Answer:
Benzoic acid: 1.288g
Sodium benzoate: 124.48g
Explanation:
Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:
pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)
Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.
As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:
2.5L ₓ (0.35mol / L) = 0.875moles of buffer.
And you can write:
0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)
Replacing (2) in (1)
pH = pKa + log [C7H5O2⁻] / [HC7H5O2]
6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]
1.913 = log [C7H5O2⁻] / [HC7H5O2]
81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]
81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]
82.846 [HC7H5O2] = 0.875mol
[HC7H5O2] = 0.01056 molesAnd moles of the benzoate, [C7H5O2⁻]:
[C7H5O2⁻] = 0.875mol - 0.01056mol =
[C7H5O2⁻] = 0.8644molUsing molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:
Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g
Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g
When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver chloride and magnesium nitrate. When enough AgNO3 is added so that 34.3 g of MgCl2 react, what mass of the AgCl precipitate should form
Answer:
103.62 g of AgCl.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2
Step 2:
Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.
This is illustrated below:
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol
Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g
Thus, from the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Step 3:
Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.
The mass of AgCl produced from the reaction can be obtained as follow:
Form the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.
Therefore, 103.62 g of AgCl were produced from the reaction.
Stote 4 ways in which excesine alcohol conscuption is
harmful to humans
Answer:
An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.
Explanation:
Answer:
Excessive alcohol is harmful because you could get addicted.Alcohol can affect your nervous system.Your sugar levels will not be good.Parts of your body and organs will become inflamed.You can get a larger amount of muscle cramps.Also you will not be able to get enough vitamins in your body.Accidents that lead to deaths could occur.You would do crazy actions with things such as theft or breaking into a house which could get you fined or arrested.Too much alcohol can lead to high blood pressure, disease and even strokes.You can have birth defectsWith excessive alcohol you can get osteoporosis.You can also get your immune system weakened.Finally, alcohol can lead to cancer.Hope this helped,
Kavitha
Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Answer:
The correct answer is 2.75 grams of HCl.
Explanation:
The given balanced equation is:
CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)
Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams
The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,
= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61M1. (2 pts) How does the wavelength of a radio wave relate to its frequency? a.) The wavelength gets longer as the frequency increases b.) The wavelength gets shorter as the frequency increases c.) There is no relationship between wavelength and frequency d.) The wavelength depends on the bandwidth of the signal 2. (2 pts) List the following EMR in the order of increasing wavelength starting with the lowest: Infrared radiation Ultraviolet radiation X-rays Visible light 3. (3 pts) Green light has a wavelength of 5.0 x 102 nm. What is the energy, in joules, of ONE photon of green light? What is the energy, in joules of 1.0 mol of photons of green light?
Answer:
1. b.) The wavelength gets shorter as the frequency increases
2. X-rays < Ultraviolet radiation < Visible light < Infrared radiation
3. 2 × 10⁵ J
Explanation:
1. Wavelength vs frequency
fλ= c
f = c/λ
Thus, frequency and wavelength are inversely proportional.
The wavelength increases (gets longer) as the frequency decreases.
2. Order of increasing wavelength
X-rays < Ultraviolet radiation < Visible light < Infrared radiation
3. Energy of green light
(a) Energy of 1 photon
λ = 5 × 10² nm = 5 × 10² × 10⁻⁹ m = 5 × 10⁻⁷ m
fλ = c
f = c/λ = (2.998 × 10⁸ m·s⁻¹)/(5 × 10⁻⁷ m) = 6 × 10¹⁴ s⁻¹
E = hf = 6.626 × 10⁻³⁴ J·s × 6 × 10¹⁴ s⁻¹ = 4 × 10⁻¹⁹ J
(b) Energy of 1.0 mol of photons
[tex]\text{Energy} = \text{1.0 mol photons} \times \dfrac{6.022 \times 10^{23}\text{ photons }}{\text{1 mol photons }} \times \dfrac{4 \times 10^{-19}\text{ J}}{\text{1 photon }} = \mathbf{2 \times 10^{5}} \textbf{ J}\\\\\text{The energy of 1.0 mol of photons of green light is $\large \boxed{\mathbf{2 \times 10^{5}}\textbf{ J}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the wavelength of the light.
A mercury manometer is used to measure pressure in the container illustrated. Calculate the pressure exerted by the gas if atmospheric pressure is 751 torr and the distance labeled is 176 mm.
