what is the molarity of 10.0 g of kcl in 0.500 l of solution

Answers

Answer 1
Molarity is defined as number of moles of the solute in 1 liter of solution.

Molarity = Moles of solute (mol) / Volume of the solution (L)

moles of KCl = mass (g) / molar mass (g/mol)
                      = 10.0 g / 74.5 g/mol
                      = 0.134 mol

Hence molarity = 0.134 mol / 0.500 L
                         = 0.268 mol/L

Related Questions

A compound consisting of atoms of large atomic mass is more likely to require:

Answers

Answer:

A) a higher temperature to liquefy

BRAINLIST?

Explanation:

Answer:

The answer is

A: a higher temperature to liquefy

Phosphorus-32 has a half-life of 14.3 days. how many grams remain from a 10.0 gram sample after 30.0 days?

Answers

when the nuclear half-life of the radioactive isotope is showing the time needed for the isotope to be half of its initial value of mass.

so with each half-life, the isotope will be halved of its initial value as example:

after the first half-life, the isotope will lose 50 % of its initial value

and after the second half-life, the isotope will lose 25% of its initial value 

and after the third half-life, the isotope will lose 12.5 % of its initial value

and so on,

So here to get how many numbers of half-lives we will use this formula:

numbers of half-lives = total time passed / the half-life of the isotope

                                    = 30 days / 14 days

                                    =2 days

∴remainig mass = initial mass / 2^numbers of half-lives

                            = 10 g / 2^2

                            = 2.5 g

 

Answer: 2.23 grams

Explanation:

Radioactive decay follows first order kinetics.

Half-life of Phosphorus-32 = 14.3 days

[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{14.3}= 0.05days^{-1}[/tex]

[tex]N=N_o\times e^{-\lambda t}[/tex]

N = amount left after time t= ?

[tex]N_0[/tex] = initial amount  = 10.0 g

[tex]\lambda[/tex] = rate constant= [tex]0.05days^{-1}[/tex]

t= time  = 30 days

[tex]N=10\times e^{- 0.05 days^{-1}\times 30days}[/tex]

[tex]N=2.23g[/tex]

Why is it important to know what temperature scale is being used in a given situation?

Answers

It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.

Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius 

Identify each of the following mixtures as either homogeneous or heterogeneous and as a solution, a suspension, or a colloid.
Blood
homogeneous heterogeneous solution colloid suspension
Salad dressing
homogeneous heterogeneous solution colloid suspension
PLEASE HELP!!!!

Answers

With reference to the solutions, following things may be noted
1) Homogeneous solutions comprises of single phase.
2) Heterogeneous solution comprises of more than one phase.
3) in colloidal solution, solute is evenly distributed in solvent. in such system, it is difficult to separate solute and solvent.
4) in suspension, there is an uneven distribution of solute in solvent. In such system, solute and solvent gets separated on standing long.  

Blood is a homogeneous and colloidal solution,
Salad dressing is a heterogeneous and suspension solution.   

blood: heterogeneous and suspension

salad dressing: heterogeneous and suspension

How many times greater is the rate of effusion of molecular fluorine than that of molecular bromine at the same temperature and pressure

Answers

According to Graham's Law of Diffusion,"the rates of diffusion of two gases are inversely proportional to the square root of their Molar masses or Densities at the same pressure and temperature".

                                           r₁ / r₂  =  [tex] \sqrt{M2 / M1} [/tex]
Where,
            r₁  =  Rate of Fluorine

            r₂  =  Rate of Bromine

            M₂  =  Molar mass of Bromine  =  159.8 g/mol

            M₁  =  Molar mass of Fluorine  =  37.98 g/mol

Putting values,

                                           r₁ / r₂  =  [tex] \sqrt{159.8 / 37.98} [/tex]

                                           r₁ / r₂  =  [tex] \sqrt{4.20} [/tex]

                                           r₁ / r₂  =  2.04

Result:
          Fluorine effuses 2 times faster than Bromine gas.

