Compound B, C6H12O2, was found to be optically active, and it was oxidized to an optically active carboxylic acid A, by Ag (aka, Tollens Reagent). Oxidation of B by PCC gave an optically inactive compound X that reacted with Zn amalgam/HCl to give 3-methylpentane. With Na2Cr2O7/H2SO4, compound B was oxidized to an optically inactive dicarboxylic acid C, C6H10O4. Provide the structures of A, B, and C (ignore specific configuration of any stereocenters).
Answers
Answer:
Check the explanation
Explanation:
Acidipic Acid (which is an essential dicarboxylic acid for manufacturing purposes with about 2. 5 billion kilograms produced per year. It is mainly used for the production of nylon and its related materials.)
Going by the question, since the [tex]H_{2} CrO_{4}[/tex] is comparatively mild oxidizing agent than the [tex]CrO_{3}[/tex], it only oxidizes the carbon group
Kindly check the attached image below for the full explanation to the question above.
Related Questions
This reaction involves a conjugate addition reaction followed by an intramolecular Claisen condensation. The steps involved are as follows: 1. Conjugate addition of methyl carbanion (from the Gilman reagent) to form enolate ion 1; 2. Cyclization to form tetrahedral intermediate 2; 3. Collapse of the tetrahedral intermediate and expulsion of methoxide ion completes the reaction to form the final product. Write out the reaction on a separate sheet of paper, and then draw the structure of tetrahedral intermediate 2.
Answers
Answer:
Explanation:
Check below for the answer in the attachment.
A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1. A solution is prepared by dissolving some urea ((NH2)2CO) in 550. g of Χ. This solution freezes at-0.9 °C. Calculate the mass of urea that was dissolved. Be sure your answer has the correct number of significant digits.
Answers
Answer:
27.60 g urea
Explanation:
The freezing-point depression is expressed by the formula:
ΔT= Kf * mIn this case,
ΔT = 5.6 - (-0.9) = 6.5 °CKf = 7.78 °C kg·mol⁻¹m is the molality of the urea solution in X (mol urea/kg of X)
First we calculate the molality:
6.5 °C = 7.78 °C kg·mol⁻¹ * mm = 0.84 mNow we calculate the moles of urea that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
0.84 m = mol Urea / 0.550 kg Xmol Urea = 0.46 molFinally we calculate the mass of urea, using its molecular weight:
0.46 mol * 60.06 g/mol = 27.60 g ureaMicrowave ovens work by irradiating food with microwaves, which are absorbed by the water molecules in the food and converted to heat. Assuming that microwave ovens emit radiation with a wavelength of 29.7 cm, calculate how many photons are emitted per second in a 600-Watt microwave oven. (1 W = 1 J/s)
Answers
Answer:
8.969×10²⁶ photons
Explanation:
From the question,
Using,
E = hc/λ........................... Equation 1
Where E = Energy of the photon, h = Planck's constant, c = speed of light, λ = wave length of the photon.
Given: h = 6.626×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 0.297 m
Substitute into equation 1
E = 6.626×10⁻³⁴(3×10⁸)/0.297
E = 6.69×10⁻²⁵ J.
The energy of the photon in one seconds = 6.69×10⁻²⁵ J/s
If the power of the microwave oven = 600 J/s
Then,
Number of photons emitted per seconds = 600/(6.69×10⁻²⁵)
Number of photons emitted per seconds = 8.969×10²⁶ photons
Tarnish on tin is the compound SnO. A tarnished tin plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge. Part A Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Sn2++2e−→An;E∘=−0.140 V
Answers
Answer:
1.52V
Explanation:
Oxidation half equation:
2Al(s)−→2Al^3+(aq) + 6e
Reduction half equation
3Sn2^+(aq) + 6e−→3Sn(s)
E°cell= E°cathode - E°anode
E°cathode= −0.140 V
E°anode= −1.66 V
E°cell=-0.140-(-1.66)
E°cell= 1.52V
Final answer:
The overall cell potential for the redox reaction between tin and aluminum in a galvanic cell is calculated as +1.52 V, indicating a spontaneous reaction with tin being oxidized at the anode and aluminum reduced at the cathode.
Explanation:
To calculate the standard cell potential for a redox reaction involving tin (Sn) and aluminum (Al), we apply the reduction potentials of their respective half-reactions. The half-reaction for tin is as follows: Sn(s) ightarrow Sn2+(aq) + 2 e - , with an associated standard reduction potential (E & Ocirc ;) of - 0.140 V, however, its oxidation potential is actually +0.140 V.
For aluminum, the half-reaction is: Al3+(aq) + 3 e - ightarrow Al(s), with an E & Ocirc ; of - 1.66 V. In a galvanic cell, the aluminum will oxidize, and since it’s a reduction potential, for oxidation, we take the negative of this value, which would make it +1.66 V.
To find the overall cell potential, we use the equation Ecell = Ecathode - Eanode. In this reaction, Sn(s) is our anode, and Al(s) is our cathode. However, since aluminum's E & Ocirc ; is already negative (signifying oxidation), we reverse its sign for use in the cell potential equation.
Ecell = Ecathode - Eanode = (+1.66 V) - (+0.14 V) = +1.66 V - 0.14 V = +1.52 V
The standard cell potential is positive, indicating that the redox reaction is spontaneous. Tin is oxidized at the anode, and aluminum is reduced at the cathode, forming the basis for electric current flow in the cell.
Ethylenediamine is a bidentate ligand. The oxalate group, used in this experiment was also a didentate ligand. The structure of ethylenediamine is NH2CH2CH2NH2. The oxalate group formed a coordinated compound using the negative charges on the oxygen. Explain how the ethylenediamine compound will bond.