Answer:
Pressure exerted by the gas is 574.85 torr
Explanation:
Atmospheric pressure = 751 torr
but 1 torr = 1 mmHg
therefore,
atmospheric pressure = 751 mmHg
1 mmHg = 133.3 Pa
therefore,
atmospheric pressure = 751 x 133.3 = 100108.3 Pa
distance labeled (tube section with mercury) = 176 mm
the pressure within the tube will be
[tex]P_{tube}[/tex] = ρgh
where ρ is the density of mercury = 13600 kg/m^3
h is the labeled distance = 176 mm = 0.176 m
g is acceleration due to gravity = 9.81 m/s^2
[tex]P_{tube}[/tex] = 13600 x 9.81 x 0.176 = 23481.216 Pa
The general equation for the pressure in the manometer will be
[tex]P_{atm}[/tex] = [tex]P_{tube}[/tex] + [tex]P_{gas}[/tex]
where [tex]P_{atm}[/tex] is the atmospheric pressure
[tex]P_{tube}[/tex] is the pressure within the tube with mercury
[tex]P_{gas}[/tex] is the pressure of the gas
substituting, we have
100108.3 = 23481.216 + [tex]P_{gas}[/tex]
[tex]P_{gas}[/tex] = 100108.3 - 23481.216 = 76627.1 Pa
This pressure can be stated in mmHg as
76627.1 /133.3 = 574.85 mmHg
and also equal to 574.85 torr
Which of the following functional groups is formed from the condensation of carboxylic acids???
a. acid anhydride
b. acid halide
c. amide
d. ester
e. ether
Answer:
a
Explanation:
its made up of carbon and hydrogen
An oxide has a chemical formula with the form X2O3. Which group is element X more likely to be a member of? Select the correct answer below: group 12 group 13 group 14 group 2
Answer:
Group 13
Explanation:
You know X has 3 valence electrons, as oxygen has a subscript of 3. This means X has an ionic charge of +3. Group 13 consists mainly of metalloids but it also has metals such as aluminum, which has a +3 charge. If you use aluminum as an example, you know that when combined with oxygen, it forms Al2O3. Group 12 has transition metals that don't have +3 ionic charges, group 14 has metalloids, metals that don't have ionic charges of +3, and nonmetals, and group 2 has metals with ionic charges of +2. Group 13 is the answer.
Write a balanced equation for: capture of an electron by cadmium-104
Answer:
104 48 Cd + 0 -1 e ---------> 104 47 Ag
Explanation:
In the process of electron capture, the nucleus captures an electron and thus converts a proton into a neutron with the emission of a neutrino. This process increases the Neutron/Proton ratio, the captured electron is usually from the K shell. An electron from a higher energy level now drops down to fill the vacancy in the K shell and characteristic X-ray is emitted. This process usually occurs where the Neutron/proton ratio is very low and the nucleus has insufficient energy to undergo positron emission.
For 104 48 Cd, the balanced equation for K electron capture is;
104 48 Cd + 0 -1 e ---------> 104 47 Ag
A diode has IS = 10−17 A and n = 1.05. (a) What is the diode voltage if the diode current is 70 μA? (b) What is the diode current for VD = 0.1 mV?
Answer:
(a) The diode voltage, [tex]V_D =[/tex] 0.776 V
(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A
Explanation:
Given;
saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A
nonideality factor, n = 1.05
(a) the diode voltage
Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A
Diode voltage is calculated as;
[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]
Where;
[tex]V_T[/tex] is thermal voltage at 25°C = 0.025
[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]
b) the diode current for VD = 0.1 mV
[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]
What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate
The question is incomplete, the complete question is;
What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate? more than one answer is possible
A) The release of a water molecule that acts as an electrophile in the reaction with methyl benzoate.
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
C)The formation of bisulfate (hydrogen sulfate), which acts as an electrophile in the reaction with methylbenzoate.
D)The release of a water molecule that acts as a nucleophile in the reaction with methyl benzoate.
Answer:
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
Explanation:
The benzene ring is known to be stable hence it can only undergo a substitution reaction with the aromatic ring still intact. When the substitution reaction involves an electrophile we refer to the process as electrophillic aromatic substitution. Electrophilic aromatic substitution is a useful synthetic route for many organic compounds.
In the electrophilic substitution of methyl benzoate using the 1:1 volume ratio mixture of H2SO4/HNO3, the nitronium ion (NO2+) is the electrophile generated in the test tube. It is this NO2+ that now reacts with the methyl benzoate to yield the reaction product.
Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].
One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.