The coordinate plane is separated into four quadrants as shown. let p: x < 0 let q: y < 0 what is represented by p ∨ q?
quadrant 1 because both x and y are positive coordinates

quadrant 3 because both x and y are negative coordinates

quadrants 1, 2, and 4 because in these quadrants x, y, or both are positive coordinates

quadrants 2, 3, and 4 because in these quadrants x, y, or both are negative coordinates

Answers

Answer:

The correct option is C which is it is the third coordinate because p and q both are negative.

Explanation:

If we look at the four quadrants of coordinate plane:

Quadrant 1: If x and y both are positive.Quadrant 2: If x is negative and y is positive.Quadrant 3: If both x and y are negative.Quadrant 4: If x is positive and y is negative.

So according to given data, x is less than 0 means xis negative. y is also negative because y is less than 0.

So both x and y are negative and it is in quadrant 3.

Answer:

Quadrant II, because the slope is positive and the y intercept is negative

Explanation:

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Answers

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

Explanation:

The given equation is as follows.

       [tex]PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^{-}[/tex]

                          s       2s

It is given that,

       [tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}[/tex]

Let the solubility of given ions be "s".

Since, KBr on dissociation will given bromine ions.

Hence,  [tex]K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}[/tex]

        [tex]6.60 \times 10^{-6}[/tex]  = [tex]s \times (2s)^{2}[/tex]

                        = [tex]1.18 \times 10^{-2}[/tex] M

Therefore, solubility of [tex][PbBr_{2}][/tex] is [tex]1.18 \times 10^{-2}[/tex] M  in KBr.

Now, we will calculate the molar solubility of [tex]PbBr_{2}[/tex] in 0.5 M KBr solution as follows.

           [tex]K_{sp} = (s) \times (2s + 0.5)^{2}[/tex]

  [tex]6.60 \times 10^{-6}[/tex]  = [tex]4s^{3} + 0.25s + 2s^{2}[/tex]

                         s = [tex]2.63 \times 10^{-5}[/tex]

Thus, we can conclude that molar solubility of [tex][PbBr_{2}][/tex] in 0.500 m KBr solution is [tex]2.63 \times 10^{-5}[/tex].

Atomic weight of gold is 197.2 ami
Calculate no of gram atoms in 7.5g of gold

Answers

the atomic weight is numerically equivalent to the molar mass.
therefore molar mass of gold is - 197.2 g/mol
avagadros number states that 1 mol of any substance is made of 6.022 x 10²³ units. These units could be atoms making up elements or molecules making up compounds. 
1 mol of gold is made of 6.022 x 10²³  atoms of gold
since 1 mol weighs 197.2 g 
therefore if 197.2 g of gold contains - 6.022 x 10²³  atoms of gold
then 7.5 g contains - 6.022 x 10²³  x 7.5/ 197.2 = 2.29 x 10²² atoms of gold 
answer is 2.29 x 10²² atoms of gold

To find the number of gram atoms in 7.5g of gold, divide the mass by the molar mass of gold, which gives approximately 0.03807 moles of gram atoms.

The student was asked to calculate the number of gram atoms in 7.5g of gold where the atomic weight of gold is given as 197.2 amu (atomic mass unit). We use the concept of molar mass to answer this question. The molar mass of gold (Au) is about 197.0g/mol, and we know that one mole of any substance contains Avogadro's number of atoms, which is 6.022 x 1023 atoms/mol.

To find the number of moles (or gram atoms) in 7.5g of gold, we use the formula:

Number of moles = Mass (g) / Molar mass (g/mol)

Substituting the values, we get:

Number of moles = 7.5g / 197.0g/mol ≈ 0.03807 mol

Therefore, there are about 0.03807-gram atoms in 7.5g of gold.

When preparing the diazonium salt, the solution is tested with potassium iodide-starch paper. a positive test is the immediate formation of a blue color. what is the ki-starch paper testing for ? explain?

