Final answer:
To find the molality of a KBr solution, divide the mass of KBr (2.10 g) by its molar mass (119.0 g/mole) to get moles, and then divide by the mass of the solvent in kilograms (0.897 kg of water).
Explanation:
The concentration of KBr in a solution prepared by dissolving 2.10 g of KBr in 897 g of water is calculated using the molality formula. Molality is defined as the number of moles of solute per kilogram of solvent. First, we need to find the number of moles of KBr by dividing the mass (2.10 g) by its molar mass (119.0 g/mole), which gives us the number of moles. Then, we divide the number of moles by the mass of the solvent (water) in kilograms to obtain the molality of the solution.
The calculation is as follows:
Number of moles of KBr = 2.10 g / 119.0 g/moleMolality (m) = Number of moles of KBr / 0.897 kg (mass of water)By performing these calculations, we can determine the molality of the KBr solution.
Phosphorus-32 has a half-life of 14.3 days. how many grams remain from a 10.0 gram sample after 30.0 days?
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]
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?
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.
a part of the periodic table is shown below. which of the following elements is less reactive then the others?
Explanation:
its the third one down :)
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)
What solution has the lowest pH?
1.0 M HNO2
1.0 M CH3COOH
1.0 M HCOOH
1.0 M HNO3
1.0 M HPO4-
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
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 massC = 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 smallestC = 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
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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?
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.
The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution
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].
When hydrochloric acid is added to sodium hydrogen carbonate?
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
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:
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?
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!!!!
blood: heterogeneous and suspension
salad dressing: heterogeneous and suspension
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
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?
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
how many grams of salt are in 5 L of a solution with a concentration of 3 g/L
What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy
Answer:
19.4
Explanation:
Answer:
19.4
Explanation:
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.
A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 15.0 ∘c is 2.15 atm . assuming ideal gas behavior, how many grams of ammonia are in the flask? express your answer to three significant figures and include the appropriate units.
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.
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.
A solution has [oh−] = 2.3×10−2 m . use the ion product constant of water kw=[h3o+][oh−] to find the [h3o+] of the solution.
Answer:
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
Explanation:
Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]
Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]
[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The ionic product of water is given as:
[tex]K_w=[H_3O^+][OH^-][/tex]
The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]
[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]
[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]
[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
The concentration of hydronium, ion [H₃O⁺] in the solution containing 2.3×10⁻² M concentration of hydroxide ion, [OH⁻] is 4.35×10⁻¹³ M
How to determine the concentration of hydronium, ion [H₃O⁺]?The following data were obtained from the question given above:
Concentration of hydroxide ion, [OH⁻] = 2.3×10⁻² MConcentration of hydronium, ion [H₃O⁺] = ?[H₃O⁺] × [OH⁻] = 10¯¹⁴
[H₃O⁺] × 2.3×10⁻² = 10¯¹⁴
Divide both side by 2.3×10⁻²
[H₃O⁺] = 10¯¹⁴ / 2.3×10⁻²
= 4.35×10⁻¹³ M
Thus, we can conclude that the concentration of hydronium ion, [H₃O⁺] in the solution is 4.35×10⁻¹³ M
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How many times greater is the rate of effusion of molecular fluorine than that of molecular bromine at the same temperature and pressure
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
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
Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
a. treatment of compound a with m−chloroperoxybenzoic acid (mcpba) gave compound b, which on reduction with lithium aluminum hydride gave compound
c. oxidation of compound c with chromic acid gave compound d (c9h14o). identify compounds a through d in this sequence, and do not indicate stereochemistry.
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.
Compared to a solution with a pH value of 7 a solution with a thousand times greater hydronium ion concentration has a pH value of what
The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.
The pH has been defined as the negative log of hydronium ion concentration in the solution. The pH has been expressed as:
[tex]\rm pH=-\;log\;[H_3O^+][/tex]
Computation for the pH of the solutionThe pH of the given solution has been 7. The hydronium ion concentration has been given as:
[tex]\rm pH=\;-log\;[H_3O^+]\\\\ 7=\;-\;log[H_3O^+]\\\\ H_3O^+=10^-^7\;M[/tex]
The concentration of the new solution has been 1000 times greater than the previous solution.
The previous solution has hydronium ion concentration of [tex]\rm 10^-^7\;M[/tex]. The concentration of new solution will be:
[tex]\rm New\;solution=10^-^7\;\times\;1000\;M\\ New\;solution=10^-^4\;M[/tex]
The concentration of the new solution has been [tex]\rm 10^-^4\;M[/tex]. The pH of the solution has been given as:
[tex]\rm pH=-log\;[H_3O^+]\\ pH=-log\;[10^-^4]\\ pH=4[/tex]
The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.
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Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. CH3OH + O2 → CH2O + H2O
Answer:
look at the screenshot below
Explanation:
The main interactions between molecules of hydrogen chloride are examples of
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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Which part of an atom is most directly involved in chemical bonding?
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.
A compound consisting of atoms of large atomic mass is more likely to require:
Answer:
A) a higher temperature to liquefy
BRAINLIST?
Explanation:
Answer:
The answer is
A: a higher temperature to liquefy
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.
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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