3. What is the mass of an object with a volume of 4 L and a density of 1.25 g/mL?

Answers

Answer 1

Answer:

5000g

Explanation:

mass= density × volume

Since the unit of density here is g/mL, we need to convert the volume to mL.

1L= 1000mL

4L= 4 ×1000 = 4000 mL

Mass of object

= 1.25 ×4000

= 5000g

Answer 2

Answer:

5,000 grams

Explanation:

The mass of an object can be found by multiplying the volume by the density.

mass= volume * density

The density is 1.25 g/mL and the volume is 4 L.

First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.

There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.

4 * 1,000 = 4,000

The volume is 4,000 mL.

Now, find the mass of the object.

mass= volume * density

volume = 4,000

density= 1.25

mass= 4,000 * 1.25 = 5,000

Add the appropriate units for mass, in this case, grams, or g.

mass= 5,000 g

The mass of the object is 5,000 grams.


Related Questions

21. What are the two main ways of working with clay?

Answers

Answer:

Diferentes tipos de arcilla

ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...

ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...

ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...

ARCILLAS “BALL CLAY” O DE BOLA. ...

CAOLIN. ...

ARCILLA REFRACTARIA. ...

BENTONITA.

Explanation:

Answer:

Coil method and the slab method.

Explanation:

Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?

Answers

Answer:

7.50 L

Explanation:

The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 6.25 atm × 1.20 L / 1.00 atm

V₂ = 7.50 L

The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding

Answers

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

Answer:

The intermolecular forces present in CH_3NH_2 includes

II. (ion-dipole) and IV. (hydrogen bonding)

Explanation:

The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)

It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.

Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.

For more information on intermolecular forces, visit

https://brainly.com/subject/chemistry

What chemical bonds hold atoms?

Answers

bond: A link or force between neighboring atoms in a molecule or compound.
ionic bond: An attraction between two ions used to create an ionic compound. This attraction usually forms between a metal and a non-metal.
covalent bond: An interaction between two atoms, which involves the sharing of one or more electrons to help each atom satisfy the octet rule. This interaction typically forms between two non-metals.

f a substance has a half-life of 8.10 hr, how many hours will it take for 75.0 g of the substance to be depleted to 3.90 g?

Answers

Answer:

35 hrs

Explanation:

half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr

initial amount [tex]N_{0}[/tex] = 75 g

The final amount [tex]N[/tex] = 3.9 g

The time elapsed [tex]t[/tex] = ?

we use the relationship

[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]

substituting values, we have

3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

take the log of both side

log 0.052 = log  [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2

-1.284 =  [tex]\frac{t}{8.1}[/tex]  x -0.301

1.284 = 0.301t/8.1 =

1.284 = 0.0372t

t = 1.284/0.037 = 34.5 ≅ 35 hrs

11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. If a chicken weighs 3kg, how many grams of cyanide would it need to ingest to kill 50% of domestic chickens?

Answers

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]?=\frac{3kg*11mg}{1kg}\\ \\?=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M

Answers

Answer:

[tex][NH_2NO_2]=0.0868M[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction, the first-order rate law is:

[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]

Which integrated is:

[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]

Thus, the concentration after 31642.0 s for a 0.384-M solution is:

[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]

[tex][NH_2NO_2]=0.0868M[/tex]

Best regards.

Answer:

[A] = 0.0868 M

Explanation:

Rate constant = 4.70×10-5 s-1

First order reaction

Initial concentration, [A]o = 0.384 M

Final concentration, [A] = ?

Time, t = 31642.0 s

All these variables are related by the following equation;

[A] = [A]o e^(-kt)

[A] = 0.384  e^(-4.70×10-5 x  31642.0)

[A] = 0.384 e^(-1.4872)

[A] = 0.384 * 0.2260

[A] = 0.0868 M

What happens to the rate of dissolution as the temperature is increased in a gas solution?

A.
The rate stays the same.
B.
The rate decreases.
C.
The rate increases.
D.
There is no way to tell.

Answers

Answer:

The rate decreases

Explanation:

When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.

Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.

Hence increase in temperature decreases the rate of solubility of a gas in water.

Answer:

B.

The rate decreases.

Explanation:

When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced

Answers

Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O

place the following substances in Order of decreasing boiling point H20 N2 CO

Answers

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

discuss four factors of learning ​

Answers

Answer:

plz mark as BRAINLIEST plz...

Explanation:

●Intellectual factor: The term refers to the individual mental level. ...

●Learning factors: ...

●Physical factors: ...

●Mental factors: ...

Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrofluoric acid and sodium hydroxide are combined. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If needed, use H for the hydronium ion.)

Answers

Answer:

The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)

Explanation:

The reaction between aqueous solutions of hydrofluoric acid and sodium hydroxide is an example of a neutralization reaction.

A neutralization reaction is a reaction between and acid and an abase to produce salt and water only.

Hydrofluoric acid is the acid while sodium hydroxide is the base. during the reaction the hydrofluoric acid will produce hydrogen and fluoride ions, while sodium hydroxide will produce hydroxide and sodium ions. The hydroxide and hydrogen ions will combine to produce water while the sodium and fluoride ions remain in solution as ions.

The equation of the reaction is as follows:

H⁺F⁻ (aq) + Na⁺OH⁻ (aq)  ----> Na⁺F⁻ (aq) + H₂O (l)

Since the sodium and fluoride ions appear on both sides of the equation, they are known as spectator ions and are cancelled out to give the net ionic equation.

The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)

The front curve of a spectacle lens is called?

Answers

Answer:

Corrective lense or just lens.

Explanation:

Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.

