Answer:
57------6.9
330------.87
Explanation:
Answer:
The required volume of air is 3.64 L
Explanation:
Ideal gas equation:-
The relation between pressure, volume, and temperature of the gas is known as the ideal gas equation and it is given as,
PV=RT
Where R=gas constant
Now,
Let the volume of air is, V
According to the question we have
Temperature, T= 60°C= (60+273) K= 333 K
Atmospheric pressure, P= 760 mm
And gas constant, R= 8.314 J/mole K
Substitute values in ideal gas equation and we get
760V= 8.314(333)
V= 2768.562/760
V= 3.642 L
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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.
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.
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
4.If 15.00 mL of 3.00 M potassium iodide is needed to reach the equivalence point with 10.00 mL of lead (Il) nitrate, determine the molarity of the lead (Il) nitrate solution
Answer:
2.25 M
Explanation:
The reaction that takes place is:
2KI + Pb(NO₂)₃ → PbI₂ + 2KNO₃First we calculate how many potassium iodide moles reacted, using the given volume and concentration:
15.00 mL * 3.00 M = 45 mmol KIThen we convert 45 mmoles of KI into mmoles of Pb(NO₂)₃, using the stoichiometric coefficients of the balanced reaction:
45 mmol KI * [tex]\frac{1mmolPb(NO_3)_2}{2mmolKI}[/tex] = 22.5 mmol Pb(NO₂)₃Finally we calculate the molarity of the Pb(NO₂)₃ solution, using the calculated number of moles and given volume:
22.5 mmol Pb(NO₂)₃ / 10.00 mL = 2.25 MHow 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!
Physical and psychological dependence, high-risk behaviour, and chronic high blood
pressure can result from excessive use of
Answer:
Physical and psychological dependence is high, and withdrawal symptoms include watery eyes, runny nose, loss of appetite, irritability, tremors, panic, abdominal cramps and diarrhea, nausea, chills, and sweating. Use of contaminated syringes/needles to inject drugs may result in serious blood borne infections such as HIV-AIDS and hepatitis.
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
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.
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]
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.
Chemical 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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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
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)=-4475kJSodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it reacts with exhaled CO2 to form Na2CO3 and O2. How many liters of respired air can react with 96.7 g of Na2O2 if each liter of respired air contains 0.0755 g of CO2
Answer:
725.15 L
Explanation:
The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:
Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂
From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.
Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:
Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol
moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂
Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂
Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:
1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂
In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:
1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L
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
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
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:
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.
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]
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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An electron moved from shell n = 2 to shell n = 1. What most likely happened during the transition?
When electron goes from a higher shell to lower shell then it loses energy .
So, when an electron moved from shell n = 2 to shell n = 1 then a photon of energy is released.
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⁻).
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)
The overall order of an elementary step directly corresponds to its molecularity.
a. True
b. False
Answer:
true
Explanation:
A rock originally has 2200 atoms of 235U and no 207Pb. It now has 800 atoms of 235U. Assuming the only chemical process is the radioactive decay of uranium to lead, what is the age of the rock
Answer:
[tex]Age=1040.55 million\ years[/tex]
Explanation:
From the question we are told that:
Initial Rock atoms [tex]a_1=2200atoms of 235^U[/tex]
Final Rock atom [tex]a_2=800 of 235^U[/tex]
Age of half life 713 million years
Generally the equation for Age is mathematically given by
[tex]\frac{1}{2^n}=\frac{800}{2200}[/tex]
[tex]\frac{1}{2^n}=\frac{1}{2.750}[/tex]
[tex]n=1.46[/tex]
Therefore
[tex]Age=713*1.46[/tex]
[tex]Age=1040.55 million\ years[/tex]
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.
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.
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.
If 0.650 mL of benzaldehyde reacts with enough of the Grignard reagent, calculate the theoretical yield (in grams) of the alcohol product. Show calculation with units for full credit.
Answer:
Theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
Explanation:
A Grignard reagent is any of the numerous organic derivatives of magnesium (Mg) which are commonly represented by the general formula RMgX (where R is a hydrocarbon radical e.g. methyl, ethyl, propyl, etc.; and X is a halogen atom, e.g. chlorine, bromine, or iodine)
A Grignard reaction converts an aldehyde to a secondary alcohol. In the grignard reaction involving benzaldehyde as in this experiment, the grignard reagent used is ethyl magnesium bromide, EtMgBr, and the resulting product is 1-Phenyl-1-propanol, a secondary alcohol. The reaction is shown in the figure attached below.
Mass of benzaldehyde in 0.650 mL = density × volume
Density of Benzaldehyde = 1.044 g/mL
Mass of benzaldehyde = 1.044 g/mL × 0.650 mL = 0.6786 g
Molar mass of benzaldehyde = 106 g/mol
Molar mass of 1-Phenyl-1-propanol = 136 g/mol
Mass of = mass of benzaldehyde × mole ratio of 1-Phenyl-1-propanol and benzaldehyde
Mass of 1-Phenyl-1-propanol = 0.6786 g × (136 g/mol)/(106 g/mol) = 0.871 g
Therefore, the theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g