Answers
Answer:
Ethylene diamine will bond to the Central metal via a lone pairs of electrons on nitrogen
Explanation:
Complexes are formed by coordinate bond formation. Before a coordinate bond is formed, one of the species must have a lone lair of electrons available for donation into empty orbitals on the central metal.
Ethylene diammine contains nitrogen which has a lone pair of electrons. The two lone pairs on the two nitrogen atoms can bond with the central metal. This makes ethylene diammine a bidentate ligand (two bonding atoms).
Ethylenediamine will bond to a metal ion through both of its nitrogen atoms, forming a chelate complex.
Ethylenediamine (en) is a bidentate ligand, which means it has two donor atoms that can simultaneously bond to a central metal ion. In the case of ethylenediamine, the donor atoms are the nitrogen atoms, each of which has a lone pair of electrons available for donation. When ethylenediamine bonds to a metal ion, it does so through both nitrogen atoms, creating a five-membered ring structure known as a chelate. The term chelate comes from the Greek word chele, which means claw. This chelation results in a more stable complex because the metal ion is bound to two sites on the ligand, reducing the likelihood of the ligand dissociating from the metal center.The formation of a chelate complex with ethylenediamine can be represented as follows: [tex]\[ \text{M}^{n+} + \text{en} \rightarrow \text{M(en)}^{n+} \][/tex] Here, M represents the metal ion, n+ is its oxidation state, and en is the ethylenediamine ligand. The resulting complex M(en)n+ has the metal ion bound within the chelate ring formed by the ethylenediamine ligand. In contrast, the oxalate group is also a bidentate ligand but with two oxygen atoms as the donor atoms. The oxalate group typically bonds to a metal ion through its negatively charged oxygen atoms, forming a similar chelate complex but with a different donor atom and potentially different coordination geometry. The stability of chelate complexes is often discussed in terms of the chelate effect, which states that metal complexes with chelating ligands are more stable than those with similar non-chelating (monodentate) ligands. This increased stability is due to the entropy effect, where the formation of a chelate ring results in the release of fewer solvent molecules compared to the formation of separate bonds with monodentate ligands.
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Mn 2+(aq) + 2 e−→Mn(s); E° = –1.180 V Pb2+(aq) + 2 e−→ Pb(s); E° = –0.130 V Group of answer choices Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); = 1.050 V Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −1.050 V Pb 2+(aq) + Mn2+(aq) →Pb(s) + Mn(s); = –1.310 V Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); =0.525 V Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −0.525 V
Answers
Answer:
Explanation:
the question has been solve below
The balanced overall reaction is Mn(s) + Pb²⁺ (aq) → Mn²⁺ (aq) + Pb(s) with a standard cell potential of 1.050 V, making option a) the correct choice.
To answer this, let's first identify the standard cell potential (E°) for the given half-reactions:
Mn²⁺ (aq) + 2 e⁻ → Mn(s); E° = –1.180 V
Pb²⁺ (aq) + 2 e⁻ → Pb(s); E° = –0.130 V
In an electrochemical cell, the half-reaction with the more positive reduction potential acts as the cathode (reduction), and the one with the less positive potential acts as the anode (oxidation). Here, E° for Pb²⁺/Pb is more positive (-0.130 V) than E° for Mn²⁺/Mn (-1.180 V).
Therefore, the cell setup will be:
Manganese will be oxidized: Mn(s) → Mn²⁺ (aq) + 2 e⁻ (oxidation at the anode)
Lead will be reduced: Pb²⁺ (aq) + 2 e⁻ → Pb(s) (reduction at the cathode)
The balanced overall reaction:
Mn(s) + Pb²⁺ (aq) → Mn²⁺ (aq) + Pb(s)
Calculating the Cell Potential:
The standard cell potential (E°cell) is calculated as follows:
E°cell = E°cathode - E°anode
E°cell = (-0.130 V) - (-1.180 V) = 1.050 V
So, the correct answer is:
a) Pb²⁺(aq) + Mn(s) → Pb(s) + Mn²⁺(aq); E° = 1.050 V
Complete question:
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Mn 2+(aq) + 2 e−→Mn(s); E° = –1.180 V Pb2+(aq) + 2 e−→ Pb(s); E° = –0.130 V Group of answer choices
a) Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); = 1.050 V
b) Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −1.050 V
c) Pb 2+(aq) + Mn2+(aq) →Pb(s) + Mn(s); = –1.310 V
d) Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); =0.525 V
e) Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −0.525 V
Consider the three ligand field spectra corresponding to octahedral complexes A, B, and C, all formed from the same metal ion.
From the following list, Ti3 , Ni2 , Pt4 , Cu2 , to which metal ions could the spectra correspond and to which would it be very unlikely
Answers
Answer:
Ni^2+ is most likely
Ti^3+ is very unlikely
Explanation:
The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.
Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.
During your reaction, you added 0.3 mL concentrated H2SO4 (18.4 M) as the catalyst. At the end of the reaction, you need to add base to neutralize it. How much volume (in mL) of 10% Na2CO3 to neutralize all the acid present
Answers
Answer:
58.72 mL
Explanation:
The chemical equation for the neutralization reaction is :
H₂SO₄(aq) + Na₂CO₃(s) --------------> Na₂SO₄(aq) + H₂O(l) + CO₂(g)
where;
M₁ = Molarity of H₂SO₄
M₂= Molarity of Na₂CO₃
V₁= Volume of H₂SO₄
V₂ = Volume of Na₂CO₃
Given that :
M₁ = 18.4 M
V₁= 0.3 mL
10% Na₂CO₃ means 100 g of solution contain 10 g of Na₂CO₃
i.e. 10 g Na₂CO₃ dissolved and diluted to 100 mL water.
