The given question is incomplete, the complete question is:
A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.
Answer:
The correct answer is 32 grams.
Explanation:
Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.
Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,
= 0.52/1000 × 200 = 0.104 moles
The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole
So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,
= 304.23 g/mol × 0.104
= 31.639 grams or 32 grams.
A mercury manometer is used to measure pressure in the container illustrated. Calculate the pressure exerted by the gas if atmospheric pressure is 751 torr and the distance labeled is 176 mm.
Answer:
Pressure exerted by the gas is 574.85 torr
Explanation:
Atmospheric pressure = 751 torr
but 1 torr = 1 mmHg
therefore,
atmospheric pressure = 751 mmHg
1 mmHg = 133.3 Pa
therefore,
atmospheric pressure = 751 x 133.3 = 100108.3 Pa
distance labeled (tube section with mercury) = 176 mm
the pressure within the tube will be
[tex]P_{tube}[/tex] = ρgh
where ρ is the density of mercury = 13600 kg/m^3
h is the labeled distance = 176 mm = 0.176 m
g is acceleration due to gravity = 9.81 m/s^2
[tex]P_{tube}[/tex] = 13600 x 9.81 x 0.176 = 23481.216 Pa
The general equation for the pressure in the manometer will be
[tex]P_{atm}[/tex] = [tex]P_{tube}[/tex] + [tex]P_{gas}[/tex]
where [tex]P_{atm}[/tex] is the atmospheric pressure
[tex]P_{tube}[/tex] is the pressure within the tube with mercury
[tex]P_{gas}[/tex] is the pressure of the gas
substituting, we have
100108.3 = 23481.216 + [tex]P_{gas}[/tex]
[tex]P_{gas}[/tex] = 100108.3 - 23481.216 = 76627.1 Pa
This pressure can be stated in mmHg as
76627.1 /133.3 = 574.85 mmHg
and also equal to 574.85 torr
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
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Part C: complete the third column
Part D: complete the fourth column
Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
pls help me
Which statement best explains the relationship between the electric force between two charged objects and the distance between them?
a.As the distance increases by a factor, the electric force increases by the square of that factor.
b.As the distance increases by a factor, the electric force increases by twice that factor.
c.As the distance increases by a factor, the electric force decreases by twice that factor.
d.As the distance increases by a factor, the electric force decreases by the same factor. e.As the distance increases by a factor, the electric force decreases by the square of that factor.
Explanation:
The electric force between two charged particles is given by the formula as follows :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here,
k is electrostatic constant
[tex]q_1,q_2[/tex] are charges
r is the distance between charges
The above formula shows that the electric force is inversely proportional to the square of the distance between charges. It means that as the distance increases by a factor, the electric force decreases by the square of that factor. Hence, the correct option is (e).
when air molecules collide with things around us, it produces _______ (2 words), which is measured with a _______.
✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.
Rank the following substances in order from most soluble in water to least soluble in water: ethane, C2H6; 1-pentanol, C5H11OH; potassium chloride, KCl; and propane, C3H8.
Rank from most to least soluble in water. To rank items as equivalent, overlap them.
Most soluble Least soluble
Answer:
Explanation:
The substances are:
-) Ethane, [tex]C_2H_6[/tex]
-) 1-pentanol, [tex]C_5H_1_1OH[/tex]
-) Potassium chloride, [tex]KCl[/tex]
-) Propane, [tex]C_3H_8[/tex]
For this question, we have to remember the structure of water. Due to the electronegativity difference between oxygen and hydrogen in this structure, we will have the formation of dipoles. The dipoles interact better with net charges, due to this, the Potassium chloride is the compound with highest solubility (due to the formation of a cation and an anion):
[tex]KC~l->~K^~+~Cl^-[/tex]
Then, in 1-pentanol we an "OH". This structure due to the presence of the hydroxyl group can form hydrogen bonds. Therefore, this compound would be the second more soluble.
Finally, the difference between propane and ethane is a carbon. In propane, we have an additional carbon. If we have more carbons we will have more area of interaction. If we have more area we will have more solubility therefore propane is more soluble than ethanol.
In conclusion, the rank from most soluble to least soluble is:
1) Potassium chloride, [tex]KCl[/tex]
2) 1-pentanol, [tex]C_5H_1_1OH[/tex]
3) Propane, [tex]C_3H_8[/tex]
4) Ethane, [tex]C_2H_6[/tex]
I hope it helps!
