Calculate the mass percent composition of O in each compound.

a. calcium nitrate
b. Iron(II) sulfate
c. Carbon dioxide

For exothermic reactions, ΔH is always negative, energy is considered a product, and energy is released in the reaction.

An exothermic reaction is one in which energy, usually in the form of heat, is released to the environment from the reaction. In other words, the final temperature of an exothermic reaction would always be more than the initial temperature.

This also means that heat energy is a product of exothermic reactions and this heat is released to the environment from the reaction.

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Answer:

a, d, and e is correct on edge of the nuity

Explanation:

have a good day

Answer:

has 5, ans 3 decimals.

thats as simple as i can put it.

Answer:

151 g/mol

Explanation:

In order to solve this problem we need to keep in mind the formula for the boiling point elevation:

Where:

Input the data and calculate m:

We now can calculate the number of moles of the substance, using the definition of molarity:

In this case kg of solvent = 90.0 g / 1000 = 0.090 kg

Finally we calculate the molar mass, using the number of moles and the mass:

There will be no oxygen in the product. Some of the oxygen will evaporate into the air.

Answer:

a

Explanation:because i did the test

Answer:

plastic

........................

Answer: [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Explanation:

A chemical compound is defined as a chemical substance that is formed by the combination of two or more atoms of different elements which cannot be separated by any physical means but when chemically treated, they decompose into their parent elements.

For example, water is made up of hydrogen and oxygen. This compound is a liquid and its individual components are gases. When water is decomposed, it forms hydrogen and oxygen gas.

For the given options:

He(Helium) is an element formed by the combination of only type of atoms.

[tex]O_2[/tex] and [tex]P_4[/tex] are molecules of same element.

[tex]Fe_2O_3[/tex] is a compound fomed by the combination of iron and oxygen atoms.

[tex]C_2H_4O_2[/tex] is a compound fomed by the combination of carbon, hydrogen, and oxygen atoms.

Hence, [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.

Answer:

pOH = - log[OH-]

[OH-] = 0.6M

[tex]pOH \: = - log(0.6) \\ = 0.2218487496 \\ pH \: + pOH \: = 14 \\ pH \: + 0.221848749 = 14 \\ pH = 14 - 0.221848749 \\ = 13.77815125 \\ 13.8[/tex]

Answer:

A wave character exhibited by all particles

Explanation:

The central idea in quantum mechanics is the paradox of wave-particle duality. In quantum mechanics, all particles are believed to also exhibit wavelike characters.

The electron is assumed to behave as a wave hence its position can not be precisely determined according the Heisenberg's uncertainty principle.

These are the underlying postulates that informed Erwin Schrödinger's wave mechanical model of the atom.

Hence, the basic postulate of quantum mechanics is that a wave character is exhibited by all particles.

Answer:

jhhhhhhhhh

Explanation:

dffffffffffffg

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

Answer: My sacrifica has been made.

Explanation:

Answer:

A. Glucose

Explanation:

Glucose is a monomer and not a polymer. So, option (A) is not a polymer.

Glucose is not a polymer because it is a kind of molecule while Starch , cellulose and DNA are polymers.

The correct answer is option A. Glucose.

The more finely divided the solid is, the faster the reaction happens. A powdered solid will normally produce faster reaction than if the same mass is present as a single lump. The powdered solid has a greater surface than the single lump

Explanation:

Mark as brainlist

Answer:

d. none of the above

Explanation:

An electrochemical cell consists of two half-cells, with each cell containing an electrode.

The potential of a single electrode in a half-cell is called the Single electrode potential.

The emf of a cell that consists of two half-cells can be determined by connecting them to a voltmeter.

However, there is no way of measuring the emf of a single half-cell directly.

The only way to determine the emf of a single electrode is to combine it with a standard hydrogen electrode (SHE) and measure it with a voltmeter.

Therefore, the correct option is D "none of the above"

An electrochemical cell consists of two half-cells, with each cell containing an electrode.

An electrochemical cell is a device that can generate electrical energy from the chemical reactions occurring in it, or use the electrical energy supplied to it to facilitate chemical reactions in it.

