The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 8.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M

Answer:

well, first off. the formula for carbon tetrachloride is CCl4

We need to find the molar mass of carbon tetrachloride

1(Mass of C) + 4(mass of chlorine)

1(12) + 4(35.5)

12 + 142

154 g/mol

Number of moles of CCl3 in 543.2g CCl3

n = given mass / molar mass

n = 543.2/153

n = 3.53 moles

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Answer:

3.53 moles

Explanation:

Explanation:

Entropy refers to the degree of disorderliness of a system.

a. A snowman melts on a spring day.

Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.

b. A document goes through a paper shredder.

Entropy increases because random, disorganized bits of paper are left.

c. A water bottle cools down in a refrigerator.

Entropy  decreases because temperature is directly proportional to entropy.

d. Silver tarnishes

Entropy increases because random bits of the sliver particles are formed.

e. Dissolved sugar precipitates out of water to form rock candy.

Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.

Answer:

gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata

Explanation:

vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs

Answer:

CaCl2 has ionic bond because here calcium gives its electron to the chlorine atom and becomes positivetly charged ion.

Answer:

Explanation:

Answer:

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]

In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:

[tex]T_2=T_1+\frac{Q}{mCp}[/tex]

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]

Thereby, answer is B. 58°C .

Regards.

Answer:

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

Explanation:

Given that:

the number of moles of  silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol

The volume of the volumetric flask = 300 mL  = 300 × 10⁻³ L

In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;

[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]

[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]

concentration   in    mmol/L = 0.8966   mmol/L

concentration   in    mmol/L = 8.97 × 10⁻¹ mmol/L

answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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Answer:

[tex]Ksp=1.2[/tex]

Explanation:

Hello,

In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):

[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]

Next, as its dissociation reaction is:

[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]

And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):

[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]

Best regards.

Answer:

b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table

Explanation:

The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).

Among the options, the incorrect option is option B.

This is because transition metals d not have  a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.

The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.

Learn more about the ionic compound containing a transition metal.

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Explanation:

they are called hydrocarbyls

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Answer:

The first carbon atom that attaches to the functional group is referred to as the alpha carbon.

Answer:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]

Explanation:

Hello,

In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]

They are have one-sensed arrow, since sulfuric acid is a strong acid.

Regards.

The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.

Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.

In water in the following way ionization of sulphuric acid takes place:

        H₂SO₄ → HSO₄⁻ + H⁺

        HSO₄⁻ → SO₄⁻ + H⁺

Hence, two steps are shown above.

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Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

where ;

[tex]P_1[/tex] is the vapor pressure at temperature 1

[tex]P_ 2[/tex] is the vapor pressure  at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

[tex]P_1[/tex] = 1 atm

[tex]P_ 2[/tex]  = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

[tex]T_1[/tex] = 282 °C  = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values  into the Clausius - Clapeyron equation, we have:

[tex]\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }[/tex]

[tex]\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }[/tex]

[tex]\dfrac{1}{T_2} = 0.002108048708[/tex]

[tex]T_2 = \dfrac{1}{0.002108048708}[/tex]

[tex]\mathbf{T_2 }[/tex] =  474.37 K

To °C ; we have [tex]\mathbf{T_2 }[/tex] =   (474.37 - 273)°C

[tex]\mathbf{T_2 }[/tex] =  201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

The temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

The given parameters;

The temperature of the liquid will be determined by applying Clausius- Clapeyron Equation;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )[/tex]

where;

[tex]ln(\frac{P_2}{P_1} ) = \frac{\Delta H}{R} (\frac{T_2 -T_1}{T_1T_2} )\\\\ln(\frac{0.35}{1} ) = \frac{28.5 \times 10^3}{8.314} (\frac{T_2 - 555}{555T_2} )\\\\-1.049 = 6.176- \frac{3427.95}{T_2} \\\\\frac{3427.95}{T_2} = 6.176 + 1.049\\\\\frac{3427.95}{T_2} = 7.225\\\\T_2 = \frac{3427.95}{7.225} \\\\T_2 = 474.5 \ K\\\\T_2 = 474.5 - 273 = 201.5 \ ^0C[/tex]

Thus, the temperature of the liquid at the given vapor pressure is 201.5 ⁰C.

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Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]

Whose equilibrium expression is:

[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

Answer:

gamma ray

Explanation:

Explanation:

Antacids are medications used to manage the symptoms of indigestion and heartburn. Antacids contain active ingredients that are bases. These allow antacids to neutralize any stomach acid which could be causing digestive discomfort.

Answer:

C.

Explanation:

I believe that it should be A and B.

Answer:

Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.

The concentration of a solution is the amount of solute in a solution.

We have from the question that he mass of manganese = 1.584 g

Hence;

Amount of [tex]Mn^2+[/tex] = 1.584 g/55g/mol = 0.0288 moles

Recall that;

Number of moles = concentration * volume

Let the concentration of the solution be C

0.0288 moles = C * 1 L

C = 0.0288 moles/ 1 L

C= 0.0288 mol/L

Hence concentration of stock solution = 0.0288 mol/L

For solution A

From the dilution formula;

C1V1 = C2V2

where;

C1 = initial concentration

C2 = final concentration

V1 =  initial volume

V2= final volume

C1 = 0.0288  mol/L

V1 = 50.00 mL

C2 = ?