Molar mass of Na₂CO₃ = 106 g/mol
106 g Na₂CO₃ dissolved in 100 mL will give 0.1 M Na₂CO₃ solution.
However;
If, 106 g Na₂CO₃ ≡ 0.1 M Na₂CO₃
Then, 10 g Na₂CO₃ ≡ 'A' M of Na₂CO₃
By cross multiplying; we have:
106 × A = 10 × 0.1
106 × A = 1
A = (1/106) M/100 mL
A = 10 x (1/106)) M/L
A = (10/106) M
A = 0.094 M
Therefore,the molarity of 10% Na₂CO₃ solution is 0.094 M.
For the Neutralization equation, we have:
M₁V₁ = M₂V₂
18.4×0.3 = 0.094×V₂
Making V₂ the subject of the formula;we have:
[tex]V_2 = \frac{18.4*0.3}{0.094}[/tex]
V₂ = 58.72 mL
A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas
Answers
Answer:
T= 257.36 k
Explanation:
using the ideal gas law
pv=nRt
first, convert pressure from mmhg to kpa
312 x (101.3\760)= 41.58 kpa
R is constant= 8.31
to get n(number of moles)
n=m\M (m is mass, M is molar mass)
molar mass of argon is 39.948
n= 2.33\39.948
n=0.0583
substitute;
41.58 x 3 = 0.0583 x 8.31 x T
T= (41.58 x 3)\ (0.0583 x 8.31)
T= 257.36 k
The temperature of the gas under ideal conditions is 257.36K
In order to get the temperature of the gas, we will use the ideal gas equation expressed according to the formula:
[tex]PV = nRT[/tex]
P is the pressure of the gas = 312mmHg = 41.58KPa
V is the volume of the gas = 3.00L
n is the moles of the gas = 0.1165moles
R is boltzmann constant = 8.31
T is the required temperature
Mole = mass/molar mass
Mole = 2.33/40
Mole of argon = 0.05825moles
Substitute the given parameters into the formula
[tex]T=\frac{PV}{nR}\\ T=\frac{41.58 \times 3}{0.05825\times8.31}\\T=257.36K[/tex]
Hence the temperature of the gas under ideal conditions is 257.36K
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2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g CO 2 and 1.943 g H 2O. What is the empirical formula of the compound?
Answers
Answer:
The empirical formula is C3H6O
Explanation:
Step 1: Data given
Mass of the sample =2.088 grams
The mass contains carbon, hydrogen, and oxygen
Mass of CO2 produced = 4.746 grams
Mass of H2O produced = 1.943 grams
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 4.746 grams/ 44.01 g/mol
Moles CO2 = 0.1078 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.1078 moles CO2 we'll have 0.1078 moles C
Step 4: Calculate mass C
Mass C: moles C * atomic mass C
Mass C: 0.1078 moles * 12.01 g/mol
Mass C= 1.295 grams
Step 5: Calculate moles H2O
Moles H2O = 1.943 grams / 18.02 g/mol
Moles H2O = 0.1078 moles
Step 6: Calculate moles H
For 1 mol H2O we'll have 2 moles H
For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H
Step 7: Calculate mass H
Mass H = 0.2046 moles * 1.01 g/mol
Mass H = 0.218 grams
Step 8: Calculate mass O
Mass O = 2.088 grams - 1.295 grams - 0.218 grams
Mass O = 0.575 grams
Step 9: Calculate moles O
Moles O = 0.575 grams / 16.0 g/mol
Moles O = 0.0359 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.1078 moles / 0.0359 moles = 3
H: 0.2156 moles / 0.0359 moles = 6
O: 0.0359 moles / 0.0359 moles =1
The empirical formula is C3H6O
The empirical formula of the compound is C₃H₆O
We'll begin by calculating the mass of C, H and O in the compound. This can be obtained as follow:
For C:Mass of CO₂ = 4.746 g
Molar mass of CO₂ = 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?Mass of C = 12/44 × 4.746
Mass of C = 1.294 gFor H:Mass of H₂O = 1.943 g
Molar mass of H₂O = 18 g/mol
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H =?Mass of H = 2/18 × 1.943
Mass of H = 0.216 gFor O:Mass of C = 1.294 g
Mass of H = 0.216 g
Mass of compound = 2.088 g
Mass of O =?Mass of O = (Mass of compound ) – (mass of C + mass of H)
Mass of O = 2.088 – (1.294 + 0.216)
Mass of O = 0.578 gFinally, we shall determine the empirical formula of the compound. This can be obtained as follow:C = 1.294 g
H = 0.216 g
O = 0.578 g
Empirical formula =?Divide by their molar mass
C = 1.294 / 12 = 0.108
H = 0.216 / 1 = 0.216
O = 0.578 / 16 = 0.036
Divide by the smallest
C = 0.108 / 0.036 = 3
H = 0.216 / 0.036 = 6
O = 0.036 / 0.036 = 1
Therefore, the empirical formula of the compound is C₃H₆O
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An accident happens in the lab of Professor Utonium, and a radioactive element X is released in the form of a gas at around 4:00 am. Element X has a short half-life (25 min), and the lab would be considered safe when the concentration of X drops by a factor of 10. Considering the decomposition of element X is of first-order, what is the earliest time Professor Utonium can come back to do experiments in the lab
Answers
Answer:
5:22 am
Explanation:
The gas X decays following a first-order reaction.
