changes have in common? dew appearing on grass in the morning water freezing into ice Select all that apply.

Answers

The similarities which exist between a freezing water and a morning dew on a plant surface is that:

Both conserve massBoth are changes of state

The correct answer choices are options b and d.

How is freezing of water a change of state of matter?

Matter means any thing or substance that has mass and occupies space. There are four different phases of matter including the solid, liquid, gas and plasma.

When a matter changes from one form or state to another, a change of state is said to have taken place. For instance when we place a water ( liquid ) into a freezer such that it turns to ice, this is known as what we call freezing. Here, the water is the liquid and the ice formed is the solid.

So we say freezing means the changing of liquid to solid.

So therefore, it can be deduced from above that there is a change of state of matter when water is freezed into ice and when dews appears of leaves.

Complete question:

What do these two changes have in common? dew appearing on grass in the morning water freezing into ice Select all that apply.

a. Both are chemical changes.

b. Both conserve mass.

c. Both are only physical changes.

d. Both are changes of state.

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ELEMENT #2
I am a non-metal.
I belong to the halogen family.
I am not the largest or smallest atom in my group.
My first ionization energy is greater than that of iodine.
I am not a gas at room temperature.

What element am I? Write my symbol and electron configuration using standard notation.
*dont have to give me the standard notation at all i can find it

Answers

I’m thinking Bromine

Symbol is Br

Election Configuration [Ar] 4s² 3d¹⁰ 4p⁵

Which of the following is NOT a true statement?
Two or more atoms held together with bonds make up a molecule..
O Pure Substances are made of only one type of atom.
O At least two types of atoms are required to make a compound.
Mixtures can be made of two elements, two compounds, or an element and a
compound.

Answers

Two or more atoms held together with bonds make up a molecule is not the true statement

Bonds arise from the electrostatic forces between positively charged atomic nuclei and negatively charged electrons and covalent bond is defined as  a chemical bond that includes the sharing of electron pair between atoms and the pair of electrons are know as bonding pairs and the stable balance of attractive and repulsive forces between atoms

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How will you corelate the measures used in agriculture with acids and bases?

Answers

Answer:

by the fertilizers that we use the soil or land to increase the fertility.

Explanation:

Which of the following is NOT an example of matter?
O All of these are examples of matter
O air
O gold
O apple juice
O a rose bush

Answers

Answer: All of these are examples of matter

Explanation:
Matter
is any substance that has mass and takes up space by having volume. All examples have mass (including air) and take up space, so the answer is, "All of these are examples of matter."
Hope this helps.

a reaction occurring in a calorimeter absorbs 850.0 j of energy. the initial temperature of 200.0 g of water is 24.5 °c. do you expect the temperature of the water to increase or decrease? explain?

Answers

The temperature of the system in equilibrium is 25.51⁰ C.

The equilibrium temperature of the system depends on the heat released from both gold and water. The total heat received by the system will equal to total heat released by objects. It should follow

Q released = Q received

The heat can be defined by

Q = m . c . ΔT

where Q is heat, m is mass, c is the specific heat constant and ΔT is the change in temperature.

The given parameters are

m = 200 g = 0.2 kg

T1 = 24.5⁰ C

c = 4200 J/kg⁰ C

Q received = 850 J

Lets assume that the value of equilibrium temperature is T. Hence,

Q released = Q received

m . c . ΔT = 850

0.2 . 4200 . ΔT = 850

ΔT = 1.01

T - T1 = 1.01

T - 24.5 = 1.01

T =  25.51⁰ C

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a certain hydrate has the formula mgso4 ∙ x h2o. a 54.2g sample of the compound is heated in an oven to drive off the water. if the steam generated exerts a pressure of 24.8atm in a 2.00l contained at 120.°c, calculate x.

Answers

The molecular formula of the given compound is [tex]MgSO_4.7H_2O[/tex]

The number of atoms in each element inside a single chemical molecule is expressed by the molecular formula. The definition of a molecular formula is the formula that shows the exact number of atoms in a molecule. The empirical formula is used to derive the Molecular Formula when the molar mass value is known. The molecular formula specifies the number of distinct types of atoms contained in a chemical molecule.

