Answers
Answer:
there are 37,8541 liters in 10 gallons
Related Questions
Write the IUPAC and common names, if any, of the carboxylate salts produced in the reaction of each of the following carboxylic acids with NaOH: 2-methylhexanoic acid
Part A
2-bromopropanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Part B
2-methylhexanoic acid Spell out the full name of the compound. If there is a common name separate your answers by a comma.
Answers
Answer:
Following are the explanation to this question:
Explanation:
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Following are the description of the given reaction:
In reaction A:
2-Bromopropanoic acid= [tex]C_3H_5BrO_2[/tex]
[tex]C_3H_5BrO_2+NaOH[/tex]⇄ [tex]C_3H_4BrNaO_2 +H_2O[/tex]
The IUPAC name is Sodium-2-Bromopropanate
In reaction B:
2-Methylhexanoic acid= [tex]C_7H_{14}O_2[/tex]
[tex]C_7H_{14}O_2+NaOH[/tex]⇄ [tex]C_7H_{13}NaO_{2}+H_2O[/tex]
The IUPAC name is Sodium-2-Methyl hexanoate
The IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the compound carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.
Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.
Thus, the IUPAC name of compound is Sodium-2-Bromopropanate in reaction 1 and in reaction 2 it is 2-Methylhexanoic acid.
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The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answers
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best regards.
Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.
1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)
2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)
3. Mg (s) + Cl2(g) - MgCǐ2(s)
4. SO3(g) + H2O(I) - H2SO4(I)
A. ΔSsys greater than
B. ΔSsys smaller than
Answers
Answer:
Answers are in the explanation.
Explanation:
In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:
Entropy of gases >>> entropy of liquid > entropy of solids.
The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.
In the reactions:
1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)
As 1 gas is produced, entropy of products is higher than entropy of reactants. That means ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)
2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.
3. Mg(s) + Cl₂(g) → MgCǐ₂(s)
You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
4. SO₃(g) + H₂O(I) → H₂SO₄(I)
1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:
1. ΔSsys > 0.
2. ΔSsys < 0
3. ΔSsys < 0
4. ΔSsys < 0
Recall:
The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.The entropy of gasses is greater than the entropy of liquid and solids.The entropy of solids is less than the entropy of liquid and gasses.Gasses have the highest entropy, while solids have the least.Thus:
In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.
In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.
In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.
In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.
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What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?
Answers
Answer:
Following are the solution to this equation:
Explanation:
In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:
In the given question "Option (iii)" is correct, which is defined in the attachment file.
When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbonate that has twice the mass of the first sample
Answers
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answers
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Answers
Answer:
H₂O
Explanation:
The empirical formular of the compound is obtained using the following steps;
Step 1: Divide the percentage composition by the atomic mass
Hydrogen = 11.21 / 1 = 11.21
Oxygen = 88.79 / 16 = 5.55
Step 2: Divide by the lowest number
Hydrogen = 11.21 / 5.55 = 2.02 ≈ 2
Oxygen = 5.55 / 5.55 = 1
This means the ratio of the elements is 2 : 1
The empirical formular (simplest formular of a compound) of the compound is;
H₂O
Answer:Empirical formula ======== H₂O
Explanation:The empirical formula of a compound shows the whole number ratio for each atom in a compound.
To find empirical formula. we follow the below steps
The total mass of the compound here is 100 grams, that is (11.21% of hydrogen + 88.79% of oxygen) we can then assume 11.21 grams of hydrogen and 88.79grams of oxygen
Hydrogen Oxygen
1.composition by mass 11.21 88.79
molecular weight 1.007g/mol 15.990g/mol
2.Divide composition by mass 11.21/1.007 88.79/15.99
by each molecular weight to get 11.13 5.553
no of moles
3 Divide by the least number of moles
to get atomic ratio 11.13/5.553 5.553/5.553
2.004 1.00
4.Convert to whole numbers 2 1
Empirical formula ======== H₂O
If 100-mL of 1.0 M Sr(OH)2 is added to 100 mL of 1.0 M HCl, the pH of the mixture would be _____. Group of answer choices
Answers
Answer:
pH = 13.7
Explanation:
A strong acid (HCl) reacts with a strong base Sr(OH)₂ producing water and a salt, thus:
2HCl + Sr(OH)₂ → 2H₂O + SrCl₂
To solve this problem, we need to find initial moles of both reactants and, with the chemical equation find limiting reactant and moles in excess to find pH as follows:
The initial moles of HCl and Sr(OH)₂ are:
100mL = 0.1L ₓ (1.0mol / L) = 0.100 moles of both HCl and Sr(OH)₂
As 2 moles of HCl reacts per mole of Sr(OH)₂, moles of Sr(OH)₂ that reacts with 0.100 moles of HCl are:
0.100 moles HCl ₓ (1 mol Sr(OH)₂ / 2 mol HCl) = 0.050 moles Sr(OH)₂
That means HCl is limiting reactant and after reaction will remain in solution:
0.100 mol - 0.050mol =
0.050 moles of Sr(OH)₂
Find pH:
1 mole of Sr(OH)₂ contains 2 moles of OH⁻, 0.050 moles contains 0.050×2 = 0.100 moles of OH⁻. In 200mL = 0.2L:, molar concentration of OH⁻ is:
0.100 moles / 0.2L =
[OH⁻] = 0.5M
As pOH of a solution is -log[OH⁻],
pOH = -log 0.5M
pOH = 0.301
And knowing:
pH = 14 - pOH
pH = 14 - 0.301
pH = 13.7How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
Answers
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
How many unit cells share an atom that is located at the center of a cube edge of a unit cell?
