A soda manufacturing company is experimenting with changing the taste of its product as the concentration of carbon dioxide changes. To track their results, they must determine how concentration changes with pressure. The concentration of CO2 under a partial pressure of 0.719 atm is 429.7 ppm. At what pressure (in atm) would the CO2 need to be so that the concentration of CO2 is 235.3 ppm at the same temperature

Answer:

0.250 mol Mg²⁺

0.500 mol Cl⁻

Explanation:

Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:

MgCl₂  ⇒  Mg²⁺ + 2 Cl⁻

1 mol      1 mol   2 mol

1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂.  If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:

0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺

0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻

Answer:

HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/

5/5 all the way.

Explanation:

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

Answer:   3.87M  of HI remains after 27.3 s

Explanation:

Using the Second order decomposition equation of

1/[H]t =K x t +1/[A]o

Given initial concentration ,[A]o = 4.78M

time, t = 27.3 s

rate of constant , k= 1.80 x 10^-3 M-1s-1

1/[H] t= 1/[A] t= concentration after time, t=?

SOLUTION

1/[A] t =kt +1/[A]o

1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78

0.04914+0.2092=0.2583

1/[A] t =0.2583

[A] t =1/0.2583= 3.87M

Answer:

1.7 × 10⁹ L

Explanation:

Step 1: Write the balanced equation

CO(g) + 2 H₂(g) → CH₃OH(l)

Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]

Step 3: Calculate the theoretical yield of CH₃OH

The real yield of CH₃OH is 3.1 × 10⁷ mol  and the percent yield is 40%. The theoretical yield is:

[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]

Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH

The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol

Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP

The volume of 1 mole of CO at STP is 22.4 L.

[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]

Answer:

The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.

Explanation:

1s^2

2s^2

2p^6

3s^2

3p^6

4s^2

3d^10

4p^5

Answer:

Ethyl cinnamate

Explanation:

For this question, we have to start with the IR info. If we have a peak at 1710 this indicates the presence of a carbonyl group in the molecule (C=O). Additionally, if we calculate the I.H.D (index of hydrogen deficiency), we will have a value of "6". We already know that we have a C=O group, so, this counts for 1 of the 6 additionally, we can have a benzene ring so, this counts for 4, so far we have 5. Finally, we will have a double bond outside of benzene and we will have a total of 6, so:

Benzene: 4

Carbonyl group: 1

Double bond: 1

For a total of six (that fits with the I.H.D calculation). So, so far we know that we have a benzene ring, a double bond, and a carbonyl group. In the formula we have 2 oxygens, therefore we can have a carboxylic acid or an ester. In this case, the IR info doesn't give any additional info, so our best option is the ester group.

The 1H NMR info give is:

Signal A= 1.3 (3 H, triplet)

Signal B= 4.3 (2 H, quartet)

Singal C= 6.5 (1 H, doublet)

Signal D= 7.4-7.6 (5 H, multiplet)

Signal E= 7.7 (1 H, doublet)

The molecule that fits with this NMH spectrum and the info given by the I.H.D is "ethyl cinnamate".

See figure 1

I hope it helps!

Answer:

0.89kg

Explanation:

Q=mL L=specific latent heat

Q=energy required in J

m=mass in Kg

Q=mL

m=Q/L

m=2000000J/2.25 x 10^6 J kg-1

m=0.89kg

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Answer:

Amount of HCL = 0.00318 L  of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH  = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH  = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed =  0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795  / 2.5

Amount of HCL = 0.00318 L  of 3.18 ml

Answer:

Explanation:

In weight/volume (w/v) terms,

   1 ppm = 1g m-3 = 1 mg L-1 = 1 μg mL-1

200 mL = 0.2 L

15 / 0.2 mg L-1 =75 ppm

The concentration in ppm of a solution which is prepared by dissolving in 15mg of NaCl in 200ml water is 75 mg/.,

ppm stand for 'part per million' and it is used to define the concentration of any substance as mass of any substance present in per liter of volume of solution, its unit for measurement is mg/L.

