At elevated temperatures, hydrogen iodide may decompose to form hydrogen gas and iodine gas, as follows:

2HI(g) ⇌ H2 (g) + I2 (g)

In a particular experiment, the concentrations at equilibrium were measured to be [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, and [H2] = 0.27 mol/L. What is Kc for the above reaction?

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to:

Answer:

The two elements with atomic number 9 and 12 are represented by letter Q and R respectively, where Q represents fluorine atom and R represents magnesium atom.

1. Electronic configuration of R that is magnesium (atomic number 12) is:

1s2 2s2 2p6 3s2

2. Q represents fluorine atom, which belongs to group 17 in periodic table that is the most reactive and lightest member of the group.

3. Q and R that is fluorine and magnesium combinely form magnesium fluoride or MgF2.

Answer:

duw8 Wert gsi

Explanation:

pues pues y 8y1rcuecisscfjfj3eoeu xv cihskdkkd HD jekifeuifkeñ elijo eh fh FC eh SSH DJ djdvheshdhs

The mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.

Mass percent is defined as a way of expressing a concentration or describing a component in a particular mixture.

To calculate the mass percent of an element in a compound, divide the mass of the element in 1 mole of the compound by the molar mass of the compound and multiply the result by 100.

Mass percent = Mass of chemical / Total mass  of compound x 100

Mass of ore = 10 kg

% antimony in ore = 10.6%

                              = 10.6 / 100 = 0.106 g

Mass of antimony = 0.106 g

Mass percent = Mass of antimony / Mass of stibnite x 100

71.7% = 0.106 / X x 100

X = 0.106 x 1000 / 71.7 x 1000

  = 106 / 717 = 0.148 g

Thus, the mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.

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Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.

Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud

B. Plant vegetation on the slops that expreience landslides.

E. Reduce the sloped where landslides occur.

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Attached are the Lewis structures:

Hope it helps...

[tex] \: \: \: [/tex]

Answer:

Calcium is a chemical element with the symbol Ca and atomic number 20.

Calcium is a mineral that is necessary for life. In addition to building bones and keeping them healthy, calcium enables our blood to clot, our muscles to contract, and our heart to beat. About 99% of the calcium in our bodies is in our bones and teeth.

Answer:

2 Methyl pent 2 ene

Answer:

C

Explanation:

This is because oxygen has the lowest temperature, thereby collision rate is least

Answer:

one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.

Explanation:

H-N=N-H

The answer to this question is approximately equal to 57.8

Answer:

[tex]K^{2000K}=0.774\\\\K^{3000K}=12.56[/tex]

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

[tex]\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}[/tex]

Thus, at 2000 K:

[tex]\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol[/tex]

And at 3000 K:

[tex]\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol[/tex]

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

[tex]K=exp(-\frac{\Delta _rG}{RT} )[/tex]

Thus, at each temperature we obtain:

[tex]K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56[/tex]

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

Answer:

D / 15.0 g

Explanation:

3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions

.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle

Answer:

Volume percent is defined as: v/v % = [(volume of solute)/(volume of solution)] x 100%

Explanation:

Answer:

condensation

Explanation:

What led to the formation of oceans after the water on the Earth's surface evaporated is a condensation reaction of water vapour.

The water present on the surface of the earth was able to evaporate as a result of the hot condition of the primitive earth. As the earth cools down the water vapour present in the atmosphere began to condense, gradually forming puddles of water and eventually leading to the formation of various oceans as we currently have it on the earth.

Answer:

as earth cooled, water in the atmosphere condensed.

Explanation:

Answer:

CO2

Explanation:

Mass of carbon = 27.29

Mass of oxygen = 72.21

Step 1:

We have to first convert these masses to moles

Carbon = 2.29/2.01 = 2.274

Oxygen = 72.71/16 = 4.544

Step 2:

We have to divide the mols by the smallest mol to get simplest ratio of whole number.

The smallest mol is 2.274. we have to divide the mols by this.

2.274/2.274 = 1

4.544/2.274 = 2

Our empirical formula is therefore CO2

Answer:

B

Explanation:

the fluorine has an high tendency to gain electrons from other elements with lower electronegativities

Answer:

5.74M or 5.74 mol/L (to 3 sign. fig.)

Explanation:

The molar mass of NaNO3 is 85g/mol, which means that:

1 mole of NaNO3 - 85g

? moles - 122.0g

= 122/85 = 1.44 moles

Concentration in mol/L = no. of moles (moles) ÷ volume (L)

[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)

I hope the steps are clear and easy to follow.

Answer:

The interpretation of the particular subject is covered in the subsection below in detail.

Explanation:

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]

And the undergoing chemical reaction:

[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]

Next, the moles of magnesium chloride consumed by the sodium fluoride:

[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]

[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]

Thereby, the reaction quotient is:

[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

The cell potential for the electrochemical cell has been 1.40 V.

The standard reaction for the cell will be:

[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]

The half-reaction of the cells has been:

[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]

The potential for this reduction has been -0.04 V.

[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]

The potential for the reduction has been 1.36 V.

The cell potential has been: Potential of reduction - Potential of oxidation

Cell potential = 1.36 - (-0.04) V

Cell potential = 1.40 V.

The cell potential for the electrochemical cell has been 1.40 V.