Order of solubility in water will be:
KCl > C₅H₁₁OH > C₃H₈ > C₂H₆
Solubility in water:Any solvent soluble in water due to its polarity and ability to form hydrogen bonds. The presence of hydrogen bonding between molecules of a substance indicates that the molecules are polar. This means the molecules will be soluble in a polar solvent such as water.
Substances that are given:
Ethane(C₂H₆), 1-pentanol(C₅H₁₁OH), Potassium chloride(KCl) and propane(C₃H₈).
We will look at each compound one by one:
Potassium chloride is an ionic compound, it has ionic interactions between its solubility in water is highest due to the formation of potassium ([tex]K^{+}[/tex]) and ([tex]Cl^{-}[/tex]) ions.In 1-pentanol, there is presence of hydroxyl group thus it can easily form hydrogen bonds with water. Therefore it will be soluble in water and comes after potassium chloride in ranking order.In ethane and propane molecule, there is one extra carbon in case of propane due to which it leads to the more area for interactions therefore more area for interaction leads to more solubility thus propane is more soluble than ethane in water.Order of solubility in water will be:
KCl > C₅H₁₁OH > C₃H₈ > C₂H₆
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what is the difference between acidic and basic protein
Answer:
Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.
Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.
Answer:
Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.
Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10
Explanation:
Given the initial rate data for the reaction being A + B + C --> D determine the rate expression for the reaction and the (k) rate constant. (The units of [A] [B] and [C] are all moles/liter and the units of IRR is moles/liter seconds). If the [A]=[B]=[C]=.30M, what would the IRR be? [A] [B] [C] IRR 0.20 0.10 0.40 .20 0.40 0.20 0.20 1.60 0.20 0.10 0.20 .20 0.20 0.20 0.20 .80
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Complete the sentences to best explain the difference. Match the words to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer.
1. The strongest force observed at the surface of glass is:________
2. Water is___________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass
3. Hexane is________ and interacts, generating____________ adhesive interactions with the glass.
a. dipole
b. nonpolar
c. only weak
d. dispersion
e. strong
f. hydrogen bonding
g. polar
Answer:
Explanation:
1 . The strongest force observed at the surface of glass is:__DIPOLE______
2. . Water is__POLAR_________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass.
3 . Hexane is_NON-POLAR _______ and interacts, generating__ONLY WEAK __________ adhesive interactions with the glass.
Find Ecell for an electrochemical cell based on the following reaction with [MnO4−]=1.20M, [H+]=1.50M, and [Ag+]=0.0100M. E∘cell for the reaction is +0.88V. MnO4−(aq)+4H+(aq)+3Ag(s)→MnO2(s)+2H2O(l)+3Ag+(aq)
Answer:
1.01 V
Explanation:
From Nernst equation;
Ecell= E°cell- 0.0592/n log Q
Where;
Ecell= observed emf of the cell
E°cell= standard emf of the cell
n= number of moles of electrons transferred
Q= reaction quotient
Q= [Ag^+]^3/[MnO4^-] [H^+]^4
Q= [0.01]^3/[1.20] [1.50]^4
Q= 1.65×10^-7
Ecell= 0.88 - 0.0592/3 log 1.65×10^-7
Ecell= 0.88 - [0.0197×(-6.78)]
Ecell= 0.88 + 0.134
Ecell= 1.01 V
Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene
Answer:
Explanation:
a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .
C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O
O is provided by KMnO₄
b ) In this reaction isophthalic acid is formed .
C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂
c)
4-Propyl-3-t-butyltoluene
In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .
The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )
Butane gas (C4H10) burns in oxygen gas to produce carbon dioxide gas and water vapor. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction (listed in the same order as given in the problem).
Answer:
C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)
2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)
Explanation:
Butane gas (C₄H₁₀) burns in oxygen gas to produce carbon dioxide gas and water vapor. The unbalanced equation is:
C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)
First, we will balance carbon and hydrogen which are in just one compound on each side.
C₄H₁₀(g) + O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)
Finally, we will balance the oxygen atoms.
C₄H₁₀(g) + 6.5 O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)
In order to have integers, we will multiply everý compound by 2.
2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)
When a solution is diluted with water, the ratio of the initial to final
volumes of solution is equal to the ratio of final to initial molarities
Select one:
True
-
When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.
Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.
There are various methods of expressing the concentration of a solution.
Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.
Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.
