When a solute is dissolved in water, it forms……

Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

The formula of manganese (IV) oxide is MNO2

The formula for manganese (IV) oxide is [tex]MnO_4[/tex]. Therefore, the correct option is A.

A manganese atom is bonded to four oxygen atoms to produce an inorganic compound known as manganese(IV) oxide with the chemical formula [tex]MnO_4[/tex]. It is often referred to as manganese dioxide. The +4 oxidation state of a manganese compound is indicated by the Roman numeral "IV" in the name.

Pyrolusite, a naturally occurring mineral, is a dark brown or black solid consisting of manganese(IV) oxide. Due to its favorable properties, it is often employed in a wide variety of applications. It acts as a catalyst, especially when breaking down hydrogen peroxide. Additionally, it is used to make batteries, ceramics, glass and pigments.

Therefore, the correct option is A.

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Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Answer:

C-O: polar covalent

Mg-F: ionic

Cl-Cl: nonpolar covalent

Explanation:

Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.

Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.

In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.

In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.

Answer:

362g

Explanation:

heat lost by metal= heat gained by acetic acid

tfs are the same so you cando delta T

convert Cal/gc to J/gc

thectgod ig follow

Answer:

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

Answer:

water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.

Explanation:

Answer:

See explanation and image attached

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.

Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.

The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.

Answer:

I think its A maybe am not sure

Answer:

d. 60.8 L

Explanation:

Step 1: Given data

Step 2: Calculate the work (W) done by the system

We will use the following expression.

ΔU = Q + W

W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J ×  1 atm.L/101.325 J = -1.593 atm.L

Step 4: Calculate the initial volume

First, we will use the following expression.

W = - P × ΔV

ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

The initial volume is:

V2 = V1 + ΔV

V1 = V2 - ΔV

V1 = 63.2 L - 2.35 L = 60.8 L

Answer:

The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Explanation:

The structure of nucleic acid polymers is built up from monomers of nucleotides.

A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.

When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.

Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Answer:

The products will be favored at equilibrium.

Explanation:

The balanced chemical equation for the reaction is the following:

2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)

The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).

The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:

[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]

Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²

So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).

Therefore, products will be favored at equilibrium at 298 K.

Answer:

The answer is "170.9 mm Hg".

Explanation:

[tex]\text{Mass of acetone = volume} \times density[/tex]

                          [tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]

[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]

                            [tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]

[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]

                                   [tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]  

[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]

                                    [tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]

[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]

Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

Then;

Using the relation:

[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

Now, using the formula for calculating the numbers of moles which can be expressed as:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For acetone:

[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]

For ethyl acetate:

[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]

For Acetone:

[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]

[tex]\mathbf{mole \ fraction =0.5545 }[/tex]

For ethyl acetate:

[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]

[tex]\mathbf{mole \ fraction =0.4455}[/tex]

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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Answer:

0.0365 is your answer

Explanation:

Answer:

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Answer:

Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas. The ideal gas molar volume increases to 24.0 liters as the temperature increases to 20 °C (at 1 atm).

Answer:

velocity = distance/time

velocity = wavelength × frequency

Both of these are commonly known equations to calculate velocity with different variables.

Answer:

C) 5

Step-by-step explaination:

Benzene has 5 resonance structures.

Answer:

a measure of concentration equal to the gram equivalent weight per liter of solution.

Explanation:

Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

hope it helped

Answer:

radical-initiated

Explanation:

Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.

Answer:

[tex]\%diff=24.0\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:

[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]

Thus, we secondly calculate the molar volume as:

[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]

Then, we plug in the entire variables in the vdW equation to get such pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]

And the ideal gas pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]

Finally, the percent difference:

[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]

Regards!

Answer:

Neither Tech A nor B is correct

Explanation:

Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.

When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.

Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.

Answer:

It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.

The correct option for the given question about Mole Concept is Na / 2 atom of oxygen.

In 1 mole number of molecules of CO₂ = Na (Avagadro Number)

In 0.25 mole number of molecules of CO₂ = 0.25 × Na molecules

In 1 molecule number of oxygen atom = 2 atom

In 0.25 × Na molecules number of oxygen atom = 2 × 0.25 × Na atom

In 0.25 × Na molecules number of oxygen atom = 0.5 × Na atom

So in 0.25 mole of CO₂ number of oxygen atom = Na / 2 atom

Thus we can conclude that in 0.25 mole of CO₂ number of oxygen atom will be Na / 2, where Na is Avagadro Number.

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Answer:

When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.

Explanation:

Answer:

False

Explanation:

The first reaction is;

NO(g) + 1/2O2(g) ---->NO2(g)

K= [NO2]/[NO] [ O2]^1/2

The second reaction is;

2NO(g) + O2(g) ---->2NO2(g)

K'= [NO2]^2/[NO]^2 [O2]

It now follows that;

K'= K^2

Hence the statement in the question is false

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

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Answer:

ΔH = 2.68kJ/mol

Explanation:

The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:

q = m*S*ΔT

Where q is heat of reaction in J,

m is the mass of the solution in g,

S is specific heat of the solution = 4.184J/g°C

ΔT is change in temperature = 11.21°C

The mass of the solution is obtained from the volume and the density as follows:

150.0mL * (1.20g/mL) = 180.0g

Replacing:

q = 180.0g*4.184J/g°C*11.21°C

q = 8442J

q = 8.44kJ when 3.15 moles of the solid react.

The ΔH of the reaction is:

8.44kJ/3.15 mol

= 2.68kJ/mol

Answer:

Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.

Answer:

which class are you please mention