Answer:
Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Calculate mole H2 using the formula: n = m/M, where:
n = mole
m = mass (g)
M = molar mass (g/mol)
Calculate molar mass of H2.
M H2 = 2 × 1.008 g/mol = 2.016 g/mol
Calculate moles H2.
n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2
Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.
2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O
The mass of water will be calculated by rearranging the n = m/M formula to isolate m;
m = n × M
Calculate the molar mass H2O.
M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol
Calculate the mass H2O.
m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O
4.16 g H2 with excess O2 will produce 37.2 g H2O.
How many grams of sodium nitrate (NaNO3) are needed to
prepare 100 grams of a 15.0 % by mass sodium nitrate
solution?
Answer:
15.0 g
Explanation:
15.0% =0.150
100.0 g × 0.150= 15.0g
Sodium nitrate is "an inorganic compound with the formula of NaNO₃.
What is an inorganic compound?Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".
15% = 0.15
100.0 g × 0.15= 15g
Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.
To learn more about Sodium nitrate here
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How many atoms are in .45 moles of P4010
Answer:
5×6.02×1023
Explanation:
there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010
Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
In what order do electrons fill orbitals?
A. Orkjtals s, p, and then d fill in one energy level before starting the
next level.
B. Before pairing, 1 electron occupies each sand porbital.
C. Electrons fill orbitals in order of increasing orbital energy.
D. The p orbitals fill before the s orbitals in an energy level.
SUBMIT
Answer:
c......................
Answer:
electrons will fill the lowest energy orbitals first and then move up to higher energy orbitals only after the lower energy orbitals are full
What is the correct order for the reactions that produce the following transformation. a. (1) H2/Lindlar (2) CH3CO2OH b. (1) H2/Lindlar (2) O3, Zn, HCl c. (1) H2/Pd (2) CH3CO2OH d. (1) Na, NH3 (2) CH3CO2OH
Answer:
Explanation:
Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)
write balanced half-reactions describing the oxidation and reduction that happen in this reaction 2Fe(s)+3Pb(NO3)2(aq)=3Pb(s)+2Fe(NO3)3(aq)
Answer:
Oxidation half-reaction: 2 Fe (s) ----> 2 Fe³+ (aq) + 6e-
Reduction half-reaction: 3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)
Explanation:
A redox reaction reaction is one in which oxidation and reduction occur simultaneously and to the same extent.
Oxidation involves a loss of electron, hence, a positive increase in the oxidation number of the atom or ion. The oxidation half-reaction is as follows:
2 Fe (s) ----> 2 Fe³+ (aq)
The metallic element iron, Fe , having an oxidation number of zero, loses three electrons to form the Fe³+ ion with a charge of +3. Since each atom loses three electrons each, The number of moles of electrons lost is six.
2 Fe (s) ----> 2 Fe³+ (aq) + 6e-
Reduction involves a gain of electrons, hence, a decrease in the oxidation number of the atom or ion. The reduction half-reaction is given below:
3 Pb²+ (aq) ---> 3 Pb (s)
The lead (ii) ion, Pb²+ having a charge of +2 gains two electrons each to become the neutral metallic lead atom, Pb, with oxidation number of zero. Since 3 moles of Pb²+ are reacting, 6 moles of electrons are gained.
3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)
2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25
Answer:
5.25 moles of protons. Option e
Explanation:
Reaction between phosphoric acid and sodium hydroxide is neutralization.
We can also say, we have an acid base equilibrium right here:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Initially we have 5.25 moles of base.
We have data from the acid, to state its moles:
M = mol/L, so mol = M . L
mol = 1.75 moles of acid
If we think in the acid we know:
H₃PO₄ → 3H⁺ + PO₄⁻³
We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)
If we have 1.75 moles of acid, we may have
(1.75 . 3) /1 = 5.25 moles of protons
These moles will be neutralized by the 5.25 moles of base
H₃O⁺ + OH⁻ ⇄ 2H₂O Kw
In a titration of a weak acid and a strong base, we have a basic pH
A clean-burning automobile engine emits about 5 lb of C atoms in the form of CO2 molecules for every gallon of gasoline it consumes. The average American car is driven about 12,000 miles per year. Using this information, check the statement that the average American car releases its own weight in carbon into the atmosphere each year. List the assumptions you make to solve this problem.