Answers
Answer:
zero
Explanation:
In a unit cell, an atom that is located at the center of a cube edge is not involved in sharing unit cells because a central atom of a unit cell belongs to the entire cell and only to that unit cell of the lattice.
Hence, the center atom of a unit cell do not share any unit cell and the correct answer is "Zero".
A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell
Answers
Answer:
The answer is D) Electrolytic cell
Explanation:
An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.
When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.
These cells are the closest thing to a galvanic battery.
Answer:
b. voltaic cell
Explanation:
Founders Education answer. had to take this quiz 4 times
Zeros laced at the end of the significant number are...
Answers
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
1500 L has how many significants figures
Answers
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–5 M 3.0 × 10–2 M 2.4 × 10–2 M None of the above.
Answers
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
3.0x10⁻²MAccount for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. decreases
2. loss
3. Increases
4. greater
A. Disturbing the equilibrium by adding NO2Cl______Qc to a value_____than Kc.
B. To reach a new state of equilibrium, Qc therefore______which means that the denominator of the expression for Qc______.
C. To accomplish this, the concentration of reagents______, and the concentration of products_______.
Answers
Answer:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.
Explanation:
Hello,
In this case, for the equilibrium reaction:
[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]
Whose equilibrium expression is:
[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]
The proper matching is:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.
Best regards.
A rock has a mass of 15.8 g and causes the water level in a graduated cylinder to raise from 22.3 mL to 32.5 mL. What is the density of the rock in Kg/mL?
Answers
Answer:
Explanation:
mass - 15.8 g = 0.0158 kg
volume = 32.5 - 22.5 = 10.2 ml
density = mass / volume
= 0.0158 / 10.2
= 0.00154 kg/ml
hope this helps
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Consider the bond dissociation energies listed below in kcal/mol. CH3-Br 70 CH3CH2-Br 68 (CH3)2CH-Br 68 (CH3)3C-Br 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.
Answers
Answer:
The answer is "Tertiary carbon".
Explanation:
Accent to the results, the carbon-bromine bond is weak, whenever, the bromine is connected to tertiary carbon so, bonding energy is separation for methyl-carbon, which is connected to the bromine = 70 kcal/mol and for the primary energy to the secondary energy is= 68 kcal/mol, and for tertiary CO2 = 65 kcal/mol.
The stronger the energy dissociating connection and the weaker, its power dissociation connection and its weaker bond becomes connecting with a tertiary carbon, that's why "Tertiary carbon" is the correct answer.
A student is performing a Benedict’s test on an unknown substance. He adds the reagent (the chemical required to make a color change), and nothing happens. What can he conclude? A- The substance is glucose-based. B- The substance is not glucose-based. C- The test was inconclusive because he needed to also test with iodine or vinegar. D- The test was inconclusive because he forgot to add heat.
Answers
Answer:
The correct answer is : option D. The test was inconclusive because he forgot to add heat.
Explanation:
Benedict's test is a test that is used to confirm the presence of the simple carbohydrates (mono saccharides and some disaccharides). It is a reagent made by mixture of solution of CuSO4 with sodium citrate and Na2CO3.
Benedict's reagent is added to the substance to test and then heated if it turns yellow to orange or red the presence of simple sugar is confirmed.
Thus, the correct answer is : option D. The test was inconclusive because he forgot to add heat.
Answer:
The test was inconclusive because the student forgot to add heat.
Explanation:
If the test revealed it was not glucose, then the student could run these tests. The student, however, does not need these substances to run the glucose test properly.
Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.
Answers
Answer:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.
Explanation:
Hello.
In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.
Best regards.
Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(
Answers
Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answers
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene
Answers
CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
How do covalent bonds form? A Sharing valence electrons between atoms. B Donating and receiving valence electrons between atoms. C Opposite slight charges attract each other between compounds. D Scientists are still not sure how they form.
Answers
Answer:
A. Sharing valence electrons between atoms.
Explanation:
This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).
If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?
Answers
Answer:
[tex]m=0.127g[/tex]
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]
Best regards.
A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answers
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)
Answers
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
I think I'm typing it into my calculator wrong. I will give brainliest to whoever gets it right.
Answers
Answer:
36.7 mg
Explanation:
The following data were obtained from the question.
Original amount (A₀) = 65.1 mg
Rate constant (K) = 2.47×10¯² years¯¹
Time (t) = 23.2 years
Amount of substance remaining (A) =?
Thus, we can obtain the amount of substance remaining after 23.2 years as follow
ln A = lnA₀ – Kt
lnA = ln(65.1) – (2.47×10¯² × 23.2)
lnA = 4.1759 – 0.57304
lnA = 3.60286
Take the inverse of ln
A = e^3.60286
A = 36.7 mg
Therefore, the amount remaining after 23.2 years is 36.7 mg.
[tex]pH = -log10[H+ ][/tex]. If [tex][H^+][/tex] is [tex]1.2 * 10^-^6[/tex], what is the [tex]pH[/tex]? (F5 to refresh page if you can't see it)
Answers
Solution is a weakly acidic
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answers
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
whts the ph of po4 9.78
Answers
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