The potential of a single electrode in a half-cell is called the Single electrode potential.The emf of a cell that consists of two half-cells can be determined by connecting them to a voltmeter.

The only way to determine the emf of a single electrode is to combine it with a standard hydrogen electrode (SHE) and measure it with a voltmeter.

Hence the correct option is D.

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Answer:

[tex]V_{Na_2S}=368mL[/tex]

[tex]m_{Al_2S_3}=67.3gAl_2S_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it is possible to realize that the only way for us to calculate the required volume of sodium sulfide, is by calculating the moles of this substance consumed 163 mL of 2.75 mol/L aluminum sulfate by using the definition of molar concentration and the 1:3 mole ratio between these two:

[tex]n_{Na_2S}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{3molNa_2S}{1molAl_2(SO_4)_3} =1.34molNa_2S[/tex]

Now, we divide these moles by the molar concentration of sodium sulfide to obtain the required volume:

[tex]V_{Na_2S}=\frac{1.34molNa_2S}{3.65mol/L} =0.368L=368mL[/tex]

For the last part, we now use the 1:1 mole ratio of aluminum sulfate to aluminum sulfide and the molar mass of the latter (150.158 g/mol) in order to calculate the required mass:

[tex]m_{Al_2S_3}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{1molAl_2S_3}{1molAl_2(SO_4)_3} *\frac{150.158gAl_2S_3}{1molAl_2S_3} \\\\m_{Al_2S_3}=67.3gAl_2S_3[/tex]

Regards!

Answer:

D. having fur that is a similar color to tree barks with algae growing on it

Explanation:

Choice A doesnt make sense because sloths do not change colors, and hanging upside down would not protect you from prey or blend in. D matches the description of a sloth.

Answer:

Having fur that is a similar color to tree barks with algae growing on it.

Explanation:

Answer:

17.48

Explanation:

mass=density×volume

= 0.713g/ml × 24.5 ml

= 17.4685

≈ 17.47

Answer:

2.85 mol

Explanation:

Step 1: Given data

Mass of sulfur tetrafluoride (SF₄): 309 g

Step 2: Calculate the number of moles corresponding to 309 g of sulfur tetrafluoride

To convert mass to moles we need a conversion factor: the molar mass. The molar mass of SF₄ is 108.07 g/mol.

309 g × 1 mol/108.07 g = 2.85 mol

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]

     OR

[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]

where, M is the molar mass of gas

Forming an equation for the two gases:

[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex]          .....(1)

Given values:

[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]

Plugging values in equation 1:

[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]

Where, M is the molar mass of the gas

Forming an equation for the two gases:

[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex]          .....(2)

Given values:

[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]

Plugging values in equation 2:

[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]

Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Answer:

Kr- Dispersion Forces

H2O- Hydrogen Bonding

CHCI3- Dipole-Dipole Forces

HF- Hydrogen Bonding

C2H6- Dispersion Forces

HBr- Hydrogen Bonding Forces

Explanation:

Dispersion forces occurs in all substances. They are the dominant intermolecular interaction in all non polar substances such as C2H6 and Kr.

Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom such as Cl, Br, O etc. It is the dominant intermolecular interaction in HF, HBr and H2O.

Dipole-Dipole interactions occur when a permanent dipole exists in a molecule such as in CHCI3

Answer: The balanced molecular equation for the precipitation reaction is [tex]K_2CO_3(aq)+2CuF(aq)\rightarrow Cu_2CO_3(s)+2KF(aq)[/tex]

Explanation:

A precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

When potassium carbonate reacts with copper (I) fluoride, it leads to the formation of solid copper (I) carbonate and potassium fluoride solution.

The balanced chemical equation follows:

[tex]K_2CO_3(aq)+2CuF(aq)\rightarrow Cu_2CO_3(s)+2KF(aq)[/tex]

The precipitate formed is copper (I) carbonate

Answer:

I dont no ma men

Explanation:

Sorry cause a dont no

Answer:

Using Newman projections, draw the most stable conformation for each of the following compounds.

(a) 3-methyl pentane, viewed along with the C2-C3 bond.

(b) 3,3-dimethyl hexane, viewed along with the C3-C4 bond.

