Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show the formal charges of all atoms in the correct structure. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom for:SeO2 and CO2?3

Answer:

Which equation represents the reaction of a weak acid with water?  the equation is : HCl + H2O H3O+ + Cl- HCO3– + H2O H2CO3 + OH– H2O H + + OH- HCOOH + H2O H3O+ + HCOO

Explanation:

hope it helps : )

Answer:

Hey mate, here is your answer. Hope it helps you.

HCOOH + H2O ↔ H3O+ + HCOO-

Explanation:-

A strong acid is one which completely dissociates into its corresponding ions in aqueous medium.

In contrast,  a weak will only partially dissociate such that there is an equilibrium between the dissociated ions and the undissociated acid.

In the given examples:

HCl, HCO3- and H2CO3 are all strong acids. However, HCOOH i.e. formic acid is a weak acid which dissociates in water to form H3O+ and formate ion, HCOO-

HCOOH + H2O ↔ H3O+ + HCOO-

Answer:

The  pKa  of  Glu is   [tex]pK_a = 4.25[/tex]

Explanation:

From the question we are told that

  The [tex]K_a[/tex] value of  Glu is [tex]K_a = 5.62*10^{-5}[/tex]

The [tex]pKa[/tex] is mathematically represented as

         [tex]pK_a = -log (K_a )[/tex]

         substituting values

          [tex]pK_a = -log (5.62 *10^{-5} )[/tex]

          [tex]pK_a = 4.25[/tex]

Answer: EXTINCTION

Explanation:

Answer:

extinction

Explanation:

Answer:

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

Explanation:

The balanced reaction is:

SOCl₂ + H₂O ----> SO₂ + 2 HCl

By stoichiometry of the reaction they react and produce:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of mass react and are produced:

Then the following rule of three can be applied: if by stoichiometry of the reaction 118.9 grams of SOCl₂ produce 2 moles of HCl, 0.226 grams of SOCl₂ how many moles of HCl do they produce?

[tex]moles of HCl=\frac{0.226 grams of SOCl_{2}*2 mole of HCl }{118.9 grams of SOCl_{2} }[/tex]

moles of HCl= 3.80*10⁻³

3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂

Answer: CO2 and H2O

Explanation: I already took the test it's right

Answer:

solubility is 1.984x10⁻⁹M

Explanation:

When CaCO₃ is in water, the equilibrium that occurs is:

CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)

Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹

If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:

[Ca²⁺] [0.250M] = 4.96x10⁻⁹

Assuming you are adding an amount of CaCO₃:

[X] [0.250 + X] = 4.96x10⁻⁹

Where X is the amoun of CaCO₃ you can add, that means, solubility

X² + 0.250X - 4.96x10⁻⁹ = 0

Solving for X:

X = -0.25M → False answer, there is no negative concentrations.

X = 1.984x10⁻⁹M.

That means, solubility is 1.984x10⁻⁹M

Answer:

[tex]1.984x10^{-8}M[/tex]

Explanation:

Hello,

In this case, the equilibrium reaction for the solubility of calcium carbonate is:

[tex]CaCO_3(s) \rightleftharpoons Ca^{2+}(aq)+CO_3^{-2}(aq)[/tex]

In such a way, since 0.250 M sodium carbonate solution is the solvent, we assume an initial concentration of carbonate anion to be also 0.250 M since sodium carbonate is completely dissolved, for that reason the equilibrium equation turns out:

[tex]Ksp=[Ca^{2+}][CO_3^{2-}]\\\\4.96x10^{-9}=x*(0.250+x)[/tex]

Hence, solving for [tex]x[/tex] we have:

[tex]x=1.984x10^{-8}M[/tex]

Which corresponds to the molar solubility if calcium carbonate as well.

Regards.

Answer:

P = 20.1697 atm

Explanation:

In this case we need to use the ideal gas equation which is:

PV = nRT (1)

Where:

P: Pressure (atm)

V: Volume (L)

n: moles

R: universal gas constant (=0.082 L atm / K mol)

T: Temperature

From here, we can solve for pressure:

P = nRT/V  (2)

According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.