The half-life ([tex]t_{1/2}[/tex]) is 25 min. We can find the rate constant (k) using the following expression.
[tex]k = \frac{ln2}{t_{1/2}} =\frac{ln2}{25min} = 0.028 min^{-1}[/tex]
We can find the concentration of X at a certain time ([tex][X][/tex]) using the following expression.
[tex][X] = [X]_0 \times e^{-k \times t}[/tex]
where,
[tex][X]_0[/tex]: initial concentration of X
t: time elapsed
[tex]\frac{[X]}{[X]_0}= e^{-k \times t}\\\frac{1/10[X]_0}{[X]_0}= e^{-0.028min^{-1} \times t}\\t=82min[/tex]
The earliest time Professor Utonium can come back to do experiments in the lab is:
4:00 + 82 = 5:22 am
Final answer:
The earliest Professor Utonium can return to the lab after a radioactive release is approximately 5:15 am, based on the half-life of 25 minutes and the requirement for the concentration of the gas to drop by a factor of 10, corresponding to just over 3 half-lives.
Explanation:
The question asks for the earliest time Professor Utonium can return to the lab after a release of a radioactive gas, X, which has a half-life of 25 minutes, and the lab is considered safe when its concentration drops by a factor of 10. Understanding that the decay of the radioactive element follows first-order kinetics, we can calculate the time required for the concentration to drop by this factor.
First-order decay implies that the time it takes for a substance to decay to half its initial amount is constant, known as the half-life. To reduce the concentration of a substance by a factor of 10, we need to go through a certain number of half-lives. The formula for calculating the amount of substance remaining after a given time is N = N0,[tex](1/2)^{(t/t1/2)}[/tex] where N is the remaining amount, N0 is the initial amount, t is time, and t1/2 is the half-life.
To reduce the concentration by a factor of 10, we effectively need the substance to go through just over 3 half-lives (since (1/2)³ = 1/8, which is just a bit more than one-tenth). Therefore, the calculation is 3 * 25 = 75 minutes after the initial release. Since the accident happened at around 4:00 am, adding 75 minutes means the earliest Professor Utonium can return to the lab is approximately 5:15 am.
Within each ______ in the periodic table, elements have similar properties because they have the same number of valence electrons.
Period
Row
Group
Metals
Plz this is really really really urgent
Answers
Answer:
group
Explanation:
elements in the same group have the same number of valence electrons
There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. Therefore, the correct option is option C.
What is periodic table?Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.
Periodic table help a scientist to know what are the different types of elements are present in periodic table so that they can discover the new elements that are not being discovered yet.
Within each group in the periodic table, elements have similar properties because they have the same number of valence electrons.
Therefore, within each group in the periodic table, elements have similar properties because they have the same number of valence electrons. The correct option is option C.
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affects of cholera toxin on adenylyl cyclase the gram negative bacterium vibrio cholerae produces a protein cholera toxin that is respondible for th e characteristic symptoms of cholera. if body fluids and Na are not replaced, severe dehydration results. what is the effect of cholera toxin on cAMP in the intestinal cells
Answers
Answer: the effects of cholera toxin on cAMP in the intestinal cells is that it INCREASES cAMP production.
Explanation:
Vibrio cholerae is a gram negative bacterium which produces a protein cholera toxin that is responsible for the characteristic symptoms of cholera such as
watery diarrhea, vomiting, rapid heart rate, loss of skin elasticity, low blood pressure, thirst, and muscle cramps. There is need for body fluids and Na replacement to avoid severe dehydration results which may lead to death.
Cyclic adenosine monophosphate( cAMP) is a derivative used for intracellular signal transduction in organisms. The cholera toxin produced by the bacteria INCREASES the production of cAMP through its polypeptides( which consist of active protomer and binding protomer). The cholera toxin first binds to cell surface receptors, the protomer then enters the cell and bind with and activate the adenylate cyclase. Increasing adenylate cyclase activity will INCREASE cellular levels of cAMP, increasing the activity of ion pumps that remove ions from the cell. Due to osmotic pressure changes, water also must flow with the ions into the lumen of the intestinal mucosa, dehydrating the tissue. I hope this helps, thanks.
CaO(s) + H2O(l) - Ca(OH)2(s)
enthalpy of rxn= -63.7 kJ/molrxn
Calcium oxide, CaO(s), has been proposed as a substance that can be used to heat water quickly for portable heating packs or for cooking. When placed in water, Cao(s) reacts as shown by the equation above.
A student wants to design a heating pad that could heat a 150.0 g sample of water from 25.0°C to 60.0°C.
Calculate the amount of heat, in joules, that the water must absorb for its
temperature to change by this amount. (Assume that the specific heat capacity
of the water is 4.18 J/gK).
Answers
Answer:
21,976 J
Explanation:
In order to increase the temperature of a certain amount of a substance by [tex]\Delta T[/tex], the amount of heat that must be supplied to the substance must be:
[tex]Q=mC\Delta T[/tex]
where
m is the mass of the substance
C is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the increase in temperature
For the sample of water in this problem we have:
[tex]m=150.0 g[/tex] is the mass
[tex]C=4.186 J/g^{\circ}C[/tex] is the specific heat capacity of water
[tex]\Delta T=60.0-25.0=35.0^{\circ}C[/tex] is the increase in temperature
Therefore, the amount of heat that must be supplied is
[tex]Q=(150.0)(4.186)(60-0-25.0)=21,976 J[/tex]
To heat 150.0 g of water from 25.0°C to 60.0°C, approximately 21885 joules of heat must be absorbed, calculated using the formula q = mcΔT with the values m = 150.0 g, c = 4.18 J/gK, and ΔT = 35.0°C.