Given that, mass of NaCl = 15mg

Volume of solution = 200mL = 0.2L

Concentration in ppm will be calculated as:
ppm = 15/0.2 = 75mg/L

Hence ppm concentration of NaCl is 75 mg/L.

To know more about ppm, visit the below link:
https://brainly.com/question/16877061

#SPJ2

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!

The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.

We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.

We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.

Learn more: https://brainly.com/question/2510654

Answer:

Approximately [tex]24\; \rm g[/tex] (at most.)

Explanation:

Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:

[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].

Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:

Calculate the formula mass of the reactants and of the product:

Calculate the quantity (in number of moles of formula units) of each reactant:

Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].

Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].

In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.

On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].

Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].

Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].

Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

Answer:

C

Explanation:

I had this question and C is the right answer

One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

Learn more about atom,here:

https://brainly.com/question/13654549

#SPJ5

Answer:

3.97

Explanation:

pH of buffer solution = pKa+Log(Cb/Ca)

pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1

Where Ca = concentration of acid, Cb = concentration of base.

Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M

Substitute into equation 1

pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)

pH of buffer solution = 4.19+(0.22)

pH of buffer solution = 3.97.

Answer: The density of the object is 1.10 g/ml

Explanation: Density of object = ?

Mass of object = 26.5 g

Volume of object = volume of water displaced = 24.1 ml

Putting values in above equation, we get:

[tex]Density: \frac{26.5g}{24.1ml} = 1.10g/ml[/tex]

Thus density of the object is 1.10 g/ml

Answer: The equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

Using ideal gas equation : [tex]PV=nRT[/tex]

P = pressure of gas

V = volume of gas

n = no of moles

R = gas constant

T = Temperature

pressure of [tex]N_2O_4[/tex] = [tex]\frac{1\times 0.0821Latm/Kmol\times 298}{5L}=5atm[/tex]

pressure of [tex]NO_2[/tex] = [tex]\frac{2\times 0.0821Latm/Kmol\times 298}{5L}=10atm[/tex]

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

at t= 0    5 atm                                10 atm

at eqm    (5-x) atm                          (10+2x) atm

[tex]K_p=\frac{[p_NO_2]^2}{[p_N_2O_4]}[/tex]

[tex]0.1134=\frac{(10+2x)^2}{(5-x)}[/tex]

[tex]x=-4.48[/tex]

pressure of [tex]N_2O_4[/tex] at equilibrium = (5-(-4.48))= 9.48 atm

pressure of [tex]N_2O_4[/tex] = [tex]\frac{n\times 0.0821Latm/Kmol\times 298}{V}[/tex]

9.48 = [tex]{M\times 0.0821Latm/Kmol\times 298}[/tex]

[tex]M=0.379[/tex]

Thus the equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Answer:

D) 4f

Explanation:

To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.

1s

2s         2p

3s         3p          3d       3f

4s         4p          4d       4f

5s         5p           5d       5f

6s         6p          6d       6f

7s         7p           7d       7f

In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

Therefore, 4f is the correct answer.

Answer:

D) 4f

Explanation:

Answer:

A) NH3

It is the only one that is a formula!!

Answer:

Option C . CO2(g) + H2O(g)

Explanation:

When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

Thus, the product of the unbalanced combustion reaction is:

CO2(g) + H2O(g)

Thus, we can balance the equation as follow:

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)

There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:

C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)

Thus, the equation is balanced.

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Answer:

[I⁻] = 0.0352M

Explanation:

Based on the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined as:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

The system reaches the equilbrium when the ratio [I₂] [I⁻] / [I₃⁻] is equal to 0.25

In the beginning, you add 0.0401M of both [I₂] [I⁻].  When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of  [I₃⁻]. That is written as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

X is known as reaction coordinate.

