can someone tell me if I got this right? so I'm supposed to Match the vocabulary terms with their detentions ​

Can Someone Tell Me If I Got This Right? So I'm Supposed To Match The Vocabulary Terms With Their Detentions

Answers

Answer 1

Answer:

I hope that help you

Explanation:

energy is change from one form to another called    _energy conversion

energy can neither be created nor destroyed it can only be transformed from one form to another called --The law of conservation of energy

cause to change in form character or function called - convert

the last sentensce

energy  transformation


Related Questions

If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?

Answers

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

If 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

What is mass?

Mass is defined as a way to gauge how much matter there is in a substance or thing. The kilogram (kg) is the fundamental SI unit of mass, while lower masses can also be measured in grams (g). Atoms make up everyday matter. A majority of an atom's mass is contained in its nucleus.

Given Fe = 38.6 g.

Fe has a molar mass = 55.845 g/mol.

Given mass/molar mass equals 38.6g/55.845gmol-1, or 0.6912 moles of iron.

The reaction is described as Fe + Br2 FeBr2.

One mole Fe yields 1 mole of FeBr2.

FeBr2 would be produced from 0.6912 moles of Fe.

FeBr2 has a molar mass of 215.65 g/mol.

Moles of FeBr2 x Molar mass of FeBr2

= 215.65 g/mole x 0.6912 mole

= 149.06 g FeBr2 produced is the formula.

Thus, if 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

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Oil does not dissolve in water because

Answers

Explanation:

Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.

Yes thank you Great explanation

Answers

Answer:

you're welcome

Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if an additional 0.182mol of methane is added to the container under constant temperature and pressure? Give your answer in three significant figures.

Answers

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

initial volume of gas in the container, V₁ = 2.65 Linitial number of moles of gas, n₁ = 0.12 moladditional concentration, n = 0.182 mol

The total number of moles of gas in the container is calculated as follows;

[tex]n_t = 0.12 + 0.182 = 0.302 \ mol[/tex]

The final volume of gas in the container is calculated as follows;

[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L[/tex]

Thus, the final volume of the methane gas in the container is 6.67 L.

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Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 8.32 percent of its original value

Answers

Answer:

la primera va con la última

Describe your observations when the paper clip was gently crushed with the pestle

Answers

Answer:

We can do an experiment with vinegar to see what happens to the calcium in an eggshell when it is exposed to an acid.

To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.128 V ? (Assume that the partial pressure of hydrogen gas remains at 1 atm.) Express your answer using two decimal places.\

Answers

The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations of H2 and F2 gas are both 1.33x10-3 M, determine the initial concentration of HF gas assuming you only started with HF gas and no products initially.

Answers

This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:

[tex][HF]=[HF]_0-2x[/tex]

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]

[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]

Finally, the initial concentration of HF is calculated as follows:

[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]

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Use the diagram below to answer 3 questions

Answers

Answer:

The element X is sulfur.

Sulfur will gain 2 electrons to become an ion.

The charge of sulfide ion is (2-).

Explanation:

Help !

Student A uses 3N of force to move a cart 5 meters in 10 seconds. Student B uses 6N of force to move the same cart the same distance in 5 seconds. Which student did more work? Which student used more power? Use evidence and explain your answersss

Work = force x distance

Power = work/time

Answers

Work is said to be done if the force applied to a body causes the body to move through a distance.

Student B used more work and power

Workdone = Force * Distance

For student A:

Force = 3N

distance = 5meters

Time taken = 10 secs

Workdone by Student A = 3 * 5

Workdone by student A = 15Nm

Power used up = workdone/time

Power used up = 15/10

Power used up = 1.5 Watts

For student B:

Force = 6N

distance = 5meters

Time taken = 5 secs

Workdone by Student A = 6 * 5

Workdone by student A = 30Nm

Power used up = workdone/time

Power used up = 30/5

Power used up = 6 Watts

This shows that student B used more work and power

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A metal (FW 341.1 g/mol) crystallizes into a body-centered cubic unit cell and has a radius of 1.74 Angstrom. What is the density of this metal in g/cm3

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This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

 

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers blood at a depth of 200 feet and pressure of 7.00 atm

Answers

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

What does Henry's law state?

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

C is the concentration of a dissolved gas. k is the Henry's Law constant. P partial pressure of the gas.

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

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How many of sodium (Na) are needed to make 4.5 liters of a 1.5mol/L of Na solution?​

Answers

Answer:

Explanation:

First you will find the mole from the molarity and then the desired mass from the mole.

Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.

Answers

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from this result? (1 point)

The scientists do not need to collect more data because they have narrowed down the range of the results.

The scientists should not report these results until they have the exact number.

The actual boiling point is either 34°C or 40°C.

The actual boiling point is probably between 34°C and 40°C.

Answers

Answer:

Explanation:

The actual boiling point is probably between 34C and 40C.