Answers

KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which then will react with 4-nitroaniline to form a diazonium salt.

Final answer:

Potassium iodide-starch paper is used to detect the presence of nitrogen dioxide in the preparation of diazonium salts by revealing a blue-black coloration upon reacting with iodine produced from the reaction between NO2 and potassium iodide.

Explanation:

The potassium iodide-starch paper is used to test for the presence of nitrogen dioxide (NO2), which is a byproduct of the reaction used to prepare diazonium salts. During the diazo coupling reaction, excess nitrous acid can decompose and produce nitrogen dioxide. This gas can then react with the potassium iodide (KI) in the starch paper to produce iodine (I2), which subsequently forms a blue-black complex with starch. The immediate formation of the blue color on the potassium iodide-starch paper is a positive test indicating the presence of nitrogen dioxide. It's important to monitor this because the presence of NO2 suggests that the diazonium salt solution might be unsafe due to the potential release of toxic gases.

The iodine-starch test is a well-known reaction in which iodine (I2), produced by the oxidation of iodide ions by NO2, interacts with starch to produce a characteristic blue-black color. This test provides a quick and sensitive method for detecting the presence of iodine, which, in this context, indirectly indicates the generation of nitrogen dioxide in the reaction mixture.

Find the empirical and molecular formula for a molar mass of 60.10g/mol; 39.97% carbon 13.41% hydrogen: 46.62% nitrogen

Answers

Assuming we have 100g, this means that

39.97g Carbon * 1 mol / 12 g = 3.33 mol Carbon
13.41g Hydrogen * 1 mol/1 g = 13.41 mol Hydrogen
46.62g Nitrogen * 1 mol / 14 g = 3.33 mol Nitrogen
Dividing everything by 3.33, we get

1 mol Carbon, 4.03 mol Hydrogen, 1 mol Nitrogen.

Empirical formula is CH4N

The mass of the empirical formula is
12 + 4 + 14 = 30

Since the molar mass is double, we multiply all our subscripts

The molecular formula is C2H8N2

The answers to this question are an empirical formula of CH4N and a molecular formula of C2H8N2 .

13) Given the redox reaction: Cr3+ + Al Cr + Al3+ As the reaction takes place, there is a transfer of A) electrons from Al to Cr3+ B) electrons from Cr3+ to Al C) protons from Al to Cr3+ D) protons from Cr3+ to Al

Answers

Cr³⁺+ Al⁰---> Cr ⁰+ Al³⁺
Cr³⁺+ 3e⁻---> Cr ⁰ 
Cr³⁺ takes electrons to become neutral atom.
 Al⁰---> Al³⁺ + 3e⁻
Al⁰ loses electrons to become ion.

Electrons transferred from  Al to Cr³⁺.
Correct answer is  
A) electrons from Al to Cr³⁺.

Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.030 m in naclo at 25 °c?

Answers

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,030 M.
[ClO⁻] = 0,03 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻] = [HClO] = x.
2,5·10⁻⁷ = x² / (0,03 M -x).
Solve quadratic equation: x = [OH⁻] = 0.0000893 M.
pOH = -log(0.0000893 M) = 4.05.
pH = 14 - 4.05.
pH = 9.95.

The main interactions between molecules of hydrogen chloride are examples of

Answers

Answer:
            The main interactions between molecules of hydrogen chloride are examples of Dipole-Dipole Interactions.

Explanation:
                   The bond between Hydrogen and Chlorine is Polar Covalent Bond as the electronegativity difference between these two elements is 0.96 which is greater than 0.4. Chlorine being more electronegative attracts the electrons from Hydrogen making the Hydrogen partial positive and itself partial negative. The two poles on HCl makes it a dipole. Now, one HCl (a dipole) interacts with another HCl (another Dipole) through their opposite charges and creates Dipole-Dipole Interaction. The picture is as below, the green dashed lines are interactions,

Final answer:

The main interactions between molecules of hydrogen chloride are examples of dipole-dipole interactions, specifically hydrogen bonding.

