What is the particle behavior of an liquid?

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer:

2.28 atm

Explanation:

V₁ = 2.33L,  V₂ = 1.08L

T₁ = 289K,  T₂ = 304K

P₁ = 1.05 atm, P₂ = ?

Where V₁ and V₂ are initial and final volume respectively

T₁ and T₂ are initial and final temperature respectively

P₁ and P₂ are initial and final pressure respectively

The formula to be used here is the general gas equation:

P₁V₁/T₁=P₂V₂/T₂

1.05 × 2.23/289 = P₂ × 1.08/304

P₂ × 1.08 × 289 = 1.05 × 2.23 × 304

P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)

P₂ = 711.82 ÷ 312.12

P₂ = 2.28 atm

Answer:

C6H1206

Explanation:

C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.

Answer : The number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

[tex]\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}[/tex]

[tex]\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol[/tex]

Now we have to calculate the moles of precipitate, [tex]PbBr_2[/tex] formed.

The balanced chemical reaction is:

[tex]2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)[/tex]

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of [tex]PbBr_2[/tex]

So, 0.02 moles of NaBr react to give [tex]\frac{0.02}{2}=0.01[/tex] mole of [tex]PbBr_2[/tex]

Therefore, the number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaBr = 0.2 M

Mole = Molarity x Volume

Mole of NaBr = 0.2 × 0.1

2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)

From the balanced equation above,

2 moles of NaBr reacted to produce 1 mole of PbBr₂.

Therefore,

0.02 mole of NaBr will react to produce = [tex]\frac{0.02}{2} \\\\[/tex] = 0.01 mole of PbBr₂.

Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole

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Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

Answer:

A

Explanation:

Slow cooling of magma beneath the surface of earth

To solve this we must be knowing each and every concept related to rock and its formation. Therefore, the correct option is option A among all the given options.

Rock is a naturally occurring mineral aggregate that is cohesive and composed of one or even more minerals. These aggregates often take the shape of recognisable and mappable volumes and are the fundamental building block of the solid Earth.

Mineral crystals as well as the sorts of rocks that serve as their hosts cycle through many forms as geologic materials. Temperature, weight, time, and variations in the climate in the Earth's crust and on its surface all play a role in the process. Slow cooling of magma beneath the surface of Earth is the  statement that best describes the formation of rock.

Therefore, the correct option is option A.

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Answer:

B

Explanation:

I did the question before and got it right.

Answer:

Explanation:

Silk and wool

Answer:

silk and wool fibers are obtained from animals ​

Explanation:

good luck :)

Answer:

[tex]T=2.78x10^3 \°C[/tex]

Explanation:

Hello,

In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:

[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]

The pressure in atm is:

[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]

And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):

[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]

In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:

[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]

Best regards.

Explanation:

Preciptation and air pressure

Answer:

It is Temperature and humidity. C

Explanation:

I took the test and it said it was C

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

Answer:I think it would be the third one.

Explanation:

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True  

Explanation:

The particles moving faster in a substance the hotter it gets.

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Molar mass = mass/ number of moles

molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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Answer:

atom

Explanation:

the atom is the smallest form.

Answer:

major ecological grouping of plants and animals

food chain

*biome*

bacteria and fungi that break down dead matter

*decomposers*

a state of change in which the end result is equal or

balanced

*dynamic equilibrium*

the basic relationships that show how a community of plants,

animals, and bacteria live and grow and how these living

*ecosystem*

things are dependent on each other as well as the Sun, soil,

and other nonliving parts of their environment; a cycle of

relationships

line of plants and animals that shows the order in which

*Food chain*

organisms are eaten

a condition or conditions of the nonliving surroundings, such

*environmental factor*

Explanation:

Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6

Answer:

u measure how much power it has

Explanation:

for example u can power a light bulb woth it if u can it was 100eg energy

Answer:

= .7567105263

Explanation:

1 atm = 760 mmHg

575.1 mmHg (1 atm/760mmHg) = .7567105263 atm

Answer:

The theoretical yield of CaS is 6.01 g.

Explanation:

The balanced reaction is given as:

[tex]Ca+S\rightarrow CaS[/tex]

The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.

Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]

So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca

Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] mol of S or 0.0833 mol of S

According to the balanced equation:

1 mol of Ca produces 1 mol of CaS

So, 0.179 mol of Ca produces 0.179 mol of CaS

According to the balanced equation:

1 mol of S produces 1 mol of CaS

So, 0.0833 mol of S produces 0.0833 mol of CaS

As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.

So, thoretically, 0.0833 mol of CaS is produced.

The molar mass of CaS is 72.143 g/mol.

So, the mass of 0.0833 mol of CaS is [tex](0.0833\times 72.143)[/tex] g or 6.01 g

Hence, the theoretical yield of CaS is 6.01 g.

Answer:

true

Explanation:

hope it helps

Answer:

I'm pretty sure it's true

Explanation:

20 characters

Answer: a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion : Common ion effect

b) The mass of a salt in grams that will dissolve in 100 mL of water: Solubility

c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Saturated solution

d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation : Solubility product constant

e) The maximum number of moles of a salt that will dissolve in 1 L of solution : Molar solubility

Explanation:

Solubility is expressed usually as the mass of solute per 100 grams or 100 ml of solvent.

Molar solubility of a solid is expressed as the concentration of the dissolved solid in a saturated solution.

Solubility product constant is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]

The common-ion effect is the decrease in solubility of a sparingly soluble salt by the addition to the solution of a soluble compound with an ion in common with the precipitate.

Saturated solution is defined as the solution in which no more solute particles can be dissolved in the solvent.

Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:

Answer:

C. TWO HAPLOID CELLS ARE FORMED

Explanation:

I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT

Answer:

Kps = 3.07 x 10⁻⁸

pKsp= 7.51

Explanation:

First, we calculate the molar solubility of  silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):

2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s

Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:

     Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻

I                               0              0

C                            +3s           +s

E                            3s               s

Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴

Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:

Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸

The pKsp value is:

pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51

Answer:

29.8

Explanation:

The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.

Answer:

[tex]p=26683.2Pa[/tex]

Explanation:

Hello,

In this case, since the pressure is computed via:

[tex]p=h*\rho*g[/tex]

Whereas h is the 0.520-m height, [tex]\rho[/tex] is the 13600-kg/m³ density and the g the 9.81-m/s² gravity. Thus, the pressure in Pa is:

[tex]p=0.20m*13,600 \frac{kg}{m^3} *9.81\frac{m}{s^2} \\\\p=26683.2\frac{kg*\frac{m}{s^2} }{m^2} =26683.2\frac{N}{m^2}\\ \\p=26683.2Pa[/tex]

Best regards.

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.

The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.

In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.

In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.

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