V2 = 1000.0 mL

C2 = 0.0288  mol/L * 50.00 mL/1000.0 mL

C2 = 0.00144 mol/L

Hence, concentration of solution A is 0.00144 mol/L

For solution B

C1V1 = C2V2

C1 = 0.00144 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 250.0 mL

C2 = 0.00144 mol/L * 10.00 mL/250.0 mL

C2 = 0.0000576 mol/L

Hence, concentration of solution B is 0.0000576 mol/L

For solution C

C1 = 0.0000576 mol/L

V1 = 10.00 mL

C2 = ?

V2 = 500.0 mL

C2 = 0.0000576 mol/L * 10.00 mL/500.0 mL

C2= 0.000001152 mol/L

Hence, concentration of solution C is 0.000001152 mol/L

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Answer:

13.5%

Explanation:

To answer this you need to understand that percentage error can never be negative as we use absolute value in the equation

To get molar/atomic mass = mass/moles

=0.3681g/(6.514×10^-3)moles

=56.509g/moles

percentage error = (65.38-56.509)/65.38 ×100

=13.56% (this is what I'm getting without rounding off your question wasn't specific on how many d.p to use)

Answer:

ITS NOT D. ITS B. 4.52x10^-9 M

Explanation:

Answer:

4.52 ×10–9 M

Explanation:

Solution:-1

[tex]\boxed{\sf {C_3H_4N_2\atop imidazole}+{H_2O\atop water}\longrightarrow {C_3H_5N_2\atop Imidazolium}+{OH^-}}[/tex]

There can be another

[tex]\boxed{\sf {C_3H_4N_2\atop imidazole}+{H_2O\atop water}\longrightarrow {C_3HN_2\atop Dicyanomethanoide}+{H_3O^+}}[/tex]

Solution:-2:-

[tex]\boxed{\sf {SO_3^{2-}\atop Sulphate\:ion}+{H_2O\atop Water}\longrightarrow {H_2SO_4\atop Sulphuric\:acid}}[/tex]

Answer:

Ka = 6.87x10⁻⁵

Explanation:

The equilibrium of benzoic acid in water is:

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The equilibrium constant, Ka, is:

Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]

The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.

X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.

Concentrations in equilibrium are:

pH is defined as -log [H₃O⁺]. As pH = 2.270

2.270 = -log [H₃O⁺]

10^-2.270 = [H₃O⁺]

5.37x10⁻³M = [H₃O⁺] = X.

Replacing, concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M

[C₆H₅COO⁻] = 5.37x10⁻³M

[H₃O⁺] = 5.37x10⁻³M

Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]

Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

The change in freezing point is 1.02°C

The change in the freezing point (ΔTf) is calculated from the following equation:

ΔTf = i x Kf x m

Given:

i = 1 (Van't Hoff factor, is equal to 1 because the solute is a nonelectrolyte)

Kf = 1.86°C/m (Freezing point constant of the solvent)

m = molality

We have to calculate the molality (m), which is equal to the moles of solute in 1 kg of solvent:

m = moles solute/ kg solvent

The solute is C₆H₁₂O₆. Thus, we calculate the moles of solute by dividing the mass into its molecular weight (Mw):

Mw(C₆H₁₂O) = (12 g/mol C x 6) + ( 1 g/mol H x 12) + (16 g/mol O x 6) = 180 g/mol

moles solute = mass C₆H₁₂O/Mw(C₆H₁₂O) = 14.7 g/(180 g/mol) = 0.082 mol

The solvent is water. Its mass (in kg) is obtained from the volume and the density, as follows:

kg solvent = V x density = 150 mL x 1.00 g/mL x 1 kg/1000 g = 0.15 kg

Now, we calculate the molality (m):

m = moles of solute/kg solvent = 0.082 mol/(0.15 kg) = 0.546 m

Finally, we calculate the change in freezing point:

ΔTf = i x Kf x m = 1 x 1.86°C/m x 0.546 m = 1.02°C

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Explanation:

Carbon-12 atoms have stable nuclei because of the 1:1 ratio of protons and neutrons.

Carbon-14 atoms have nuclei which are unstable. C-14 atoms will undergo alpha decay and produce atoms of N-14. Carbon-14 dating can be used to determine the age of artifacts which are not more than 50,000 years old.

Answer:

[tex]Ksp=1.07x10^{-8}[/tex]

Explanation:

Hello,

In this case, the dissociation reaction is:

[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]

For which the equilibrium expression is:

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

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]

[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]

Thereby, the solubility product results:

[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]

Regards.

Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].

The dissociation reaction for lead (II) iodide

[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]

Solubility product constant at equilibrium.

[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]

The molar solubility of the substance can be calculated by using the molar mass,

[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]

Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.

Thus Ksp will be,

[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]

Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].

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Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer:

number of moles of gas increases the volume also increases.