The volume for any cylinder would be:

V = πr²h  (3)

Replacing the data here, we can solve for the volume:

V = π * (17)² * 164

V = 148,898.93 cm³

This volume converted in Liters would be:

V = 148,898.93 mL * 1 L / 1000 mL

V = 148.899 L

Now we can solve for pressure:

P = 125 * 0.082 * 293 / 148.899

Answer:

building and managing their own clients, purchasing supplies and keeping records

hope this helps you

Answer:

Yes that is why you can even see how some chemicals might react but that is if you are not considering temperature, pressure, or if catalyst are involved.

Answer:

(b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen.

Explanation:

The law of definite proportion states that a gen chemical compound always contains its constituent elements in a fixed ratio by mass, independent on the method of preparation.

The molar mass of Nitrogen and Oxygen would always remain the same, allowing for exact reactant masses (or mole ratio) irrespective of the given amount of sample.

Explanation:

H

|

H-C=C-C-OH

| | |

H H H

in this organic compound the functional groups are OH and alkene

prop-2-enol

since it has a double bond, alkene is the functional group. and also it has OH group so hydroxyl group also the other functional group

Answer:

ΔGrxn for the reaction A(g)→B(g) will most likely to be negative under the PA=10.0atm ; PB=0.050atm condition

The FIRST option is correct

Explanation:

From Thermodynamics,

Using the Gibbs Free Energy (G) formula to determine the reaction A(g)→B(g) most likely to be negative.

Gibbs Free Energy (G) which is the energy associated with a chemical reaction that can be used to do work.

CHECK THE ATTACHMENT FOR THE DETAITALED EXPLANATION

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

Hello,

In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.

Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.

Regards.

Answer:

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2

Explanation:

A substance is acting as oxidising agent when is reducing its chemical state in the reaction.

In HNO3, Nitrogen has an oxidation state of +5 (Each oxygen is -2, 3 oxygens, -6, 1 hydrogen, +1). Thus, if the oxidation state of the nitrogen in the products is < +5, the nitrogen is reducing its oxidation state acting as oxidising agent.

In the reactions:

A. CuO + 2HNO3 → Cu(NO3)2 + H2O . In Cu(NO₃)₂, Nitrogen has an oxidation state of +5. Thus, nitric acid is not acting as oxidising agent.

B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 . Also, oxidation state in Cu(NO₃)₂ does not have any change but there is another product, NO₂, where nitrogen has an oxidation state of +4. That means nitric acid is acting as oxidising agent.

C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2 . Here, in NaNO₃, oxidation state of nitrogen is +5 (Na: +1, 3 O: -6). Thus, nitrogen is not changing is oxidation state and nitric acid is not acting as oxidising agent.

D. NaOH + HNO3 → NaNO3 + H2O. Here, also, the product is NaNO₃. That means nitric acid is not acting as oxidising agent.

Answer:

The metal surface becomes more positive as electrons are lost from it.

Explanation:

Let us note that photoelectric effect refers to a phenomenon in which electrons are ejected from a clean metal surface irradiated with light of appropriate frequency. This photon must possess a frequency above the threshold frequency of the metal and its energy must exceed the work function of the metal. When these conditions are met, electrons are emitted from a clean metal surface, having a constant kinetic energy as long as the frequency of the incident photon remains constant.

However, as photoelectric effect progresses and electrons are lost from the metal surface, the metal surface becomes more positive. The more positive the surface, the greater the attraction of the positive surface for the emitted electrons. This reduces the kinetic energy of the emitted photons even though the frequency of incident photons is held constant.