Explanation:

Hydrogen chloride (HCl) is a polar molecule, which means it has a partially positive end (hydrogen) and a partially negative end (chlorine). The main interactions between molecules of hydrogen chloride are examples of dipole-dipole interactions, specifically hydrogen bonding. These interactions occur when the partially positive hydrogen atom in one molecule is attracted to the partially negative end of another molecule, creating a bond between them.

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What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Answers

Answer:

19.4

Explanation:

Answer:

19.4

Explanation:

Write a balanced nuclear equation for the beta decay of thallium-210.

Answers

Final answer:

The beta decay of thallium-210 results in the conversion of the thallium nucleus into a lead-210 nucleus, with the emission of a beta particle and an antineutrino.

Explanation:

The beta decay of thallium-210 involves the transformation of a neutron into a proton, with the emission of an electron (beta particle) and an antineutrino. The balanced nuclear equation for this process can be written as follows:




^{210}_{81}Tl →
^{210}_{82}Pb +
^-1_{0}e (or β^-) + ar{ν}_e



Here, thallium-210 (Tl-210) undergoes beta decay to become lead-210 (Pb-210), with the emission of a beta particle (β^-) and an antineutrino (ν_e).

Calculate the volume in liters of a ×4.3910−5/moll mercury(ii) iodide solution that contains 500.mg of mercury(ii) iodide hgi2 . round your answer to 3 significant digits.

Answers

the molarity of the solution is 4.39 x 10⁻⁵ mol/L
molarity is the number of moles of solute in 1 L of solution 
in this instance there are 4.39 x 10⁻⁵ mol of HgI₂ in 1 L solution 
the mass of HgI₂ in 1 L - 4.39 x 10⁻⁵ mol x 454.4 g/mol = 19.9 mg
if 19.9 mg are in 1 L 
then 500.0 mg are in - 500.0 mg / 19.9 mg/L = 25.1 L
therefore volume is 25.1 L

Final answer:

To calculate the volume of the mercury(II) iodide solution in liters, convert the mass of the solute to moles and use the Molarity equation. The volume is approximately 25.1 L.

Explanation:

To calculate the volume in liters of the mercury(II) iodide solution, we need to convert the mass of the solute (HgI2) to moles, using the molar mass of HgI2. Then, we can use the equation Molarity = moles of solute / volume of solution in liters to find the volume.

First, calculate the moles of HgI2:

Moles = mass / molar mass = 500 mg / (454.39 g/mol) = 1.101 x 10^-3 mol

Next, rearrange the equation to solve for volume:

Volume = moles of solute / Molarity = 1.101 x 10^-3 mol / (4.3910^-5 M) = 25.07 L

Rounding to three significant digits, the volume of the solution is approximately 25.1 L.

how many grams of salt are in 5 L of a solution with a concentration of 3 g/L

Answers

Three grams per liter times five liters tells us that there are fifteen grams of salt in this five liter solution.

Describe what changes occur during electron capture. describe what changes occur during electron capture. the mass number and atomic number increases. the mass number and atomic number decreases. the mass number is unchanged and the atomic number decreases. the mass number and atomic number do not change. the mass number is unchanged and the atomic number increases.

Answers

The mass number is unchanged and the atomic number decreases.
Final answer:

Electron capture is a nuclear process where an inner shell electron combines with a proton, creating a neutron and resulting in a decrease in atomic number while keeping the mass number unchanged. The process generates energy often in the form of an X-ray. An example of this is seen in potassium-40, wherein electron capture transforms it into a different nuclide.

Explanation:

Electron capture is a process during which an inner shell electron combines with a proton in the nucleus and transforms into a neutron. This results in the creation of a vacancy within the atom, which is then filled by an electron from an outer shell. This electron, as it falls into the vacancy, releases energy, often in the form of an X-ray.