The ideal gas equation is:-

PV = nRT

Here,

Pressure, P = 24.8 atm

Volume, V = 2.00 L

Temperature, T = (120 + 273.15) K = 393.15 K

No. of moles, n =??

Molar gas constant, R= 0.082 Latm/molK

Putting the values in the above Equation,

n = PV / RT

n = 24.8 x 2 / 0.082 x 393.15

n = 1.54 moles

Now molecular weight of water = 18 g/mol

So, wt. of water produced = (1.54 × 18) g = 27.7 g [tex]H_2O[/tex]

Here [tex]MgSO_4. xH_2O[/tex] used = 54.2g

So, in 54.2 g, [tex]MgSO_4. xH_2O[/tex], [tex]H_2O[/tex] present = (54.2 - 27.7) g = 26.5g

Molar mass of [tex]MgSO_4[/tex] = 120.36 g/mol.

So,

Here in 26.534 g [tex]MgSO_4[/tex], water present 27.7g

So,

In 120.36g [tex]MgSO_4[/tex], water present = 27.7 / 26.5 x 120.36 = 125.8g

So, no. of moles of [tex]H_2O[/tex] = 125.8 / 18 = 6.98 = 7 moles

Therefore, with one mole of [tex]MgSO_4[/tex], water molecule present = 7 moles.

Result:

Thus, Molecular formula is [tex]MgSO_4.7H_2O[/tex]

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Does the orbiting nucleus of an atom have a positive, negative, or neutral charge

Answers

Answer:

Explanation:

If you look at an atom as a whole, it is electrically neutral and possesses no overall charge.  The nucleus consists of Protons and neutrons, Protons have a positive charge and neutrons have no charge on them.

Hence, the nucleus of an atom is positively charged and is generally surrounded by one or more electrons.

The electrons on the other hand have a negative charge on them.  

The sign convention for proton (+1), neutron(0) and electron(-1). The nucleus is very heavy but is very small compared to the overall size of an atom.            

An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. I hope this helped.

How do we reuse material like plastic container and footwear

Answers

Answer:

Explanation:

-Soda Bottle Sprinkler

-DIY Kitchen Storage Containers

-Piggy Bottle Bank

- pencil case

- shoe pot for plants

- turn shoe into sandals

calculate molar solubilities, concentrations of constituent ions, and solubilities in grams/liter for the following compounds at 25°c. i. barium phosphate (ksp

Answers

The solubility product, Ksp of barium phosphate is helpful to calculate the molar solubility and concentration of ions.

The equilibrium constant for a solid material dissolved in an aqueous solution is the solubility product constant, Ksp. It denotes the concentration at which a substance dissolves in solution. A material's Ksp value indicates how soluble it is - larger the Ksp value, the more soluble it is. Solubility equilibrium is a sort of dynamic equilibrium that occurs when a chemical substance in solid form is in chemical equilibrium with that compound's solution. The solid may dissolve unmodified, by dissociation, or through chemical reactivity with another solution element, such as acid or alkali. The solubility product is determined by the molar concentrations of the ions in saturated solutions, whereas the ionic product is determined by any solution.

Reaction:

[tex]Ba_3(PO_4)_2 \rightleftharpoons 3Ba^{2+} + 2PO_4^{3-}[/tex]

The solubility product of barium phosphate, Ksp =  1.3 x [tex]10^{-29}[/tex]

The expression of solubility product for barium phosphate is given as:

Ksp = [tex](3S)^3[/tex] x [tex](2S)^2[/tex] = 108 [tex]S^5[/tex]

1.3 x [tex]10^{-29}[/tex] x [tex](2S)^2[/tex] = 108 [tex]S^5[/tex]

[tex]S^5[/tex] = 1.3 x [tex]10^{-29}[/tex] / 108

S = 6.5479 x [tex]10^{-7}[/tex]M

Concentration of barium ions = [tex][Ba^{2+}][/tex] = 3 x 6.5479 x [tex]10^{-7}[/tex] = 1.9643 x [tex]10^{-6}[/tex]M