Answer:

ΔH° = -1255.8 kJ

Explanation:

Step 1: Data given

ΔHf∘ C2H2 = 227.4 kJ/mol

ΔHf∘ O2 = 0 kJ/mol

ΔHf∘ CO2 = -393.5 kJ/mol

ΔHf∘ H2O = -241.8 kJ/mol

Step 2: The balanced equation

2C2H2(g)+5O2(g) ⟶ 4CO2(g)+2H2O(g)

C2H2(g)+5/2O2(g) ⟶ 2CO2(g)+H2O(g)

Step 3: Calculate ΔH° of the reaction

ΔH° = (2*ΔHf∘ CO2 + ΔHf∘ H2O) - (ΔHf∘ C2H2)

ΔH° = (2* -393.5 kJ/mol + (-241.8) kJ/mol) - 227.4 kJ/mol

ΔH° = -787 - 241.8 - 227. kJ/mol

ΔH° = -1255.8 kJ

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618 [tex]\times[/tex] 0.067302) + ( ? [tex]\times[/tex] (1-0.067302)

39.0983 = 2.756811064 + ( ? [tex]\times[/tex] 0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ? [tex]\times[/tex] 0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618  0.067302) + ( ?  (1-0.067302)

39.0983 = 2.756811064 + ( ?  0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ?  0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

that is why co2 is in the power of 2ik

Answer:

1. Increase in the temperature of the water

2. Increasing the surface area of the lithium

Explanation:

1. Increase in the temperature of the water

The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.

2. Increasing the surface area of the lithium

As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.

Answer: Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]heat_{released}=heat_{absorbed}[/tex]

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex]

Q = heat absorbed or released

[tex]m_1[/tex] = mass of mercury= 452 g

[tex]m_2[/tex]= mass of water = 145 g

[tex]T_{final}[/tex]  = final temperature = ?

[tex]T_1[/tex]   = temperature of mercury = [tex]85.0^0C[/tex]

[tex]T_2[/tex] = temperature of water = [tex]23.0^0C[/tex]

[tex]c_1[/tex]   = specific heat of mercury  = [tex]0.138J/g^0C[/tex]

[tex]c_2[/tex]   = specific heat of water= [tex]4.184J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex]-[452\times 0.138\times (T_f-85.0)^0C]=145\times 4.184\times (T_f-23.0)^0C[/tex]

[tex]T_f=28.8^0C[/tex]

Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]

Image is not given in the question, so the image for the question is given below.

Answer:

Fischer projection is a two-dimensional representation of a three-dimensional organic molecule by projection.

Wedge and dash bonds are used to represent the three-dimensional structure of a molecule, in which wedges indicates bonds towards the viewer, solid lines indicates bonds in the plane of the image and dashed lines indicates bonds away from the viewer.

Wedge and dash bonds structure of D-glyceraldehyde is attached below.

Answer:

Calcium hydride

Explanation:

hey there!

atomic mass cobalt = 58.9332 amu

therefore:

1 mol Co ------------ 58.9332 g

moles ------------ 50 g

moles = 50 x 1 / 58.9332

moles  = 50 / 58.9332

 = 0.8484 moles of Co

Hope this helps!

0.8484 moles are in 50 grams of cobalt.

A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Atomic mass cobalt = 58.9332 amu

Therefore:

1 mol Co ------------ 58.9332 g

moles ------------ 50 g

moles = 50 x 1 ÷ 58.9332

moles  = 50 ÷ 58.9332

= 0.8484 moles of Co

Learn more about moles here:

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Answer:

Yes it is true

Explanation:

This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.

When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.

Answer:

ai) Rate law,  [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]

If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:  

[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]

Using the information provided in lines 3 and 4 of the table and insering the value of a:

[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]

[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]

The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]

The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:

[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]

b) Overall order of reaction =  a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Let's consider the following reaction.

CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)

The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.

The  rate law for this reaction is:

r = k [CH₃Cl]ᵃ [Cl₂]ᵇ

where,

If we write the ratio r₂/r₁, we get:

r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ

r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ

(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ

a ≈ 1

If we write the ratio r₃/r₂, we get:

r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ

r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ

(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ

b ≈ 0.5

So far, the rate law is:

[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]

Let's use the values of the first experiment to find the value of the rate constant.