Replacing in Kc:

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Solving for X:

X = 0.0049M → Right solution

X = 0.3252M → False solution. Produce negative concentrations

Replacing, equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

Since

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Here Kc should be defined

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

Also, The system finished the equilibrium at the time when the ratio [I₂] [I⁻] / [I₃⁻] is equivalent to 0.25.

Also,

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

Also,

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Now

X = 0.0049M → Right solution

X = 0.3252M → False solution

Now equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

[I⁻] = 0.0352M

Hence, The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

Learn more about molarity here: https://brainly.com/question/14782192

Answer:

1. REDOX

2. None of the above

3. Precipitation

4. Preicipitation

5. Acid base neutralization

Explanation:

Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.

Option 4 and 3.

3) CaO (s) + CO₂ (g) →  CaCO₃(s)

4) KOH (aq)  + AgCl (aq)  →  KCl (aq)  + AgOH(s)

Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.

5) Ba(OH)₂ (aq)  +  2HNO₂(aq)  →  Ba(NO₂)₂ (aq) + 2H₂O(l)

Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.

1) Ba(ClO₃)₂ (s)  → BaCl₂ (s) +  3O₂(g)

Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).

2.  2NaCl(aq)  +  K₂S(aq)  Na₂S (aq)  + 2KCl (aq)

None of the above

Answer:

8.00L of ammonia can be produced

Explanation:

The reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g)

Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.

As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:

1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia

Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:

12L H₂(g) ₓ (2L NH₃(g)  /  3L H₂(g)) =

Answer:

[tex]w/v\%=3.02\frac{g}{mL} \%[/tex]

Explanation:

Hello,

In this case, we first define the formula for the calculation of weight/volume percentage considering cobalt (II) fluoride as the solute, water the solvent and the both of them as the solution:

[tex]w/v\%=\frac{mass_{solute}}{V_{solution}}*100\%[/tex]

In such a way, since the mass of the solute is given as 6.04 g and the final volume of the solution 200 mL, the weight/volume percentage turns out:

[tex]w/v\%=\frac{6.04g}{200mL}*100\%\\\\w/v\%=3.02\frac{g}{mL} \%[/tex]

Regards.

Answer:

See figure 1

Explanation:

In this case, we have to start with the ionization reaction of HBr to produce the hydronium ion ([tex]H^+[/tex]) and the bromide ion ([tex]Br^-[/tex]). Then the double bond in the alkene can attack the hydronium ion to produce a carbocation. The most stable carbocation would be the tertiary one, therefore we have to put the positive charge in the tertiary carbon. Then, the bromide attacks the carbocation to produce the final halide.

See figure 1

I hope it helps!

Answer:

22.4 L H2

Explanation:

There is a better explanation https://brainly.com/question/9562878

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

Answer:

1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis

2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis

3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis

4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis

5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis

Explanation:

Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.

Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.

Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.

The following shows the various regulatory methods and their effects on both processes:

1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.

Increased levels of fructose-2,6-bisphosphatase  activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.

2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.

3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.

4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.

5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the  activation of gluconeogenesis and inhibition of glycolysis.

Answer:

a) 1.5 x 10^-3 mol/L

b) 1.35×10^-8

c) decrease

Explanation:

The solubility of lead II iodide is given by the equation;

PbI2(s) -----> Pb^2+(aq) + 2I^-

By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L

Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L

Ksp= [Pb^2+] [2I^-]^2 =

Let the molar solubility of each ion be x, therefore;

Ksp= 4x^3

Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8

Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.

a) The molar solubility of PbI₂ is  [tex]1.5 * 10^{-3} mol/L[/tex]

b) The solubility constant is [tex]1.35*10^{-8}[/tex]

c) The molar solubility of lead (II) will decrease.

The solubility of lead II iodide is given by the equation;

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

By looking at the ICE table,

[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex]  [tex]1.5 * 10^{-3} mol/L[/tex]

Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]

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

Let the molar solubility of each ion be x, therefore;

[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]

The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.

Find more information about Solubility product here:

brainly.com/question/9807304