Answers

Answer:

7.23 J

Explanation:

Step 1: Given data

Mass of graphite (m): 566.0 mgInitial temperature: 5.2 °CFinal temperature: 23.2 °CSpecific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows: The reaction is second order with a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration in molarity remaining after 10.0 min? Report your answer to 3 decimal places.

Answers

Answer:

[tex]C_{C_4H_6}=0.179M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]

And the rate law is:

[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]

Which integrated is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]

In such a way, the concentration after 10.0 min is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]

Regards.

a. Name a chemical or product that was once considered safe but is now considered
harmful. (1 point)
-

Answers

Answer:

Bisphenol A (BPA)

Explanation:

Bisphenol A (BPA) is a chemical additive commonly found in resins and plastics, such as water bottles or food containers. It can also be found in household electronics, medical devices, dental fillings and sales receipts, just to name a few other applications.

What's the name for the part of Earth made of rock?
A. Geosphere
B. Atmosphere
C. Hydrosphere
D. Biosphere
SUBMIT

Answers

Answer:I think it's Geosphere

Explanation:

Answer:

A

Explanation:

Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. Each function is characterized by 3 quantum numbers: n, l, and ml. If the value of n = 3 ... The quantum number l can have values from ? to ? . ... The total number of orbitals possible at the n = 3 energy level is ? . If the value of l = 3 ... The quantum number ml can have values from to ? . ... The total number of orbitals possible at the l = 3 sublevel is ?? .

Answers

Answer:

1) The quantum number l can have values from

2 to 0

2)The total number of orbitals possible at the n = 3 energy level is 3'2=9

3) If the value of l = 3 ... The quantum number ml can have values from 3 to -3

The quantum number l determines the shape of the orbital.  In this case, if the value of n is 3, then the quantum number l can have values from 0 to (3-1), which is 2.

The total number of orbitals possible at the n = 3 energy level can be determined using the formula 2l + 1. So, for l = 0, there is 1 orbital. For l = 1, there are 3 orbitals. And for l = 2, there are 5 orbitals. Therefore, the total number of orbitals possible at the n = 3 energy level is 1 + 3 + 5 = 9.

On the other hand, the quantum number ml represents the magnetic quantum number. It specifies the orientation of the orbital in space. The value of ml ranges from -l to +l. So, if the value of l is 3, then the quantum number ml can have values from -3 to +3.

The total number of orbitals possible at the l = 3 sublevel can be determined using the formula 2ml + 1. So, for ml = -3, there is 1 orbital. For ml = -2, there is 3 orbitals. For ml = -1, there is 5 orbitals. For ml = 0, there is 7 orbitals. For ml = 1, there is 5 orbitals. For ml = 2, there is 3 orbitals. And for ml = 3, there is 1 orbital.

Therefore, the total number of orbitals possible at the l = 3 sublevel is 1 + 3 + 5 + 7 + 5 + 3 + 1 = 25.

Learn more about quantum number,here:

https://brainly.com/question/32773003

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An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)

Answers

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.

Answers

Answer:

[tex]pH=11.12[/tex]

Explanation:

Hello,

In this case, ammonia dissociation is:

[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]

So the equilibrium expression:

[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

That in terms of the reaction extent and the initial concentration of ammonia is written as:

[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]

Thus, solving by using solver or quadratic equation we find:

[tex]x=0.00133M[/tex]

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]

And the pH from the pOH is:

[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]

Best regards.


One gram is approximately the same as half the mass of a new U.S.
A) penny.
B) dime.
C) quarter.
D) dollar.

Answers

Answer:

b) dime

Explanation:

a dime is approximately 2.2g

half of this is 1.1g, which can be rounded down to one gram.

hope this helps

Which of the following represents a compound made of five molecules? CO 5 C 2O 5 C 5O 5CO 2

Answers

Answer:

Co5

Please mark me brainliest so that I get encouraged to make more great answers like this one!

Answer:

GUYS ITS 5CO 2

Explanation:

How are pH and pOH ?

A. pH = 14 + pOH
B. pOH = 14 - pH
C. pOH = 14 + pH
D. pH = 14 - pOH

Answers

Answer:

B. pOH = 14 - pH and D. pH = 14 - pOH.

Explanation:

Hello,

In this case, we must remember that pH and pOH are referred to a measure of acidity and basicity  respectively, since pH accounts for the concentration of H⁺ and pOH for the concentration of OH⁻ in a solution. In such a way, since the maximum scale is 14, we say that the addition between the pH and pOH must be 14:

[tex]pH+pOH=14[/tex]

Therefore, the correct answers are B. pOH = 14 - pH and D. pH = 14 - pOH since the both of them are derived from the previous definition.

Best regards.

Answer:

D: by subtracting the pOH from 14.

Explanation:

A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3

Answers

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]

Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the data and mechanism. a. the mechanism for this reaction is E2 b. an increase in 1-butene was observed when t-butoxide was used c. an increase in 1-butene was observed when methoxide was used d. the mechanism for this reaction is E1 e. no significant difference was observed

Answers

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.
A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4

Answers

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C  and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to

Answers

Answer:

Zinc- anode

Copper- cathode

Sodium sulphate- salt bridge

Explanation:

A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.

In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e

Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)

Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.

Answer:

zinc=anode

copper=cathode

Explanation:

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answers

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true of the resulting conjugate base H2PO−4? Select the correct answer below: H2PO−4 can act as an acid.

Answers

Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.

What is acid - conjugate base pair?

An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.

Given chemical reaction is:

H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺

In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.

Hence, H₂PO₄⁻ is a conjugate base.

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To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.

Answers

Answer:

a. the Kb of NO₂⁻ to find the hydroxide concentration.

Explanation:

When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.

NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)

Na⁺ comes from NaOH (strong base) so it doesn't react with water.

NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to  find the hydroxide concentration.

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