[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]

The final rate law is:

[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

It is the sum of the individual orders of reaction.

a + b = 1 + 0.5 = 1.5

The rate law is for this reaction is    [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]

where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.

Learn more about the rate law here: https://brainly.com/question/14945022

Answer:

(D) - 16.6 kcal

Explanation:

Hello,

In this case, the Gibbs free energy for the given reaction is computed in terms of the Gibbs free energy of formation of each species involved in the chemical reaction:

[tex]\Delta _rG=2\Delta _fG_{NO_2}-2\Delta _fG_{NO}-\Delta _fG_{O_2}[/tex]

Thus, it is found for nitrogen monoxide, oxygen and nitrogen dioxide the following Gibbs free energies of formation: 87.6, 0 and 51.3 kJ/mol respectively, therefore we compute:

[tex]\Delta _rG=2(51.3kJ/mol)-2(87.6kJ/mol)=-72.6kJ*\frac{1kcal}{4.184kJ} \\\\\Delta _rG=-17.35kcal[/tex]

The closest result is (D) - 16.6 kcal, as such difference is noticed when different sources for thermochemical data are used, in this case, the NIST data were used.

Best regards.

Answer:

THE MOLECULAR FORMULA FOR THE COMPOUND IS C2 Cl2

Explanation:

mass = 0.04380 g

Volume = 10 mL = 10 / 1000 L = 0.010 L

Pressure = 1.10 atm

R = 0.082 L atm mol^-1 K^-1

Temperature = 290.5 K

Carbon = 25.305 %

Chlorine = 74.695 %

Atomic mass of carbon = 12

Atomic mass of chlorine = 35.5

First calculate the empirical formula by following these steps:

1. Write the percentage composition of the elements involved and divide by its atomic mass

Carbon = 25.305 / 12 = 2.10875

Chlorine = 74.695 /35.5 = 2.1040

2. Divide each by the smaller value

Carbon = 2.10875 / 2.1040 = 1.002

Chlorine = 2.1040 / 2.1040 = 1

3. Round up to the whole number

The empirical formula is C Cl

Next is to calculate the molar mass of the compound using ideal gas equation

PV = mRT/ MM

Mm = mRT / PV

Mm = 0.04380 * 0.082 * 290.5 / 1.10 * 0.010

Mm = 1.0433 / 0.011

Mm = 94.845 g/mol

Mm = 94.85 g/mol

Now that we know the molar mass, we can go on to calculate the molecular formula:

(C Cl) n = Molar mass

( 12 + 35.5) n = 94.85

(47.5)n = 94.85

n = 94.85 / 47.5

n = 1.9968

n ~ 2

The molecular formula can then be written as C2Cl2.

Answer:

Explanation:

2 HgCl₂ + C₂O₄²⁻   =  2 Cl⁻ + Hg₂Cl₂ + 2CO₂

1 )

Rate of reaction [tex]= k[HgCl_2]^m[C_2O_4^{-2}]^n[/tex]

             [HgCl₂]        [C₂O₄²⁻ ]           Rate  

1 .              .124             .115               1.61 x 10⁻⁵

2 .             .248             .115             3.23 x 10⁻⁵

3 .              .124             .229              6.4 x 10⁻⁵

4 .              .248             .229            1.28 x 10⁻⁴

comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻  becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻  .

Hence n = 2

comparing 1 and 2 , when concentration of HgCl₂ becomes twice  and concentration of C₂O₄²⁻  remains constant  , rate becomes 2 times so rate is proportional to simply  concentration of C₂O₄²⁻  .

Hence m = 1

Putting the data of  1 in the rate equation found

 1.61 x 10⁻⁵ = k x .124 x  .115²

k = 11.3 x 10⁻⁴ M⁻² s⁻¹

Answer:

ATGGCCTA

Explanation:

For this we have to keep in mind that we have a specific relationship between the nitrogen bases:

-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.

-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.

Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:

TACCGGAT

ATGGCCTA

I hope it helps!

Answer:

d

Explanation:

it's called the hydrocarbon

Answer:

A hydrocarbon

Explanation: