Answers
In above case, following is the balanced reaction.
2 CH3OH + O2 → 2 CH2O + 2 H2O
Answer:
look at the screenshot below
Explanation:
Related Questions
Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.
Answers
rate of effusion of gas1 : rate of effusion of gas = 16 : 1.
rate of effusion of gas1 = 1/√M(gas1).
rate of effusion of gas2 = 1/√M(gas2).
rate of effusion of gas1 = rate of effusion of gas2 · 16.
1/√M(gas1) = 1/√M(gas2) · 16 /².
1/M(gas1) = 1/M(gas2) · 256.
M(gas1) · 256 = M(gas2).
The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution
Answers
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].
elaborate on the difference in natural occurrences between nuclear fission and nuclear fusion reactions. A) Neither nuclear fission nor nuclear fusion reactions naturally occur on earth or outer space. B) Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars. C) Both nuclear fission and nuclear fusion reactions readily naturally occur in the pressure of the earth's core. D) Nuclear fusion reactions do not naturally occur while nuclear fission reactions occur in the volcanoes.
Answers
Option B: Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
Nuclear fission reaction is defined as formation of two or more atoms by splitting of large atoms. On the other hand, nuclear fusion reaction is formation of large atom from small atoms.
The splitting of large atoms into small atoms (nuclear fission) generally does not occur naturally because it requires high speed neutrons and critical mass of the substance undergoing fission.
Nuclear fusion reaction occurs in stars such as sun because it requires extremely high energy to bring two more proton closer to each other by overcoming electronic repulsion.
Therefore, nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
What are the boiling point and freezing point of a 3.47 m solution of naphthalene in benzene? (the boiling point and freezing point of benzene are 80.1°c and 5.5°c respectively. the boiling point elevation constant for benzene is 2.53°c/m, and the freezing point depression constant for benzene is 5.12°c/m.)?
Answers
ΔTb = Kb X m
where Kb = boiling point elevation constant = 2.53°c/m (for benzene)
m = molality of solution = 3.47 m (given)
∴ ΔTb = 2.53 X 3.47
= 8.779 oC
But, boiling point of benzene = 80.1 oC
∴ Boiling point of solution = 88.879 oC
Now, Depression in freezing point = ΔTf = Kf X m
where, Kf = freezing point depression constant = 5.12°c/m (for benzene)
∴ ΔTf = 5.12 X 3.47
= 17.766 oC
But freezing point of benzene = 5.5 oC
∴ Freezing point of solution = -12.266 oC
The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.
The correct answer is -12.266.
What is the boiling point?The boiling point of a liquid varies according to the applied pressure; the normal boiling point is the temperature at which the vapor pressure is equal to the standard sea-level atmospheric pressure (760 mm [29.92 inches] of mercury). At sea level, water boils at 100° C.Elevation in boiling point is mathematically expressed as
[tex]Tb = Kb * m[/tex]
Where
Kb = boiling point elevation constant = 2.53°c/m (for benzene)m = molality of solution = 3.47 m (given)ΔTb[tex]= 2.53 * 3.47 = 8.779 oC[/tex]
But, the boiling point of benzene = 80.1 oC
Boiling point of solution = 88.879 oC
Now, Depression in freezing point = ΔTf[tex]= Kf * m[/tex]
where,
Kf = freezing point depression constant = 5.12°c/m (for benzene)ΔTf =[tex]5.12 X*3.47 = 17.766 oC[/tex]
But the freezing point of benzene = 5.5 oC.
Freezing point of solution = -12.266 oC.
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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⁺+ 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⁺.
how many grams of salt are in 5 L of a solution with a concentration of 3 g/L
Answers
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] = 6.3×10−4m , and [hno2] = 0.21m ? be sure to express your answer in units of kj in the box below. answers without units will not be given credit.
Answers
By applying the modified Gibbs Free Energy formula with given values for equilibrium constants, atom concentrations, and other parameters, we find ΔG for nitrous acid at 25°C to be around 27.94 KJ/mol.
Explanation:In chemistry, the Gibbs free energy (ΔG) is calculated using the equation ΔG = -RTlnK, where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. However, since we're given Ka, the equation must be adapted. Therefore, we use ΔG = -RTlnKa + RTlnQ, where Q is the reaction quotient given by [NO₂⁻][H⁺] / [HNO₂].
Inserting the given values, such as Ka = 4.5×10⁻⁴, R (in appropriate units) as 0.0083145 KJ/(mol.K), T as 298.15K (25°C in Kelvin), and Q = ([NO₂⁻][H⁺]) / [HNO₂] = (6.3×10⁻⁴ × 5.9×10−2) / 0.21, we can now solve for ΔG. Doing the math, we find that ΔG ≈ 27.94 KJ/mol.
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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
Answers
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.
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
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.
Write the balanced nuclear equation for β− decay of sodium−26. include both the mass numbers and the atomic numbers with each nuclear symbol. use the sup-subscript button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says -select−.