Solution :
1 lb = 453.592 g
1 gallon = 3785 g
For every 5 gallon gasoline = 5 lb of C is found
= 5 x 453.592 g of C atoms
= 2267.96 g of C atoms
Assume the consumption of car = 15 miles per kg of gasoline
The amount of gasoline used per year [tex]$=\frac{12000}{15}$[/tex]
= 800 kg
In gallons = [tex]$\frac{800}{3.785} = 211.36\text{ gallons}$[/tex]
5 gallons will produce = 2267.96 g of C atoms
Therefore,
211.36 gallons will produce = [tex]$\frac{211.36 \times 2267.96}{5}$[/tex]
= 95871.21 g
= 95.87 kg
or = 25.32 gallon
How are all atoms similar?
Answer:
Atoms are similar in the way that their nuclei contain only protons and neutrons.
Answer:
All things are made of atoms, and all atoms are made of the same three basic particles - protons, neutrons, and electrons. But, all atoms are not the same. The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.
I hope this helps :)
A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?
Answer:
Likely [tex]\rm In[/tex] (indium.)
Explanation:
Number of atoms: [tex]N = 2.241 \times 10^{21}[/tex].
Dividing, [tex]N[/tex], the number of atoms by the Avogadro constant, [tex]N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}[/tex], would give the number of moles of atoms in this sample:
[tex]\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}[/tex].
The mass of that many atom is [tex]m = 0.4272\; \rm g[/tex]. Estimate the average mass of one mole of atoms in this sample:
[tex]\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The average mass of one mole of atoms of an element ([tex]114.82\; \rm g \cdot mol^{-1}[/tex] in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass [tex]114.82[/tex]. Indium, [tex]\rm In[/tex], is the closest match.
The symbol of the element is In
StoichiometryFrom the question, we are to determine the identity of the element
First, we will determine the number of moles of sample present
Using the formula
[tex]Number \ of\ moles = \frac{Number\ of \ atoms }{Avogadro's\ constant}[/tex]
Number of moles of the sample = [tex]\frac{2.241\times 10^{21} }{6.022\times 10^{23} }[/tex]
Number of moles of the sample = 0.003721355 mole
Now, we will determine the Atomic mass of the sample
From the formula,
[tex]Atomic\ mass = \frac{Mass}{Number\ of\ moles}[/tex]
Therefore,
Atomic mass of the sample = [tex]\frac{0.4272}{0.003721355}[/tex]
Atomic mass of the sample = 114.8 amu
The element that has an atomic mass of 114.8 amu is Indium. The symbol of Indium is In.
Hence, the symbol of the element is In.
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During the postabsorptive state, metabolism adjusts to a catabolic state.
a. True
b. False
Answer:
The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).
The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.
Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.
So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.
What biological molecules provide energy and structural support to living organisms?
A. Lipids
B. Nucleic acids
C. Proteins
D. Carbohydrates
Answer:
Option D: Carbohydrates
Explanation:
Carbohydrates are a group of macromolecules that are a vital energy source for the cell, provide structural support to many organisms, and can be found on the surface of the cell as receptors or for cell recognition.
What is the molecular formula of a compound that contains 38.6% C, 45% N, and 16.4% H, if 0.158 g of the compound occupies 125 mL with a pressure of 0.982 atm at 22oC?
Answer:
Molecular formula => CH₅N
Explanation:
We'll begin by calculating the number of mole of the compound. This can be obtained as follow:
Pressure (P) = 0.982 atm
Volume (V) = 125 mL = 125 / 1000 = 0.125 L
Temperature (T) = 22 °C = 22 + 273 = 295 K
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =?
PV = nRT
0.982 × 0.125 = n × 0.0821 × 295
0.12275 = n × 24.2195
Divide both side by 24.2195
n = 0.12275 / 24.2195
n = 0.0051 mole
Next, we shall determine the molar mass of the compound.
Mass of compound = 0.158 g
Mole of compound = 0.0051 mole
Molar mass of compound =?
Mole = mass /molar mass
0.0051 = 0.158 / molar mass
Cross multiply
0.0051 × molar mass = 0.158
Divide both side by 0.0051
Molar mass = 0.158 / 0.0051
Molar mass of compound = 31 g/mol
Next, we shall determine the empirical formula of the compound. This can be obtained as follow:
C = 38.6%
H = 16.4%
N = 45%
Divide by their molar mass
C = 38.6 / 12 = 3.217
H = 16.4 / 1 = 16.4
N = 45 / 14 = 3.214
Divide by the smallest
C = 3.217 / 3.214 = 1
H = 16.4 / 3.214 = 5
N = 3.214 / 3.214 = 1
Empirical formula => CH₅N
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Empirical formula = CH₅N
Molar mass of compound = 31 g/mol
Molecular formula = Empirical × n = molar mass of compound
[CH₅N]ₙ = 31
[12 + (5×1) + 14]n = 31
[12 + 5 + 14]n = 31
31n = 31
Divide both side by 31
n = 31 / 31
n = 1
Molecular formula => [CH₅N]ₙ
Molecular formula => [CH₅N]₁
Molecular formula => CH₅N
The overall order of an elementary step directly corresponds to its molecularity.
a. True
b. False
Answer:
true
Explanation:
In order to promote the common ion effect, the concentration of the common ion must first: _____________
a. increase
b. decrease
c. be equal to its equilibrium value
d. depends on the equilibrium
Answer:
a. increase.