Concentration of phosphate ions = [tex][PO_4^{3-}][/tex] = 2 x 6.5479 x [tex]10^{-7}[/tex] = 1.3096 x [tex]10^{-6}[/tex]M

Molar solubility of [tex][Ba_3(PO_4)_2] \frac{[Ba^{2+}]}{3}[/tex] = 1/3 x 1.9643 x [tex]10^{-6}[/tex] = 6.5479 x [tex]10^{-7}[/tex]M

Solubility product of [tex]Ba_3(PO_4)_2[/tex] = 6.5479 x [tex]10^{-7}[/tex] x 601 = 0.0003935g/L

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the monomer's concentration remains the same the monomer may be completely used to create the polymer and it may partly degrade during the synthesis. the monomer may partly degrade during the synthesis the monomer may be completely used to create the polymer

Answers

Depolymerization; Macromolecule-initiated cleavage; Thermal; Photochemical; Mechanochemical; Oxidative; Polymer-burning; Kinetics of cleaving Macromolecules when Chain Depolymerization is Negligible; Degradation in Polymer Recycling; Protection of Polymers Against Degradation.

Catalytic oxidation, hydrogenation, and pyrolysis are some of the methods that can be used to depolymerize lignins all the way down to phenilics.68 Little has been done to exploit these processes in the direction of monomers; instead, the products have been thought of in the context of fuels and commodity or fine chemicals.

Nevertheless, it does not seem improbable that certain degradation pathways followed by functionalization will become feasible in order to synthesize aromatic monomers with original structures that are not easily available from petrochemistry, such as vinyl phenols, p-methoxystyrene for polyaddition reactions, and difunctional aromatic compounds for polycondensations that lead to polyesters, degradation, etc.

Several of these potential uses for lignin-derived monomers have been thoroughly covered in two recent publications.

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Calculate the mass in AMU of a sample that contains 25 potassium atoms

Answers

Answer:

1.52x10^-21 grams/atom K

Explanation:

The molar mass of potassium (K) is 39.1 g/mole.  One mole = 6.02x10^23 atoms of K.

 (25 K atoms)*(39 g/mole)(1 mole/6.023x10^23 atoms) = 1.52x10^-21 grams/atom K

flowback waste water is disposed of in a process called deep well injection which plumps large quantities of waste water down into porous sandstone and limestone rock formations underground. what potential problems could result from this?

Answers

The potential problems associated with a deep well injection is that it can result in polluting underground water.

If there are many wells nearby, injecting wastewater into subsurface rock strata can be problematic. Consider porous sandstone, which contains minute openings. Water under high pressure, such as wastewater from fracking, can penetrate the sandstone and travel with underground water.

An injection well is employed to inject fluid underground into porous geologic formations. These subterranean structures might be anything from a modest soil layer to thick sandstone or limestone. Water, wastewater, brine (salt water), and water that has been combined with chemicals are all examples of injected fluids.

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The planet Venus is surrounded by a thick layer of gases. In fact, the atmospheric pressure on Venus is over 90 times greater than the atmospheric pressure on Earth. Which statements are true about boiling water on Venus? Choose the two statements that apply.

Answers

The two statements that apply about boiling water on Venus are:

The tempurature will remain constant while it boilsWater will boil at a higher tempurature on Venus than on Earth

What impact does pressure have on boiling point?

Because the boiling point is the point at which the vapour pressure equals or exceeds the atmospheric pressure, the amount of energy needed to boil a liquid increases as pressure increases.

Explanation:

According to the definition given earlier, anything reaches its boiling point when its vapour pressure reaches or exceeds that of the atmosphere. More and more particles have the energy to transition into the gas phase as temperature rises. As a result, the liquid boils as the vapour pressure rises to equal or higher than atmospheric pressure.

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Question:

The planet Venus is surrounded by a thick layer of gases. In fact, the atmospheric pressure on Venus is over 90 times greater than the atmospheric pressure on Earth. Which statements are true about boiling water on Venus? Choose the two (2) statements that apply.

A. The temperature of water will remain constant while it boils.

B. Water will boil at a higher temperature on Venus than on Earth.

C. Water will boil at a lower temperature on Venus than on Earth.

D. The temperature of water will increase as it boils.

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