Explanation:

(a) The structure of 3-methyl pentane is shown below:

In Newman projection, the most stable conformation is staggard conformation.

In staggard conformation, the torsional strain is very less compared to eclipsed conformation.

(b)3,3-dimethyl hexane, viewed along with the C3-C4 bond.

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

[tex]PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)[/tex]

Lo cual hace que la expresión de equilibrio se calcule como:

[tex]Ksp=[Pb^{2+}][Cl^-]^2[/tex]

Y que en términos de la solubilidad molar, s, se resuelve como:

[tex]1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L[/tex]

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

[tex]s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L[/tex]

¡Saludos!

Answer: The number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].

Explanation:

The reaction equation is as follows.

[tex]FeCl_{3}(aq) + 3NaOH(aq) \rightarrow Fe(OH)_{3}(s) + 3NaCl(aq)[/tex]

Therefore, moles of [tex]Fe(OH)_{3}[/tex] are calculated as follows.

Moles = Molarity of [tex]Fe(OH)_{3}[/tex]  [tex]\times[/tex] Volume (in L)

= 0.654 M [tex]\times[/tex] 0.197 L  

= 0.128 mol

Now, according to the given balanced equation 1 mole of [tex]FeCl_{3}(aq)[/tex] reacts with 3 moles of NaOH(aq). Hence, moles of [tex]Fe(OH)_{3}[/tex]  reacted are calculated as follows.

3 [tex]\times[/tex] 0.128 mol = 0.384 moles of NaOH

As moles of NaOH present are as follows.

Moles of NaOH = Molarity of NaOH [tex]\times[/tex] Volume (in L)

0.384 mol = 0.587 M [tex]\times[/tex] Volume (in L)

Volume (in L) = 0.654 L (1 L = 1000 mL) = 654 mL

Thus, we can conclude that the number of milliliters of 654 mL for 0.587 M NaOH required to precipitate all of the [tex]Fe^{3+}[/tex] ions in 197 mL of 0.654 M [tex]FeCl_{3}[/tex] solution as [tex]Fe(OH)_{3}[/tex].

Answer: The factor that lead to cyclopropane being less stable than the other cycloalkanes is the presence of a RING STRAIN.

Explanation:

In organic chemistry, the end carbon atoms of an open aliphatic chain can join together to form a closed system or ring to form cycloalkanes. Such compounds are known as cyclic compounds. Examples include cyclopropane, cyclobutane, cyclopentane and many among others.

Cyclopropane is less stable than other cycloalkanes mentioned above because of the presence of ring strain in its structural arrangement. The ring strain is the spatial orientation of atoms of the cycloalkane compounds which tend to give off a very high and non favourable energy. The release of heat energy which is stored in the bonds and molecules cause the ring to be UNSTABLE and REACTIVE.

The presence of the ring strain affects mainly the structures and the conformational function of the smaller cycloalkanes. cyclopropane, which is the smallest cycloalkane than the rest mentioned above, contains only 3 carbons with a small ring.

Answer: The instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]

Explanation:

From the graph,

Initial rate of the [tex]Br_2[/tex] = 1.0 M

Time when the concentration of [tex]Br_2[/tex] is 0.5 M (half the concentration ) = 60 sec

For first order reaction:

Calculating rate constant for first order reaction using half life:

[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)

[tex]t_{1/2}[/tex] = half life period = 60 s

k = rate constant = ?

Putting values in equation 1:

[tex]k=\frac{0.693}{60s}\\\\k=0.01155s^{-1}[/tex]

For the given chemical reaction:

[tex]H_2(g)+Br_2(g)\rightarrow 2HBr(g)[/tex]

Rate of the reaction = [tex]-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]

Negative sign represents the disappearance of the reactants

From the above expression:

[tex]k[Br_2]=-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]

At 50 seconds, [tex][Br_2]=0.6 M[/tex]

Plugging values in above expression, we get:

[tex]\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=0.01155\times 0.6\\\\\frac{\Delta [HBr]}{\Delta t}=2\times 0.01155\times 0.6=0.01386=1.4\times 10^{-2}M/s[/tex]

Hence, the instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]