Explanation:
Formula for each of the speciesStart by finding the formula for each of the compound.
Both chlorine [tex]\rm Cl[/tex] and bromine [tex]\rm Br[/tex] are group 17 elements (halogens.) Each On the other hand, potassium [tex]\rm K[/tex] is a group 1 element (alkaline metal.) EachTherefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between
The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].
Balanced equation for the reactionWrite down the equation using these chemical formulas.
[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.
Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:
[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.
There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.
Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.
In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.
One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].
Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.
One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].
Hence:
[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:
[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:
[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].
If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final velocity is 8 m/s, what was the contact time? t= s
Answer:
t = 2 seconds
Explanation:
It is given that,
Mass of a boy, m = 50 kg
Initial speed of boy, u = 0
Final speed of boy, v = 8 m/s
Force exerting by another boy, F = 200 N
Let t is the time of contact. The force acting on an object is given by :
F = ma
a is acceleration
So,
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times 8}{200}\\\\t=2\ s[/tex]
So, the contact time is 2 seconds.
Answer:
t=2 s
Explanation:
Convert cm/S^2 to km/h^
2
Answer:
The answer to this question is 0.072km/h
A student mixes 2.83 mL of benzoyl chloride with excess 15 M NH4OH to produce 1.95 g of benzamide. What is the percent yield of this student's experiment
Answer:
Explanation:
The reaction of benzoyl chloride with NH₄OH to produce benzamide is:
Benzoyl chloride + ammonia → Benzamide + NH₄Cl
Molar mass of benzoyl chloride: 140.57 g/mol. Density 1.21g/mL
Molar mass benzamide: 121.14g/mol.
To know percent yield you must know the theoretical yield of the reaction (How many grams are produced assuming a yield of 100%). Percent yield will be (Actual yield / Theoretical Yield) ₓ 100
Moles of 2.83mL of benzoyl chloride are:
2.83mL ₓ (1.21g/mL) ₓ (1mol / 140.57g) = 0.02436 moles of benzoyl chloride.
As 1 mole of benzoyl chloride produce 1 mole of benzamide (Theoretical yield), theoretical moles of benzamide produced are 0.02436. In mass:
0.02436 moles ₓ (121.14g / mol) = 2.95g of benzoyl chloride
As there are produced just 1.95, percent yield is:
(1.95g / 2.95g) ₓ 100 = 66.1%
Compare strontium with rubidium in terms of the following properties:
a. Atomic radius, number of valence electrons, ionization energy.
b. Strontium is smaller than rubidium.
c. Rubidium is smaller than strontium.
d. Strontium has more valence electrons.
e. Rubidium has more valence electrons.
f. Strontium has a larger ionization energy.
g. Rubidium has a larger ionization energy.
Answer:
Strontium is smaller
Strontium has the higher ionization energy
Strontium has more valence electrons
Explanation:
It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table
While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)
Since they are in the same period, periodic trends would be useful in evaluating their properties
In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size
Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius
Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has
In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case
A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.
Answer:
[tex]Kp=0.0386[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]
For which the equilibrium expression is:
[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]
Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:
[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]
And the total pressure:
[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]
Yet it is 748 torr, for which the extent is:
[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]
Therefore, Kp turns out:
[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]
Best regards.
Zn + 2 HCl --> H2 + ZnCl2 If 1.70 g of Zn are reacted, how many grams of ZnCl2 can be created? Show work and process and I will give brainliest
Explanation:
first find the the number of moles of of zinc .
as the number of moles of zinc and ZnCl2 is same we can calculate the mass of ZnCl2.
1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents. What are these two solvents? (2 pts)
Answer:
Water
Explanation:
Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.
The solvent for KHT is water.
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.) Express your answer to two significant figures and include the appropriate units. nothing nothing
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 32.3 mg produced 87.7 mg of CO2 and 18.0 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Answer:
Empirical formula: C₅H₅O
Molecular formula: C₁₀H₁₀O₂
Explanation:
When a compound containing C, H and O elements is combusted, the general reaction is:
CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O
Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.
Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =
2.0x10⁻³ moles of CO₂ = moles C
Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =
1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H
The mass of the moles of C and H are:
2x10⁻³ moles C ₓ (12g / mol) = 0.024g C
2x10⁻³ moles H ₓ (1g / mol) = 0.002g H
Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O
Moles are:
0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O
Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:
C: 2.0x10⁻³ / 4x10⁻⁴ = 5
H: 2.0x10⁻³ / 4x10⁻⁴ = 5
O: 4x10⁻⁴ / 4x10⁻⁴ = 1
Thus, empirical formula is:
C₅H₅OThe molar mass of the empirical formula is:
12×5 + 1×5 + 16×1 = 81g/mol
As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:
C₁₀H₁₀O₂Content attribution
QUESTION 2 • 1 POINT
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
The given question is incomplete. The complete question is :
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
a) [tex]O^{2-}[/tex]
b) [tex]F^{-}[/tex]
c) [tex]N^{3-}[/tex]
d) [tex]S^{2-}[/tex]
Answer: b) [tex]F^{-}[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here potassium is having an oxidation state of +1 called as cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to give neutral ionic compound.
Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]
Using the periodic table provided, identify the atomic mass of sodium (Na) . Your answer should have 5 significant figures. Provide your answer below: __ amu
Answer:
Your answer will either be 22.9897 or 22.990 !!
Explanation:
The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glucose, C6H12O6?
Answer:
Vapor pressure of water = 23.14torr
Explanation:
When you made a solution, vapor pressure decreases following Raoult's law:
[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]
Where P is vapor pressure and X mole fraction
As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).
1000g of H₂O are, in moles (Molar mass: 18.02g/mol):
1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.
As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:
[tex]X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737[/tex]
Replacing in Raoult's law, pressure of water above the solution is:
[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]
[tex]P_{solution} = 0.9737*23.77torr[/tex]
Vapor pressure of water = 23.14torrYou are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concentration of the unknown acid was 0.2230 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid
Answer:
[tex]pKa=3.70[/tex]
Explanation:
Hello,
In this case, given the information, we can compute the concentration of hydronium given the pH:
[tex]pH=-log([H^+])\\[/tex]
[tex][H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M[/tex]
Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:
[tex]HA\rightleftharpoons H^++A^-[/tex]
We can write the law of mass action for equilibrium:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:
[tex][HA]=0.2230M-6.61x10^{-3}M=0.2164M[/tex]
As the concentration of hydronium also equals the reaction extent ([tex]x[/tex]). Thereby, the acid dissociation constant turns out:
[tex]Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}[/tex]
And the pKa:
[tex]pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70[/tex]
Regards.
Rank the compounds in each set in order of increasing acid strength.
(a) CH3CH2COOH CH3CHBrCOOH CH3CBr2COOH
(b) CH3CH2CH2CHBrCOOH CH3CH2CHBrCH2COOH CH3CHBrCH2CH2COOH
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).
With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.
See figure 1
I hope it helps!
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What is the molarity of this solution? Express your answer to four significant figures and include the appropriate units.
Answer:
Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].
Explanation:
Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.
Formula mass of strontium hydroxideLook up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.
[tex]\rm Sr[/tex]: [tex]87.62[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:
[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].
Number of moles of strontium hydroxide in the solution[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].
The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.
How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?
[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].
Molarity of this strontium hydroxide solutionThere are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:
[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].
The molarity of a solution measures its molar concentration. For this solution:
[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
(Rounded to four significant figures.)
when air molecules collide with things around us, it produces _______ (2 words), which is measured with a _______.
✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.
A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The balloon was released to an altitude with a pressure of 530 torr. What was the volume (L) of the weather balloon
Answer:
4.33 L
Explanation:
Step 1: Given data
Initial volume of the balloon (V₁): 3.00 L
Initial pressure of the balloon (P₁): 765 torr
Final volume of the balloon (V₂): ?
Final pressure of the balloon (P₂): 530 torr
Step 2: Calculate the final volume of the balloon
If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L[/tex]
Draw the Lewis structure of ethyne (C₂H₂) and then choose the appropriate pair of molecular geometries of the two central atoms. Your answer choice is independent of the orientation of your drawn structure.
A) linear / linear
B) trigonal/pyramidal
C) pyramidal/trigonal
D) trigonal pyramidal/trigonal pyramidal
E) planar / linear
Answer:
A) linear / linear
Explanation:
In this case, we have a triple bond beetween the atoms (See figure 1). If we have this triple bond we will have an Sp hybridization (in both carbons). We have to remember the relationship between the geometry and the hybridization:
-) Sp3 = Tetrahedral
-) Sp2 = Trigonal
-) Sp = Linear
Due to the hybridization, we will have a linear structure between the atoms. The angle between the atoms is 180º (See figure 2).
So, if we have a hybridization Sp for both carbons, we will have a linear geometry in each carbon. Therefore, the answer is A.