Explanation:The amount of heat required to raise the temperature of 150.0 g of water from 25.0°C to 60.0°C can be calculated using the formula q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given values, the mass m = 150.0 g, specific heat capacity c = 4.18 J/gK, and the temperature change ΔT = (60.0 - 25.0)°C = 35.0°C, we get:
q = (150.0 g)(4.18 J/gK)(35.0 K)
q = 150.0 g * 4.18 J/g°C * 35.0°C
q = 21885 J
Therefore, the water must absorb approximately 21885 joules of heat for the temperature to increase from 25.0°C to 60.0°C.
Enter your answer in the provided box.A mixture of helium and neon gases is collected over water at 28°C and 791 mmHg. If the partial pressure of helium is 381 mmHg, what is the partial pressure of neon? (Vapor pressure of water at 28°C is 28.3 mmHg.)
Answers
Answer:
Explanation:
Using Dalton's law of partial pressure
P total pressure = Pressure of helium + Pressure of neon + Vapor pressure of water
P = 28.3 mmHg, Pressure of helium = 381 mmHg, Vapor pressure of water at 28°C
791 mmHg - 381 mmHg - 28.3 mmHg = Pressure of neon
Pressure of neon = 381.7 mmHg
What is the pH of a 0.0538 M solution of Sr(OH)2?
Answers
Answer:
pH= 13.03
Explanation:
Since dissociation of Sr(OH)2= Sr2+ + 2[OH-]
[OH-]=2×0.0538 = 0.1076M
pOH= -log[0.1076]= 0.97
pH= 14-pOH = 14-0.97= 13.03
What is the solubility in moles/liter for magnesium hydroxide at 25 oC given a Ksp value of 1.1 x 10-11. Write using scientific notation and use 1 or 2 decimal places (even though this is strictly incorrect!)
Answers
Answer:
S = 0.00014 moles /L = 1.4 * 10^-4 moles/L
Explanation:
Step 1: Data given
Temperature = 25.0 °C
Ksp = 1.1 * 10^-11
Step 2: The balanced equation
Mg(OH)2(s) ⇆ Mg^2+(aq) + 2OH-(aq)
Step 3: Define Ksp
[Mg(OH)2 = 1.11 * 10^-11 = S
[Mg^2+] = S
[OH-] = 2S
Ksp = [Mg^2+]*[OH-]²
Ksp = S * (2S)²
1.1 * 10^-11 = 4S³
S³ = 2.75 * 10^-12
S = 0.00014 moles /L
Which of the following statements applies to the E2 mechanism? Which of the following statements applies to the E2 mechanism? It occurs with inversion of stereochemistry. It occurs with racemization of stereochemistry. The C-H and C-X bonds that break must be anti. Use of a bulky base gives the more highly substituted alkene product. It proceeds through the more stable carbocation intermediate.
Answers
Final answer:
The E2 reaction is a type of elimination mechanism where a base removes a proton from a carbon adjacent to one with a leaving group, leading to the creation of a double bond. It requires the bonds to be anti-aligned and uses a bulky base to favor the formation of more highly substituted alkenes, with no intermediates formed.
Explanation:
The E2 elimination reaction is a concerted process where a base removes a proton (H) adjacent to a carbon with a leaving group (often denoted as 'X'), resulting in the formation of a double bond. The key characteristics of the E2 mechanism include the requirement for the C-H and C-X bonds to be in an anti-periplanar arrangement, ensuring they are aligned properly for the elimination to occur. Additionally, the use of a bulky base typically leads to the formation of the more highly substituted alkene product, favoring the more stable Zaitsev product. There is also no carbocation intermediate in the E2 mechanism; instead, the reaction proceeds through a single concerted step without intermediates. The E2 mechanism is typically observed with secondary and tertiary substrates where steric hindrance inhibits SN2 reactions.
his mechanism has been proposed for the reaction between chloroform and chlorine. Step 1: Cl2(g) 2Cl(g) fast Step 2: CHCl3(g) Cl(g) CCl3(g) HCl(g) slow Step 3: CCl3(g) Cl(g) CCl4(g) fast a. Write the stoichiometric equation for the overall reaction. b. Identify any reaction intermediates in this mechanism. c. Write the rate equation for the rate determining step. d. Show how the rate equation in c. can be used to obtain the rate law for the overall reaction. e. If the concentrations of the reactants are doubled, by what ratio does the reaction rate change
Answers
Answer:
a) Balanced Overall Stoichiometric Equation
CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)
b) The reaction intermediates include Cl(g) and CCl₃(g)
c) The rate of reaction of the rate determining step is:
Rate = k [CHCl₃] [Cl]
d) The rate of overall reaction is given as
Rate = K [CHCl₃] √[Cl₂]
e) The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.
Explanation:
Step 1: Cl₂(g) → 2Cl(g) fast
Step 2: CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow
Step 3: CCl₃(g) + Cl(g) → CCl₄(g) fast
a) The overall reaction is a reaction between Chloroform (CHCl₃) and Chlorine (Cl₂). It can be obtained by summing all the elementary equations and eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations).
