A leaking tap that drops water at the rate of 3 drops every second, will leak 259.2 L in a day.
We know that a leaking tap drops water at the rate of 3 drops every second and that each drop is approximately 1 ml. The milliliters of water dropped every second are:
[tex]\frac{3drop}{1s} \times \frac{1mL}{1drop} = \frac{3mL}{1s}[/tex]
We want to know the number of seconds in 1 day. We will use the following conversion factors:
1 day = 24 h1 h = 60 min1 min = 60 s[tex]1day \times \frac{24h}{1day} \times \frac{60min}{1h} \times \frac{60s}{1min} = 86400 s[/tex]
3 mL of water are dropped every second. The mL of water dropped in 86400 s are:
[tex]86400 s \times \frac{3mL}{1s} = 259200 mL[/tex]
Finally, we will convert mL to L using the conversion factor 1 L = 1000 mL.
[tex]259200 mL \times \frac{1L}{1000 mL} = 259.2 L[/tex]
Approximately 259.2 L of water will be dropped in 1 day.
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Oleic acid and elaidic acid are isomeric alkenes.
a. True
b. False
Answer:
False
Explanation:
Because Elaidic acid is an isomer of oleic acid. I really hope this helps you.
which of the following kb values represents the weakest base?
Answer:
the weakest base will have a higher Kb value since it will be closer to an acid than a base
7 kb values represents the weakest base.
What is kb value?Kb is the base dissociation constant which is a measure of how completely a base dissociates into its component ions in water. pKb is define as the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.
Ka is define as the acid dissociation constant while pKa is the -log of this constant. Kb is define as the base dissociation constant, while pKb is the -log of the constant.
The acid and base dissociation constants are usually expressed in terms of moles per liter (mol/L).
Thus, 7 kb values represents the weakest base.
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You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
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2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
Suppose that a certain atom possesses only four distinct energy levels. Assuming that all transitions between levels are possible, how many spectral lines will this atom exhibit
Answer:
Following are the response to the given question:
Explanation:
The number of shells
n = 4
Calculating the spectral line:
[tex]= \frac{n(n-1)}{2}\\\\ = \frac{4(4-1)}{2} \\\\= \frac{4\times 3}{2}\\\\ = \frac{12}{2}\\\\ = 6[/tex]
why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = [tex]m\times s\times t \ J[/tex]
then,
⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]
⇒ [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]
⇒ [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]
⇒ [tex]43903.5 = 3582.185 T[/tex]
⇒ [tex]T = 12.25^{\circ} C[/tex]
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
Write the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
How to solve this problem step by step
Answer:
[tex]V_2= 736mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?
Answer:
The right answer is "3 g".
Explanation:
Given:
Initial mass substance,
[tex]M_0=24 \ g[/tex]
By using the relation between half lives and amount of substances will be:
⇒ [tex]M=\frac{M_0}{2^n}[/tex]
[tex]=\frac{24}{2^3}[/tex]
[tex]=3 \ g[/tex]
Thus, the above is the correct answer.
Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.
Answer:
Explanation:
The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.
what are the major specials presents in a solution of a strong acid like HCl
Answer:
hydrogen ions (H⁺) and chloride ions (Cl⁻)
Explanation:
Hydrochloric acid (HCl) is a strong acid. That means that the compound dissociates completely into ions when is dissolved in water, as follows:
HCl → H⁺ + Cl⁻
The equilibrium is completely shifted to the right side (products). Thus, it is considered that the concentration of the non-dissociated compound (HCl) is negligible, and the major specials present in the solution are the hydrogen ions (H⁺) and chloride ions (Cl⁻).
Consider the following equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover.
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g) ΔHrxn = −1790kJ
If a bottle of nail polish remover contains 143 g of acetone, how much heat would be released by its complete combustion? Express your answer to three significant figures.
Molar mass of Acetone
C3H6O3(12)+6+1658g/molNow
1 mol releases -1790KJ heat .Moles of Acetone:-
143/58=2.5molAmount of heat:-
2.5(-1790)=-4475kJChemical reactions can exhibit different rate constants at differing: Select the correct answer below: initial concentrations volumes of container temperatures none of the above
Explanation:
Chemical reactions can exhibit different rate constants at differing:
i)initial concentrations
ii)volumes of container
iii) temperatures
iv)none of the above.
The rate constant of a reaction is constant at a particular temperature.
It is not depending on the initial concentration of the reactants. It varies with temperature.
Thus, among the given options the correct answer is Temperatures.
Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
What is a rate constant?
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.
For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.
The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.
On the other hand, the volume of the container, initial concentration does not affect the rate constant.
Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
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9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain
Answer:
See explanation
Explanation:
Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.
When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.
When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.
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
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)
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.
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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:
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
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
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
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.
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
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)
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!
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.