Temperature measures the average kinetic energy of particles of the substances. Therefore, the correct option is option D that is the actual boiling point is probably between 34°C and 40°C.

What is temperature?

Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.

Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.

We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:

°C={5(°F-32)}÷9

The actual boiling point is probably between 34°C and 40°C.

Therefore, the correct option is option D.

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Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Answers

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻   (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).  

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

 

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HELP!! what are the usual products of combustion reactions?

Answers

Explanation:

Carbon dioxide and water

I hope it helps

Answer:

The usual products of combustion reactions are carbon dioxide and water.

Explanation:

Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.

Doing Labs at home

I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!

Answers

Answer:

go get the stuff.

Explanation:

What is the mass in grams of one mole of sulfur? (round to 3 sig figs)

Answers

32.065
One mole of sulfur is equal to the atomic mass of sulfur

Answer:

This tells you that one mole of sulfur atoms, S , has a mass of 32 g

The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.

Answers

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Water waves in a small tank are 6.0 cm long. They pass at a given point at a rate of 4.8 waves per second. What is the speed of the wave?

Answers

Answer:,   Correct option is    0.288m/s

Explanation:    

The relationship between the velocity of the wave, its wavelength and frequency is given by the formula

Wavelengthλ=

Frequency(ν)

Speed(v)

,

where, v - velocity of the wave

           λ - wavelength of the wave

          f - frequency of the wave.

In the question it is given that the frequency is 4.8 Hz and the wavelength is 6.0 cm, that is, 0.06 meters.

The velocity of the sound is calculated as follows.

v=f×λ=4.8 Hz×0.06 m=0.288 m/s

Hence, the speed of the water wave is 0.288 m/s.

2 nitrogen atoms and five chlorine atoms what compound does that make

Answers

Answer:

dinitrogen pentachloride

Calculate the second ionization energy of the metal M (?Hion2� in kJ/mol) using the following data:


Lattice enthalpy of MO(s), ?Hl� = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

Enthalpy of sublimation of M = + 130 kJ/mol

First ionization energy of M = + 267 kJ/mol

Standard enthalpy of formation of MO(s), ?Hf� = -307 kJ/mol

Answers

From the information provided in the question, the second ionization energy of the metal is  578 kJ/mol.

From the question, we have the following information;

Lattice enthalpy of MO(s) = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

First ionization energy of M = + 267 kJ/mol

Heat of sublimation of M = + 130 kJ/mol

Standard enthalpy of formation of MO(s) = -307 kJ/mol

Using Hess law of constant heat summation;

ΔHf = ΔHs + BE + ∑IE + ∑EA + U

ΔHs  = Heat of sublimation of metal

ΔHf = Heat of formation MO

BE = Bond energy of O2

∑EA  = sum of electron affinities of Oxygen

∑IE  = Sum of the ionization energies of M

U = Lattice energy of MO

Let the second ionization energy be x

Substituting values;

(-307) = 130 + 498 + (267 + x) + 603 + (-2383)

(-307) = -885 + x

-x =   -885 + 307

-x = -578

x = 578 kJ/mol

The second ionization energy of the metal is  578 kJ/mol.

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solvent extraction explain ???​

Answers

Answer:

this the partial removal of a substance from a solution or mixture by dissolving it in another immiscible solvent in which it is more soluble.

analyze why is the result of the stage of meiosis shown below an advantage for organisms that reproduce sexually.

Answers

Answer:

Meiosis produces haploid cells (gametes), which contain single chromosomes, or on-half the number of chromosomes in diploid cells. When a sperm and an egg join, the single chromosomes pair up, which results in genetic diversity in the offspring

Suppose you have samples of three unknown solids. Explain how you could use their properties to
determine whether or not they are ionic solids.

Answers

Using melting and boiling temperature, hardness and electric current passing testing.

Ionic solids

Ionic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.

In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.

Finally, ionic solids are also excellent conductors of electricity.

So, their properties used to determine whether or not they are ionic solids are  melting and boiling temperature, hardness and electric current passing testing.

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Thin-layer chromatography explain ?????​

Answers

Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

Which stage of a fire can be described as follows: The flames are not visible and the combustible item no longer generates heat or combustion products

Answers

Fire extinguishment is referred to the stage in which flames are not visible

and the combustible item no longer generates heat.

This stage refers to when the fire has been put out through a fire

extinguisher or other compounds.   When fire is put out, there is no longer

flames present.

The absence of flames also means that there is no heat generation or

combustion products due to the absence of a heat source.

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The height of the cylinder is 30 cm. The outside pressure is 105 Pa. The temperature of the gas is kept at 250 K throughout the experiment. The volume filled by the gas is 2.0 l. Now assume that solid cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. Cylinder and piston have the same diameter. Assume that the kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Calculate the change of entropy of the gas and of the environment. Please read this text very carefully

Answers

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

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Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid or a weak acid

Answers

Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.

Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.

Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.

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