Cl₂(g) + CHCl₃(g) + Cl(g) + CCl₃(g) + Cl(g) → 2Cl(g) + CCl₃(g) + HCl(g) + CCl₄(g)
Now, eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations), we are left with
Cl₂(g) + CHCl₃(g) → HCl(g) + CCl₄(g)
CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)
b) Like I mentioned in (a), the reaction intermediates are the species that appear on both sides upon summing up all the elementary equations. They include:
Cl(g) and CCl₃(g)
c) The rate determining step is usually the slowest step among the elementary equations.
The rate of reaction expression is usually written from the slowest step.
The slowest step is step 2.
CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow
The rate of reaction is then given as
Rate = k [CHCl₃] [Cl]
d) The rate of reaction for the overall reaction is obtained by substituting for any intermediates that appear in the rate of reaction for the rate determining step
The rate of reaction of the rate determining step is:
Rate = k [CHCl₃] [Cl]
But we can substitute for [Cl] by obtaining an expression for it from the step 1.
Step 1: Cl₂(g) → 2Cl(g)
K₁ = [Cl]²/[Cl₂]
[Cl]² = K₁ [Cl₂]
[Cl] = √{K₁ [Cl₂]}
We then substitute this into the rate determining step
Rate = k [CHCl₃] [Cl]
Rate = k [CHCl₃] √{K₁ [Cl₂]}
Rate = (k)(√K₁) [CHCl₃] √[Cl₂]
Let (k)(√K₁) = K (the overall rate constant)
Rate = K [CHCl₃] √[Cl₂]
e) If the concentrations of the reactants are doubled, by what ratio does the reaction rate change
Old Rate = K [CHCl₃] √[Cl₂]
If the concentrations of the reactants are doubled, the new rate would be
New Rate = K × 2[CHCl₃] × √{2 × [Cl₂]}
New Rate = K × 2[CHCl₃] × √2 × √[Cl₂]
New Rate = 2√2 K [CHCl₃] √[Cl₂]
Old Rate = K [CHCl₃] √[Cl₂]
New Rate = 2√2 × (Old Rate)
The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.
Hope this Helps!!!!
How do ice and water on the ground affect incoming solar radiation?
Answers
Answer:
Ice and water on the ground affect incoming solar radiation by reflecting 4 percent of solar radiation that reaches the surface.
Explanation:
Answer:
ice and water on the ground affect incoming solar radiation by reflecting 4 percent of solar radiation that reaches the surface. the state of water and the sun angle are vital in determining the amount of reflection that take place. at low sun angle and at times when the surface is ice, more reflection occurs.
Explanation:
None
In the following reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq)
a. HPO42- is an acid and H2PO4- is its conjugate base.
b. H2PO4- is an acid and F- is its conjugate base.
c. HPO42- is an acid and HF is its conjugate base.
d. HF is an acid and F- is its conjugate base.
e. HF is an acid and HPO42- is its conjugate base.
Answers
d. HF is an acid and [tex]F^-[/tex] is its conjugate base.
What is acid/ base and its conjugate acid/base?
Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair.For example: HCl is an acid and its conjugate base is [tex]Cl^-[/tex].Given chemical reaction:
[tex]HF(aq) + HPO_4^{2-}(aq)----> F^-(aq) + H_2PO_4^-(aq)[/tex]
In this reaction, HF is an acid and its conjugate base is [tex]F^-[/tex].
Thus out of all the options; the correct option is d.
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In the reaction HF(aq) + HPO4^2-(aq) = F^-(aq) + H2PO4^-(aq), HF is the acid and F^- is the conjugate base, making the correct answer (d).
Explanation:In the given reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq), we need to identify the correct acid-base pairs. When looking at this reaction, HF donates a proton (H+) to form its conjugate base F-, which makes HF the acid. On the other hand, since HPO42- receives a proton (H+) to become H2PO4-, HPO42- clearly acts as the base in this reaction and H2PO4- is its conjugate acid, which shows that it is amphoteric.
Based on this explanation, the correct answer is (d) HF is an acid and F- is its conjugate base.
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A gas is placed in a storage tank at a pressure of 30.0 atm at 20.3 C. As a safety device, a small metal plug in the tank is made of a metal alloy that melts at 130 C. If the tank is heated, what is the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas?
Answers
Answer: 41.2 atm
Explanation
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=30.0atm\\T_1=20.3^0C=(20.3+273)=293.3K\\P_2=?\\T_2=130^0C=(130+273)K=403K[/tex]
Putting values in above equation, we get:
[tex]\frac{30.0}{293.3K}=\frac{P_2}{403}\\\\P_2=41.2atm[/tex]
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.2
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
From the question given above, the following data were obtained:
Initial pressure (P₁) = 30 atm
Initial temperature (T₁) = 20.3 °C = 20.3 + 273 = 293.3 K
Final temperature (T₂) = 130 °C = 130 + 273 = 403 K
Final pressure (P₂) =?The final pressure can be obtained as illustrated below:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\\frac{30}{293.3} = \frac{P_{2}}{403}\\\\[/tex]
Cross multiply
293.3 × P₂ = 30 × 403
293.3 × P₂ = 12090
Divide both side by 293.3
[tex]P_{2} = \frac{12090}{293.3} \\\\[/tex]
P₂ = 41.22 atmTherefore, the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
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Liquid nitrogen trichloride is heated in a 2.50−L closed reaction vessel until it decomposes completely to gaseous elements. The resulting mixture exerts a pressure of 818 mmHg at 95°C. What is the partial pressure of each gas in the container?
Answers
Answer:
1. Partial pressure of N2 is 204.5 mmHg
2. Partial pressure of Cl2 is 613.5 mmHg
Explanation:
Step 1:
The equation for the reaction. This is given below:
NCl3 —> N2 (g) + Cl2 (g)
Step 2:
Balancing the equation.