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A compound is found to contain 63.65 % nitrogen and 36.35 % oxygen by mass. The molar mass for this compound is 44.02 g/mol. The molecular formula for this compound is
Answer:
THE MOLECULAR FORMULA FOR THE COMPOUND IS N20
Explanation:
To calculate the molecular formula for the compound, we follow the following steps:
Write out the percentage abundance of the individual elements
N = 63.65 %
O = 36.35 %
2. Divide the percentage composition by the atomic masses of the elements
N = 63.65 / 14 = 4.546
O = 36.35 / 16 = 2.272
3. Divide the values by the lowest value
N = 4.546 / 2.272 = 2.00
O = 2.202 / 2.272 = 1
4. The empirical formula of the compound will be:
N2O
5. Calculate the molecular mass
(N2O ) x = 44.02 g/mol.
(14 * 2 + 16) x = 44.02
(28 + 16) x = 44.02
44 x = 44.02
x = 44.02 / 44
x = 1
The molecular formula for the compound is N2O
the equilibrium constant at 250C for the equation PCl5 PCl3 is kp 4.5 x 10^3 bar calculate value of grxn at 250C in which direction is reaction spontaneous when pcl3 cl2 standard conditions
Answer:
-36.6 kJ·mol⁻¹
Explanation:
1. Calculate ΔG
[tex]\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K[/tex]
T = (250 + 273.15) K = 523.15 K
[tex]\begin{array}{rcl}\Delta G & = & -8.314 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times \text{523.15 K} \ln (4.5 \times 10^{3}) \\& = & -4349\text{ J}\cdot \text{mol}^{-1} \times 8.412\\& = & \text{-36 590 J}\cdot \text{mol}^{-1}\\& = & \textbf{-36.6 kJ}\cdot \text{mol}^{-1}\\\end{array}[/tex]
2. Direction of spontaneity
ΔG is negative, so the reaction is spontaneous in the forward direction.
Calculate the mass defect in Mo-96 if the mass of a Mo-96 nucleus is 95.962 amu. The mass of a proton is 1.00728 amu and the mass of a neutron is 1.008665 amu.
Answer:
0.81167 amu
Explanation:
Number of protons=42
Number of neutrons = 54
Mass of all the 42 protons = 42× 1.00728= 42.30576 amu
Mass of all 54 neutrons= 54 × 1.008665 = 54.46791 amu
Calculated mass of protons and neutrons in Mo-96 nucleus= 42.30576 amu + 54.46791 amu= 96.77367 amu
Actual mass of Mo-96 nucleus= 95.962 amu
Therefore mass defect of Mo-96 nucleus= 96.77367 amu - 95.962 amu = 0.81167 amu
Give the name of the following molecule
Answer:
[tex]\boxed{Heptene}[/tex]
Explanation:
Double Bond => An Alkene molecule
So, the suffix will be "-ene"
7 Carbons => So, we'll use the prefix "Hept-"
Combining the suffix and prefix, we get:
=> Heptene
Answer:
[tex]\boxed{\mathrm{Heptene}}[/tex]
Explanation:
Alkenes have double bonds. The molecule has one double bond.
Suffix ⇒ ene
The molecule has 7 carbon atoms and 14 hydrogen atoms.
Prefix ⇒ Hept (7 carbons)
The molecule is Heptene.
[tex]\mathrm{C_7H_{14}}[/tex]
What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate (HgSO4) in aqueous sulfuric acid (H2O/H2SO4)
Answer:
heptan-2-one
Explanation:
In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.
The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.
See figure 1
I hope it helps!
A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 32.3 mg produced 87.7 mg of CO2 and 18.0 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Answer:
Empirical formula: C₅H₅O
Molecular formula: C₁₀H₁₀O₂
Explanation:
When a compound containing C, H and O elements is combusted, the general reaction is:
CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O
Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.
Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =
2.0x10⁻³ moles of CO₂ = moles C
Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =
1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H
The mass of the moles of C and H are:
2x10⁻³ moles C ₓ (12g / mol) = 0.024g C
2x10⁻³ moles H ₓ (1g / mol) = 0.002g H
Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O
Moles are:
0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O
Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:
C: 2.0x10⁻³ / 4x10⁻⁴ = 5
H: 2.0x10⁻³ / 4x10⁻⁴ = 5
O: 4x10⁻⁴ / 4x10⁻⁴ = 1
Thus, empirical formula is:
C₅H₅OThe molar mass of the empirical formula is:
12×5 + 1×5 + 16×1 = 81g/mol
As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:
C₁₀H₁₀O₂1. In the addition of HBr to conjugated dienes, is the product which results from 1,2-addition or that which results from 1,4-addition the product of kinetic control?