Answers
In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and in beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.
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
Answers
pH₁ = 7.
pH = -log[H₃O⁺].
[H₃O⁺]₁ = 10∧(-pH).
[H₃O⁺]₁ = 10⁻⁷ M.
[H₃O⁺]₂ = 10⁻⁷ M ·10³.
[H₃O⁺]₂ = 10⁻⁴ M.
pH₂ = -log[H₃O⁺]₂.
pH₂ = -log(10⁻⁴ M).
pH₂ = 4.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
When pH is less than seven (pH<7), solution is acidic..
When is equal seven (pH = 7), solution is neutral.
When pH is greater than seven (pH > 7), solution is basic.
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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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
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 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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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
"cas o 4 "
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate the freezing point of this solution. [kfp(benzene) = 5.12°c/m, kbp(benzene) = 2.53°c/m] (the boiling point and freezing point of benzene are 80.1°c and 5.5°c, respectively).
Answers
PV = nRT
when P is the pressure = 748 mmHg/760 = 0.984 atm
V is the volume = 4 L
R is ideal gas constant = 0.0821
T is the temperature in Kelvin = 300 K
∴ n = 0.984atm*4L/0.0821*300
= 0.1598 moles
when the concentration = moles * (1000g / mass)
= 0.1598 * (1000g / 58 g )
= 2.755 M
when the freezing point = 5.5 °C
and Kf = - 5.12 °C/m
∴ the freezing point for the solution = 5.5 °C + (Kf*m)
= 5.5 °C - (5.12°C/m * 2.755m)
= -8.6 °C
The new freezing point of the solution is -8.63 °C.
Using the ideal gas law, PV = nRT, we need to convert 4.00 L of gas at 27°C (which is 300 K) and 748 mmHg to moles of gas.
P = 748 mmHg / 760 mmHg atm = 0.984 atm
[tex]n = \frac{PV}{RT} = \frac{0.984 \, \text{atm} \times 4.00 \, \text{L}}{0.0821 \, \text{L atm / K mol} \times 300 \, \text{K}} = 0.160 \, \text{mol}[/tex]
Calculate molality:
Molality (m) = moles of solute / kg of solvent = 0.160 mol / 0.0580 kg = 2.759 m
Calculate freezing point depression (ΔTf):
ΔTf = i * Kf * m
For non-electrolytes, i = 1.
ΔTf = 1 * 5.12 °C/m * 2.759 m = 14.13 °C
Determine the new freezing point:
The freezing point of pure benzene is 5.5 °C.
New freezing point = 5.5 °C - 14.13 °C = -8.63 °C
Thus, the freezing point of the solution is -8.63 °C.
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.
Answers
PV = nRT
where
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol
number of moles of NH₃ is 0.273 mol
molar mass of NH₃ - 17.0 g/mol
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g
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.
Answers
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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Equilibrium is established in a reversible reaction when:
a) the [product] = [reactants]
b) rate of reaction of products = rate of reaction of reactants
c) all the reactants dissolve or dissociate
d) when product are no longer produced
Answers
Equilibrium in a reversible reaction is established when the rate of the forward reaction (reactants transforming into products) equals the rate of the backward reaction (products reconverting into reactants). This indicates that the amounts of products and reactants are no longer changing over time.
Explanation:In a reversible reaction, equilibrium is established when option (b) is correct: the rate of the forward reaction (reactants transforming into products) becomes equal to the rate of the backward (or reverse) reaction (products converting back into reactants). This does not necessarily mean the concentrations of the products and reactants are equal. Rather, it means the amounts of products and reactants are no longer changing over time, demonstrating a state of dynamic equilibrium. The equilibrium can shift depending on external factors and conditions such as temperature, pressure, or concentration changes, which is described by Le Chatelier's Principle.
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N aqueous solution of sodium hydroxide is standardized by titration with a 0.154 m solution of hydrochloric acid. if 17.5 ml of base are required to neutralize 17.6 ml of the acid, what is the molarity of the sodium hydroxide solution?
Answers
NaOH + HCl --> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
Number of HCl moles reacted - 0.154 mol/L x 0.0176 L = 0.00271 mol
the number of NaOH moles reacted = number of HCl moles reacted
number of NaOH moles reacted - 0.00271 mol
number of NaOH moles in 17.5 mL - 0.00271 mol
therefore NaOH moles in 1000 mL - 0.00271 mol / 17.5 mL x 1000 mL
molarity of NaOH - 0.155 M
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.
Which would be most useful if you were lost in the Sahara desert? A. Landsat satellite B. OSTM/Jason-2 satellite C. Global Positioning System D. topographic map of Africa
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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
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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When hydrochloric acid is added to sodium hydrogen carbonate?
Answers
Hope this helps!
Which part of an atom is most directly involved in chemical bonding?
Answers
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.
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.
Answers
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH₄) followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H₂CrO₄).
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
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.
Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
Answers
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive and the cathode is negative.
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-
Answers
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
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
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
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
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