One significant outcome of electron capture is that the atomic number of the atom decreases by one, while the mass number remains unchanged. This happens because the process effectively converts a proton into a neutron within the atomic nucleus. Thus, the atom moves closer to the band of stability, improving the neutron to proton (n:p) ratio.

For instance, in the case of potassium-40, the atom undergoes electron capture, transforming it into a different nuclide with an atomic number one less than the original, and an unchanged mass number.

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An industrial chemist introduces 8.1 atm h2 and 8.1 atm co2 into a 1.00-l container at 25.0°c and then raises the temperature to 700.0°c, at which keq = 0.534: h2(g) + co2(g) ⇔ h2o(g) + co(g) how many grams of h2 are present after equilibrium is established?

Answers

i reccomend searching it up

Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. CH3OH + O2 → CH2O + H2O

Answers

If the number of different atoms of elements on the reactant side is equal to that present on the product side than the reaction is said to be chemically balanced. 

In above case, following is the balanced reaction.

2 CH3OH + O2 → 2 CH2O + 2 H2O

Answer:

look at the screenshot below

Explanation:

what is the most susceptible to damage from ionizing radiation. sorft tissue, paper, wood, lead

Answers

Tissues do the most dammage to susceptible radiation
Final answer:

Ionizing radiation is most damaging to soft tissue among the materials listed (soft tissue, paper, wood, lead) due to its ability to break chemical bonds and cause cell malfunctions. However, the impact largely depends on the type of radiation and the characteristics of the material.

Explanation:

The effects of ionizing radiation on materials largely depend on the type of radiation and the characteristics of the material. Ionizing radiation is harmful as it can ionize molecules or break chemical bonds, causing malfunctions in cell processes. This can lead to somatic or genetic damage, particularly in rapidly reproducing cells.

In terms of the materials listed, soft tissue is the most susceptible to damage from ionizing radiation. Materials like paper, wood, and lead have different degrees of resistance to ionizing radiation. Paper and wood can block alpha and beta particles, low-energy forms of ionizing radiation, while metal or lead can stop gamma radiation, a high-energy form of ionizing radiation, more effectively.

However, it's crucial to note that ionizing radiation has its greatest effect on cells that rapidly reproduce. Therefore, in living organisms, areas with rapidly dividing cells, such as the skin or the lining of the stomach or intestines, are especially susceptible to damage from ionizing radiation.

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Which part of an atom is most directly involved in chemical bonding?

Answers

The outer or valence electrons

Chemical bonding is defined as the attraction between elements, ions or molecules which results in the formation of compounds. The bonding takes place due to sharing of electrons as in covalent bond or by the electrostatic force of attraction between positive charge ion (cation) and negative charge ion (anion) as in ionic bond.

Now, electrons are the negative charged particles of an atom and it is found in clouds which is surrounded by the nucleus of an atom. Electrons play a vital role in chemical bonding. In both type of bonding i.e. ionic bonding in which electrons are transferred from on atom to other atom and covalent bonding which results due to sharing of electrons between two atoms.

Thus, electron is a part of an atom is most directly involved in chemical bonding.





If you add 500 kj of heat to 500 g of water at 50.0°c, how much water is left in the container? the latent heat of vaporization of water is 2.26 ×103 j/g and its specific heat is is 4.190 j/(g∙k)

Answers

First, we will get the heat required to raise the 500 g of water from 50°C to 100°C = m*C*ΔT 

when m is the mass = 500 g 

and C is the specific heat capacity of water = 4.19

ΔT change in temperature = 50 °C

by substitution:

q = 0.5 Kg * 4.19 * 50°C

   =104.75 KJ

∴ heat left to boil the water= 500KJ - 104.75KJ = 395 KJ

the heat required to boil water from 100°C to steam = mass *latent heat of vaporization

  395KJ     = M * 2.26 x 10^3

 Mass = 0.17Kg = 170 g

∴ water remain= 500 g - 170 = 330 g


Write the electron configuration for ni2+. use the buttons at the top of the tool to add orbitals in order of orbital filling, starting at the bottom with the lowest-energy orbitals. click within an orbital to add electrons. g.com

Answers

The atomic number of Nickle is 28, so its electronic configuration in ground state is,

Ni  =  28 e⁻  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁸

So, when Nickle looses two electrons it will attain +2 charge, remaining with 26 electrons as,

Ni⁺²  =  26 e⁻  =  1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸

Ni looses electrons from 4s shell instead instead of 3d shell because the 4s shell electrons are at larger distance as compared to 3d, hence, felting less nuclear charge and easily lost.