Explanation:
Hello!
In this case, according to the given information, it turns out possible for us to tell that the common ion effect decreases the solubility of the ionic solid by firstly increasing the concentration of the common ion, which is further solved for the solubility in order to evidence the aforementioned decrease.
As an example, we can consider the solubility equilibrium for silver chloride:
[tex]Ksp=[Ag^+][Cl^-][/tex]
Which goes to:
[tex]Ksp=s^2[/tex]
Whereas s is the solubility to be solved for. However, when a silver- or chloride-containing solution is added, say 0.1 sodium chloride, the equilibrium expression changes to:
[tex]Ksp=(s)(s+0.1)[/tex]
Which turns out into a smaller value for s.
Regards!
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).
Answer:
Al^3+
Explanation:
Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.
Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.
If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;
Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)
When perchloric acid (HClO4) reacts with tetraphosphorus decaoxide, phosphoric acid and dichlorine heptaoxide are produced.
a. Trei
b. False
Answer:
я не знаю ответа :(
Explanation:
An unknown element, X, reacts with potassium to form the compound K2X. In other compounds this element also can accommodate up to 12 electrons rather than the usual octet. What element could X be
Answer:
Se
Explanation:
First of all, we must note that any element that we must choose is an element that is in group sixteen.
Elements of groups 16 have six electrons in their outermost shell which can be used for bond formation thereby yielding a total of twelve electrons on the valence shell.
However, this is only possible for the heavier members of the group 16 (from sulphur downwards) which are able to expand their octet.
Oxygen can not expand its octet hence it is not the answer.
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
enmoles of helium gas and one mole of solid argon are in thermal equilibrium with each other at 10 K. Both helium and argon are monatomic, and argon is ten times as massive as helium. How do the average speeds of the atoms in these two substances compare under the conditions specified
Answer:
Average speed of helium is higher than argon.
Explanation:
The average speed of helium is higher than argon atom under the conditions specified because of lower mass of the helium atom as compared to argon atom. Average speed of an atom is inversely proportional to mass of the atom. If mass of an atom decreases, the atom moves with higher speed while on the other hand, if the mass of an atom increases the average speed of an atom decreases.
A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mol ?
Answer:
[tex]k_2=0.504s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:
[tex]ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}[/tex]
Regards!
How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 56.0 mL of 0.671 M AgNO3 solution?
Answer:
The mass of silver carbonate precipitated is 5.18 grams.
Explanation:
Molarity of the silver nitrate solution = 0.671 M
Volume of the silver nitrate solution = 56.0 mL
[tex]1 mL = 0.001 L\\56.0 mL = 56.0\times 0.001 L=0.0560 L[/tex]
Moles of silver nitrate = n
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=\frac{n}{0.0560 L}\\n=0.671 M\times 0.0560 L=0.0376 mol[/tex]
Moles of silver nitrate used = 0.0376 mol
[tex]K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3[/tex]
According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:
[tex]=\frac{1}{2}\times 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3[/tex]
Moles of the silver carbonate formed = 0.0188 mol
Molar mass of silver carbonate = 275.7453 g/mol
Mass of silver carbonate :
[tex]=275.7453 g/mol\times 0.0188 mol=5.1840 g\approx 5.18 g[/tex]
The mass of silver carbonate precipitated is 5.18 grams.
A 250-mg sample of carbon from wood underwent 15300 carbon-14 disintegrations in 36 hours. Estimate the time since the death of the sample.
Answer:
The correct answer is - 9935 years approximately.
Explanation:
Let z be the age in years to be found:
(15300 disintegrations) x (1.0 g / 0.250 g) / (1.84×10^4 disintegrations)
= 3.3260
half life of carbon = (1/2)^(z/5730 yr)
Solve for z
3.3260 = (1/2)^(z/5730)
Take the log of both sides:
log 3.3260 = (z/5730) log (1/2)
log 3.3260 / log (1/2) = z/5730
z = 5730 log 3.3260 / log (1/2)
= 1.73378816*5730
= 9935 years approximately.
Which of the following are examples of physical properties of ethanol? Select all that apply.