NCl3 (l) —> N2 (g) + Cl2 (g)
The above equation is balanced as follow:
There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by putting 2 in front of NCl3 as shown below:
2NCl3 (l) —> N2 (g) + Cl2 (g)
There are 6 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of Cl2 as shown below:
2NCl3 (l) —> N2 (g) + 3Cl2 (g)
Now the equation is balanced.
Step 2:
Determination of the mole fraction of each gas.
From the balanced equation above, the resulting mixture of the gas contains:
Mole of N2 = 1
Mole of Cl2 = 3
Total mole = 4
Therefore, the mole fraction for each gas is:
Mole fraction of N2 = mole of N2/total mole
Mole fraction of N2 = 1/4
Mole fraction of Cl2 = mole of Cl2/total mole
Mole fraction of Cl2 = 3/4
Step 3:
Determination of the partial pressure of N2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of N2 = 1/4
Partial pressure of N2 = 1/4 x 818
Partial pressure of N2 = 204.5 mmHg
Step 4:
Determination of the partial pressure of Cl2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of Cl2 = 3/4
Partial pressure of Cl2 = 3/4 x 818
Partial pressure of Cl2 = 613.5 mmHg
Answer:
[tex]p_{N_2}=204.5mmHg\\p_{Cl_2}=613.5mmHg[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NCl_3(g)\rightarrow N_2(g)+3Cl_2(g)[/tex]
Thus, by knowing that the nitrogen trichloride is completely decomposed, one assumes there is one mole of nitrogen and three moles of chlorine (stoichiometric coefficients) as a basis to compute the partial pressures since they have the mole ratio from the nitrogen trichloride. Hence, the mole fractions result:
[tex]x_{N_2}=\frac{1}{1+3}=0.25\\ x_{Cl_2}=1-0.25=0.75[/tex]
In such a way, for the final pressure 818 mmHg, the partial pressures become:
[tex]p_{N_2}=x_{N_2}p_T=0.25*818mmHg=204.5mmHg\\p_{Cl_2}=x_{Cl_2}p_T=0.75*818mmHg=613.5mmHg[/tex]
Best regards.
Compound A has a solubility of 0.2 g/mL in toluene at toluene's boiling point and a solubility of 0.05 g/mL at 0 ºC. How much toluene would be necessary to recrystallize 3.2 g of A. What would be the maximum amount of A that could be recovered if the saturated solution was allowed to cool to 0 ºC. How much A would be recovered, if you accidentally used twice as much toluene as was necessary?
Answers
Answer:
1) 16 mL of toluene is necessary to recrystallize 3.2 g of compound A
2) 2.4 g of A is the maximum amount that could be recovered.
3) 1.6 g of compound A can be recovered if you accidentally use twice as much toluene as was necessary
Explanation:
1)
The volume of the solvent ( toluene) = [tex]\frac{starting \ amount }{solubility \ at \boiling \ point }[/tex]
= [tex]\frac{3.2 \ g}{0.2 \ g/mL}[/tex]
= 16 mL
∴ 16 mL of toluene is necessary to recrystallize 3.2 g of compound A
2)
The maximum amount A that could be recovered if the saturated solution was allowed to cool to 0 ºC is determined by the difference of the starting amount and amount left in the solution at 0 ºC
i.e
maximum amount of A = 3.2 - ( 16 mL × 0.05 g/mL)
= 3.2 g - 0.8 g
= 2.4 g
∴ 2.4 g of A is the maximum amount that could be recovered.
3)
Amount of A that would be recovered, if you accidentally used twice as much toluene as was necessary is calculated as follows;
amount of A = 3.2g - (32 mL × 0.05 g/mL)
= 3.2 g - 1.6 g
= 1.6 g
Thus; 1.6 g of compound A can be recovered if you accidentally use twice as much toluene as was necessary
Esters can be synthesized by an acid-catalyzed nucleophilic acyl substitution between an alcohol and a carboxylic acid; this process is called the Fischer esterification reaction. Because the alcohol oxygen is a poor nucleophile, the carbonyl carbon is made a better electrophile by protonation of the carbonyl oxygen. The steps of the synthesis are all reversible. The reaction is generally driven to completion by using an excess of the liquid alcohol as a solvent, or by distilling off the product as it forms. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answers
Answer:
See attachment for mechanism.
Explanation:
The Fischer esterification reaction is a nucleophilic substitution in the acyl group of a carboxylic acid, catalyzed by an acid.
1) The protonation of the oxygen of the carbonyl group activates the carboxylic acid...
2) ... against a nucleophilic attack by the alcohol, and produces a tetrahedral intermediate.
3) The transference of a proton from an oxygen atom to another produces a second tetrahedral intermediate and converts the -OH into a good leaving group.
4) The loss of a proton and the expulsion of H₂O regenerates the acid catalyzer and gives an ester as a product.
The Fischer esterification reaction involves reacting a carboxylic acid with an alcohol to form an ester. The reaction is driven to completion by using an excess of alcohol or by distilling off the product.
Explanation:
In the
Fischer esterification reaction
, an alcohol (for example, ethanol) reacts with a carboxylic acid (like acetic acid) to form an ester. The carbonyl carbon of the acid is protonated to make it a better electrophile that can react with the alcohol's oxygen. Here's a simplified version of what happens:
R-C-0-H (a carboxylic acid) + :O: (the oxygen of an alcohol) -> R-C(=O)-O-R' (an ester)
, where R and R' represent any organic groups. The reaction is balanced by distilling off the ester product or using an excess of the alcohol solvent. In a more detailed mechanism, curved arrows would be used to show the movement of electrons, but those can't be represented in text form.