The electron configuration for Ni²⁺ is [Ar] 3d⁸.

Nickel (Ni) has an atomic number of 28, which means that it has 28 protons in its nucleus. It also has 28 electrons, which are arranged in shells around the nucleus.

The electron configuration for a neutral nickel atom is:

[Ar] 3d⁸ 4s²

The argon (Ar) core is filled with 18 electrons, and the remaining 10 electrons are arranged in the 3d and 4s orbitals.

When nickel loses two electrons to form Ni²⁺, the two electrons that are lost are the two 4s electrons. This leaves the 8 3d electrons behind, which gives Ni²⁺ the electron configuration shown above.

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a part of the periodic table is shown below. which of the following elements is less reactive then the others?

Answers

The one element that is less reactive than the others is Selenium.

Explanation:

its the third one down :)

Notice that " s o 4 " appears in two different places in this chemical equation. s o 2− 4 is a polyatomic ion called "sulfate." what number should be placed in front of cas o 4 to give the same total number of sulfate ions on each side of the equation? ?cas o 4 +alc l 3 →cac l 2 +a l 2 (s o 4 ) 3 express your answer numerically as an integer.

Answers

a "3" should but put in front of
"cas o 4 "

What mass of hcl gas must be added to 1.00 l of a buffer solution that contains [aceticacid]=2.0m and [acetate]=1.0m in order to produce a solution with ph = 4.11?

Answers

pH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
 

To achieve a pH of 4.11 in the given buffer solution, approximately 15.71 grams of HCl gas need to be added.

To determine the mass of HCl gas needed to achieve the desired pH in the buffer solution, we can use the Henderson-Hasselbalch equation:

pH = pKₐ + log([A⁻] / [HA])

For acetic acid (CH₃COOH), the pKₐ is approximately 4.76. Given: [HA] = 2.0 M (acetic acid), [A-] = 1.0 M (acetate), and the target pH is 4.11, we can set up the equation:

4.11 = 4.76 + log([A⁻] / [HA])

Rearranging to solve for the ratio [A⁻]/[HA]:

4.11 - 4.76 = log([A⁻] / [HA])
-0.65 = log([A⁻] / [HA])
[A⁻] / [HA] =  [tex]10^-^0^.^6^3[/tex] ≈ 0.234

Next, let x be the number of moles of HCl gas added. HCl will convert acetate (A⁻) to acetic acid (HA):

(1.0 - X) / (2.0 + X) = 0.234

Solving for X: multiply both sides by (2.0 + X):

1.0 - x = 0.234 x (2.0 + X)
1.0 - x = 0.468 + 0.234X
1.0 - 0.448 = 0.224X + X
0.532 = 1.234X
X ≈ 0.431

Thus, moles of HCl needed are approximately 0.431 moles. Using the molar mass of HCl (36.46 g/mol), we can calculate the mass:

Mass of HCl = 0.431 mol x 36.46 g/mol ≈ 15.71g

Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen. determine the empirical formula for a compound that is 70.79 carbon, 8.91 hydrogen, 4.59 nitrogen, and 15.72 oxygen. c18h27no2 c18h27no3 c17h27no3 c17h26no3

Answers

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 70.79% = 70,79 g 
H: 8.91% = 8.91 g
N: 4.59% = 4.59 g
O: 15.72% = 15.72 g

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

[tex]C: \dfrac{70.79\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 5.89\:mol[/tex]