The boiling point is 78.37°C
It is a clear, colorless liquid
It is flammable
It is a liquid at room temperature
A beverage contains tartaric acid, H2C4H4O6, a substance obtained from grapes during wine making. If the beverage is 0.190 tartaric acid, what is the molal concentration? What is the mole fraction of tartaric acid and water? Calculate the mass percent of tartaric acid. The density of the solution is 1.016g/mL.
Answer:
1) [tex]molality = 0.19[/tex]
2) [tex]Mole\ fraction= 0.003486[/tex]
3)[tex]Mass\ percent = 2.8%[/tex]
Explanation:
Concentration of Tartaric acid=0.190mole /l
1)
Generally
[tex]Mass\ of\ tartaric\ acid = 150.087 *0.190[/tex]
[tex]Mass\ of\ tartaric\ acid = 28.5 g[/tex]
Since 1L of solution tartartic acid is
[tex]T_{1l}= density * volume[/tex]
[tex]T_{1l}= 1.016Kg / L X 1 L[/tex]
[tex]T_{1l}= 1016[/tex]
Therefore
[tex]Mass of solvent = 1016-28.5[/tex]
[tex]Mass of solvent = 987.5 g[/tex]
Generally the equation for molality is mathematically given by
[tex]molality = \frac{moles}{Kg}[/tex]
[tex]molality = \frac{0.190 * 1000}{987.5}[/tex]
[tex]molality = 0.19[/tex]
2.
Generally the equation for Moles of water is mathematically given by
[tex]Moles\ of\ water = \frac{mass}{mol wt}[/tex]
[tex]Moles\ of\ water = \frac{987.5}{18 }[/tex]
[tex]Moles\ of\ water= 54.86[/tex]
Therefore
[tex]Mole\ fraction = \frac{Moles\ of\ solute}{total\ moles}[/tex]
[tex]Mole\ fraction= \frac{0.190}{54.5}[/tex]
[tex]Mole\ fraction= 0.003486[/tex]
3
Generally the equation for Mass Percent is mathematically given by
[tex]Mass\ percent = \frac{mass\ of\ tartaric\ acid}{total mass}[/tex]
[tex]Mass\ percent = \frac{28.5* 100}{1016}[/tex]
[tex]Mass\ percent = 2.8%[/tex]
Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).
Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Explanation:
Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M
[tex]K_{a} = 1.0 \times 10^{-8}[/tex]
Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] is as follows.
[tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for [tex]K_{a}[/tex] value is as follows.
[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]
Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]
Formula to calculate pH is as follows.
[tex]pH = -log [H^{+}][/tex]
Substitute the values into above formula as follows.
[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]
Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the nitrogen atom, followed by nucleophilic addition of water. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism.
Answer:
See explanation and image attached
Explanation:
The acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the nitrogen atom. This step is shown in the image attached.
The next step is the nucleophilic addition of water. The task is to show the movement of electrons in this step of the reaction mechanism. This was clearly shown in the image attached to this answer.
In the preparation of methyl benzoate, what is the purpose of (a) washing the organic layer with sodium bicarbonate solution
Answer: The purpose of washing the organic layer with sodium bicarbonate solution is to
(to remove any of the unreacted benzoic acid).
Explanation:
Methyl benzoate is a typical example of an ester in organic chemistry. It is formed by the action of benzoic acid and methanol. Sulphuric acid is often used as a catalyst in this reaction.
The procedure for the preparation of the ester includes:
--> the acid and the alcohol are added into a round bottom flask,
--> concentrated sulphuric acid is added down the side of the flask,
--> a separatory funnel is used to extract the ester into an organic layer where sodium bicarbonate is added.
--> when the aqueous layer from the sodium bicarbonate wash is acidified with concentrated HCl, the unreacted benzoic acid should precipitate. This is how the unreacted benzoic acid is removed.
Answer:
See explanation
Explanation:
Let us remember that Sodium bicarbonate is basic in nature. It is one of the common bases used to neutralize acids.
In the preparation of methyl benzoate, H2SO4 is used as the catalyst for the esterification reaction.
Hence, when we wash the organic layer with sodium bicarbonate we are trying to remove the H2SO4 thereby neutralizing the organic layer.
Identify the solute and solvent in each solution:
a. 6mL of ethanol and 35mL of water
b. 300 g of water containing 8g of NaHCO3
c. 0.005LofCO2and2LofO2
Answer:
a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).
b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).
c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).
Explanation:
Hello there!
In this case, according to the given problem, it turns out possible for us to solve these questions by bearing to mind the fact that in a solution, we can find two substances, solute and solvent, whereas the former is in a smaller proportion in comparison to the latter; in such a way, we infer the following:
a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).
b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).
c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).
Regards!