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A hydrogen atom can be in the 1S state, whose energy we'll call 0, the 2S state, or any of 3 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total?
Answers
Answer:
-- 5.8×10⁻⁹ of the H is in 2P states at T=5900 K, a typical Sun surface temperature.
-- 3.3×10⁻¹² of the H is in 2P states at T=4300 K, a typical Sunspot temperature.
Explanation:
Answer:
Explanation:
your question seems to be uncompleted , but however i have a solution to a similar question, check the attachment and follow the format to evaluate your question.
check the attachement
A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a 900.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid sodium hydroxide and add it to the flask. Fill the flask to the mark with distilled water.
Answers
Answer:
28.58 g of NaOH
Explanation:
The question is incomplete. The missing part is:
"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"
To do this, we need to know how much of the base we have to weight to prepare this solution.
First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:
NaOH -------> Na⁺ + OH⁻
As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.
This can be done with the following expression:
14 = pH + pOH
and pOH = -log[OH⁻]
So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:
pOH = 14 - 13.9 = 0.10
[OH⁻] = 10⁽⁻⁰°¹⁰⁾
[OH⁻] = 0.794 M
Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:
n = M * V
n = 0.794 * 0.9
n = 0.7146 moles
Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:
m = 39.997 * 0.7146
m = 28.58 gTreating (CH3)3C-Cl with a mixture of H2O and CH3OH at room temperature would yield: A) CH2=C(CH3)2 B) (CH3)3COH C) (CH3)3COCH3 D) All of these choices. E) None of these choices.
Answers
Answer:
All of the choices.
Explanation:
As the reaction is involving with a mixture of H₂O and CH₃OH, these two reagents can work as nucleophyle of the reaction. Both of them, are polar and promoves a Sn1/E1 reaction. When it reacts with water it will produce product B); it will form product C) when it reacts with methanol, and product A) will be formed when the reaction undergoes an E1 reaction.
In this case, the only way to show you this, it's doing the mechanism of reaction for each product. Picture attached show the mechanism for the formation of all these products.
Treating (CH3)3C-Cl with H2O and CH3OH usually results in the formation of tert-butyl alcohol, (CH3)3COH, through a nucleophilic substitution reaction, making option B the correct answer.
Explanation:Treating (CH3)3C-Cl with a mixture of H2O and CH3OH at room temperature typically involves a nucleophilic substitution reaction where the chloride ion (Cl-) is replaced by the nucleophile. In this case, the nucleophiles are water (H2O) and methanol (CH3OH). The product of this reaction would be (CH3)3COH, also known as tert-butyl alcohol, as both water and methanol could act as nucleophiles to perform the substitution, but water is generally a better nucleophile than methanol. Hence, option B is the correct answer.
You have a 25.0 L cylinder of helium at a pressure of 132 atm and a temperature of 19 [infinity]C. The He is used to fill balloons to a volume of 2.50 L at 732 mm Hg and 27 [infinity]C. How many balloons can be filled with He? Assume that the cylinder can provide He until its internal pressure reaches 1.00 atm (i.e., there are 131 atmospheres of usable He in the cylinder).
Answers
Answer: Number of balloons that can be filled with He are 1397.
Explanation:
The given data is as follows.
V = 25 L He , P = 131 atm
T = [tex]19^{o}C[/tex] = (19 + 273) K
= 292 K
According to ideal gas equation,
PV = nRT
where, n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{131 \times 25}{0.082 \times 292}[/tex]
= 136.76 mol of He
The data for small balloons is given as follows.
T = [tex]27^{o}C[/tex] = (27 + 273) K
= 300 K
P = 732 = 0.963 atm
V = 2.50 L
Now, we will calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{0.963 \times 2.50}{0.082 \times 300}[/tex]
= 0.098 mol
So, number of balloons that can be filled with He are calculated as follows.
n = [tex]\frac{N_{1}}{N_{2}}[/tex]
= [tex]\frac{136.76}{0.098}[/tex]
= 1397.60 balloons
or, = 1397 balloons (approx)
Thus, we can conclude that number of balloons that can be filled with He are 1397.
50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol
Answers
Answer:
Approximately [tex]0.47\; \rm mol \cdot L^{-1}[/tex] (note that [tex]1\; \rm M = 1 \; \rm mol \cdot L^{-1}[/tex].)
Explanation:
The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this [tex]\rm NaNO_3[/tex] solution in water,
Let [tex]n[/tex] be the number of moles of the solute in the whole solution. Let [tex]V[/tex] represent the volume of that solution. The formula for the molarity [tex]c[/tex] of that solution is:
[tex]\displaystyle c = \frac{n}{V}[/tex].
In this question, the volume of the solution is known to be [tex]1250\; \rm mL[/tex]. That's [tex]1.250\; \rm L[/tex] in standard units. What needs to be found is [tex]n[/tex], the number of moles of [tex]\rm NaNO_3[/tex] in that solution.
The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of [tex]\rm NaNO_3[/tex] is [tex]85\; \rm g \cdot mol^{-1}[/tex] means that the mass of one mole of
[tex]\displaystyle n = \frac{m}{M}[/tex].
For this question,
[tex]\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}[/tex].
Calculate the molarity of this solution:
[tex]\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}[/tex].
Note that [tex]1\; \rm mol \cdot L^{-1}[/tex] (one mole per liter solution) is the same as [tex]1\; \rm M[/tex].
Answer:
.47
Explanation:
just did it on CK-12