[tex]H: \dfrac{8.91\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 8.91\:mol[/tex]

[tex]N: \dfrac{4.59\:\diagup\!\!\!\!\!g}{14\:\diagup\!\!\!\!\!g/mol} \approx 0.328\:mol[/tex]

[tex]O: \dfrac{15.72\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 0.9825\:mol[/tex]

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

[tex]C: \dfrac{5.89}{0.328}\to\:\:\boxed{C\approx 18}[/tex]

[tex]H: \dfrac{8.91}{0.328}\to\:\:\boxed{H\approx 27}[/tex]

[tex]N: \dfrac{0.328}{0.328}\to\:\:\boxed{N\approx 1}[/tex]

[tex]O: \dfrac{0.9825}{0.328}\to\:\:\boxed{O\approx 3}[/tex]

Thus, the minimum or empirical formula found for the compound will be:

[tex]\boxed{\boxed{C_{18}H_{27}N_1O_3\:\:or\:\:C_{18}H_{27}NO_3}}\end{array}}\qquad\checkmark[/tex]

I hope this helps. =)

The empirical formula of the compound is C₁₈H₂₇NO₃

From the question given above, the following data were obtained:

Carbon (C) = 70.79%

Hydrogen (H) = 8.91%

Nitrogen (N) = 4.59%

Oxygen (O) = 15.72%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 70.79%

H = 8.91%

N = 4.59%

O = 15.72%

Divide by their molar mass

C = 70.79 / 12 = 5.899

H = 8.91 / 1 = 8.91

N = 4.59 / 14 = 0.328

O = 15.72 / 16 = 0.9825

Divide by the smallest

C = 5.899 / 0.328 = 18

H = 8.91 / 0.328 = 27

N = 0.328 / 0.328 = 1

O = 0.9825 / 0.328 = 3

Therefore, the empirical formula of the compound is C₁₈H₂₇NO₃

Learn more: https://brainly.com/question/24818135

In the simulation, open the micro mode, then select solutions indicated below from the dropdown above the beaker in the simulation. the beaker will fill up to the 0.50 l mark with the solution at 25 ∘c. arrange the acids in increasing order of acidity. rank from lowest to highest. to rank items as equivalent, overlap th

Answers

Milk (lowest acidity) < Coffee < Orange juice < Soda pops < vomit < Battery acid (highest acidity)

explanation :

pH values of all :

battery acid pH = 1.0

vomit pH = 2.0

soda pop pH = 2.5

orange juice pH = 3.5

coffee pH = 5.0

milik pH = 6.5

pH value is lesser acidity is more . high pH indicate lesser acidic nature

Final answer:

Explanation of the increasing freezing points of aqueous solutions including acetic acid, NaCl, sucrose, and CaCl₂.

Explanation:

Arrange the aqueous solutions in increasing freezing points:

0.3 m acetic acid

0.2 m NaCl

0.2 m sucrose

0.1 m CaCl₂

A certain half-reaction has a standard reduction potential . an engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least of electrical power. the cell will operate under standard conditions. note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.

Answers

An electrochemical cell are of two types 1) Chemical 2) concentration

For the present application, since the engineer requires this half-reaction to happen at the anode of the cell, chemical cell must be preferred. Further since, the engineer required, the emf of cell equal to half cell reaction, the counter electrode/electrolyte that can be used is Pt/H2/H+. It must be noted that std reduction potential of this electrode is 0v. It must also be noted that, engineer requires this half-reaction to happen at the anode of the cell. Oxidation occurs at anode compartment. Hence, the remaining half cell should comprise of chemical, whose std. oxidation potential is higher than that of Pt/H2/H+ or std.  reduction potential is less that that of Pt/H2/H+. For, this purpose, those chemicals must be chosen, which are placed below H2/H+ in electrochemical series. If this material is of the form X/X+, the cell representation of the overall electrochemical cell is
X/X+//H+/H2/Pt
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