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Answer:

THR ANSWER IS C) _ methylpropanoate

Answer:

here's your answer

Explanation:

Molar mass of CH3CH2OH = 46.06844 g/mol

This compound is also known as Ethanol.

Convert grams CH3CH2OH to moles  or  moles CH3CH2OH to grams

Molecular weight calculation:

12.0107 + 1.00794*3 + 12.0107 + 1.00794*2 + 15.9994 + 1.00794

Percent composition by element  

Hydrogen H 1.00794 6 13.128%

Carbon C 12.0107 2 52.143%

Oxygen O 15.9994 1 34.730%

The mass percentage of C in CH₃CH₂OH is 52.14% (to two decimal places)

To calculate the mass percentage of C (Carbon) in CH₃CH₂OH (Ethanol),

First, we will determine the mass of CH₃CH₂OH

Molar mass of CH₃CH₂OH = 46.07 g/mol

Mass of C = 12.01 g/mol

Now, for the mass percentage of C in CH₃CH₂OH,

We will determine the ratio of the total mass of C to the mass of CH₃CH₂OH, and then multiply by 100%

Since we have 2C in CH₃CH₂OH

Then, total mass of C in CH₃CH₂OH = 2 × 12.01 g/mol = 24.02 g/mol

That is,

Mass percentage of C in CH₃CH₂OH = [tex]\frac{24.02}{46.07} \times 100\%[/tex]

Mass percentage of C in CH₃CH₂OH = 0.5213805 ×100%

Mass percentage of C in CH₃CH₂OH = 52.13805%

Mass percentage of C in CH₃CH₂OH ≅ 52.14%

Hence, the mass percentage of C in CH₃CH₂OH is 52.14% (to two decimal places)

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Answer:

Molar solubility of AgBr = 51.33 × 10⁻¹³

Explanation:

Given:

Amount of NaBr = 0.150 M

Ksp (AgBr) = 7.7 × 10⁻¹³

Find:

Molar solubility of AgBr

Computation:

Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr

Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150

Molar solubility of AgBr = 51.33 × 10⁻¹³

When The Molar solubility of AgBr is = 51.33 × 10⁻¹³

Given as per question:

The Amount of NaBr is = 0.150 M

Then Ksp (AgBr) is = 7.7 × 10⁻¹³

Now we Find:

The Molar solubility of AgBr

The we Computation is:

The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr

After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150

Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³

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Answer:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]

The equilibrium expression is:

[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:

[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]

Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:

[tex]Molar \ solubility=3.12x10^{-5}M[/tex]

Regards.

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

The dissociation of calcium fluoride has been given by:

[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]

The solubility constant, ksp has been given as:

[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]

From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.

The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.

The concentration of Ca in the solution has been resulted as x + 0.01 M.

The solubility product can be given as:

[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]

The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].

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Answer: A=1, B=3, C=12

Explanation:

For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.

For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.

Now that we know the unit conversion, we can directly solve.

36 ft×(1 yd/3 ft)=12 yd

Our final answer is A=1, B=3, C=12.

Answer:

A=1,000, B=1, C=5,400

Explanation:

the question was 5.4L x AmL / BmL = CmL

Answer:

a. 121 ml, b. 253 ml and c. 2.57 L.

Explanation:

The new volume can be calculated by using the Boyle's law equation:  

P1V1 = P2V2

In the equation, P1 and P2 are the initial and final pressures and V1 and V2 are the initial and final volumes for a real gas at constant temperature.  

a) Based on the given information, P1 = 755 mmHg, V1 = 125 ml, P2 = 780 mm Hg and V2 will be,  

V2 = P1V1/P2

V2 = 755 mmHg × 125 ml/780 mmHg

V2 = 121 ml

b) Based on the given information, P1 = 1.08 atm, V1 = 223 ml, P2 = 0.951 atm and V2 will be,  

V2 = P1V1/P2

V2 = 1.08 atm × 223 ml/0.951 atm

V2 = 253 ml

c) Based on the given information, P1 = 103 kPa, V1 = 3.02 L, P2 = 121 kPa and V2 will be,  

V2 = P1V1/P2

V2 = 103 kPa × 3.02 L/121kPa

V2 = 2.57 L

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

A transition occurs when an electron absorbs energy and moves from one energy level to another.

Let us recall the formula;

E = hc/λ

h= planks constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 ms-1

λ = wavelength =

We can see from the question that the exponent in the value for the wavelength was not properly rendered hence we can not be able to substitute values and obtain a numerical answer for the energy of the infrared radiation.

However, if all the values are properly substituted, then the energy of the infrared radiation can conveniently be obtained.

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Answer:

The answer is noble gases

Explanation:

Here is your explanation The vertical columns are called groups. There are eighteen groups. The last group on the far right is called the noble, or inert gases. Elements in a group have similar chemical properties. For example, elements in the noble gas group are all gases under. This is the thing from the passege bye god bless you

The last group of elements on the periodic table are called "noble gases."

The noble gases are a group of elements located in Group 18 of the periodic table. They are also known as Group 0 or the "inert gases." The noble gases include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

The noble gases are unique because they have a full complement of valence electrons in their outermost energy level. This full electron configuration gives them exceptional stability, making them chemically unreactive or inert under normal conditions. In other words, noble gases are less likely to form chemical bonds with other elements.

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