A. From 1,2-addition
B. From 1,4-addition
2. Which of the following is the strongest acid?
A. CH3CH20H
B. CHзOCH3
C. CH3CH
D. CH3COCH3
E. CH3COH
Answer:
The answer to this question can be defined as follows:
In question 1, the answer is "Option A".
In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".
Explanation:
In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:
The product From 1,2-addition as its consequence of 1,4-addition is the result of kinetic regulation by HBr in conjugated dienes.The chemical name of the [tex]CH_3COOH[/tex] is the acetic acid, it is one of the carboxylic acids quite basic. It is a major chemical production factor for use as disposable soft drinks, movies or wood glue, polyethylene terephthalate, and many plastics, fibers, and fabrics. It is also used in the storage of the water and soft drinks in the bottles.Initial mass of triphenyl methanol in g = 0.220g Initial volume of 33% HBr solution in ml = 0.60 ml Find mas of triphenyl bromide in g = 240 g
Answer:
0.792g of triphenyl bromide are produced.
Explanation:
The reaction of triphenyl methanol with HBr is:
triphenyl methanol + HBr → Triphenyl bromide.
Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.
To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first limiting reactant:
Moles triphenyl methanol (Molar mass: 260.33g/mol) =
0.220g × (1mol / 260.33g) = 8.45x10⁻³ moles Triphenyl methanol
Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =
0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = 2.45x10⁻³ moles HBr
As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is limiting reactant.
As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = 2.45x10⁻³ moles
As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:
2.45x10⁻³ moles × (323.2g / mol) =
0.792g of triphenyl bromide are produced.18. All of the following sets of quantum numbers are allowed EXCEPT a. n = 1, = 0, = 0 b. n = 2, = 2, = +1 c. n = 3, = 1, = –1 d. n = 4, = 1, = 0 e. n = 5, = 4, = –3
Answer:
b
Explanation:
it is impossible for n & l to be equal
There are 4 quantum numbers:
Principal Quantum number [tex](n)[/tex] specifies the energy of the electron in a shell.Azimuthal Quantum number [tex](l)[/tex] specifies the shape of an orbital. The value of it lies in the range of 0 to (n-1)Magnetic Quantum number [tex](m)[/tex] specifies the orientation of the orbital in space. The value of it lies in the range of -l to +lSpin Quantum number [tex](s)[/tex] specifies the spin of an electron in an orbital. It can either have a value of [tex]+\frac{1}{2}[/tex] or [tex]-\frac{1}{2}[/tex]To find the forbidden set, we need to know the following facts about quantum numbers:
The value of [tex]l[/tex] will always be less than the value of [tex]n[/tex]The value of [tex]m[/tex] and [/tex]l[/tex[ be be equal to 0 but the value of [tex]n[/tex] can never be equal to 0.Thus, we can say that the option that affirms that n = 2, l = 2, m = +1 is forbidden because the value of [tex]l[/tex] is equal to the value of [tex]n[/tex].
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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.
Answer:
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
ΔG ≅ 199.91 kJ
Explanation:
Consider the reaction:
[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]
temperature = 298.15K
pressure = 22.20 mmHg
From, The standard Thermodynamic Tables; the following data were obtained
[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :
[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]
[tex]K_p = \dfrac{1}{ (22.20)}[/tex]
[tex]K_p[/tex] = 0.045
[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]
where;
R = gas constant = 8.314 × 10⁻³ kJ
[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]
[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]
[tex]\Delta G =207.6+ (-7.687048037)[/tex]
[tex]\Delta G =[/tex] 199.912952 kJ
ΔG ≅ 199.91 kJ
Using Charles’s law, V1/T1 =V2/T2 and the information below, solve the question. If a balloon takes up 625L at 273K, what will the new volume be when the balloon is heated to 353K.
Answer:
The new volume is 808L.
Explanation:
First, you need to set up your proportion, using Charles's law.
V1 V2
=
T1 T2
Then, substitute for the ones you already know.
625 L V2
=
273 K 353 K
Then, you need to cross multiply.
625 x 353 = 273 x V2
Then, solve for V2.
220625 = 273 x V2
220625 ÷ 273 = V2
808 = V2
The final volume is 808L.
Why do you think sodium bicarbonate is included to neutralize an acidic spill rather than sodium hydroxide?
Imagine a hypothetical situation in which 250 mL of diethyl ether (SDS) has spilled inside of a chemical fume hood onto a stir plate that is plugged in and stirring. Discuss the risks associated with this situation (location, size, compound spilled, and external hazards), and then explain how this spill should be managed.
Answer:
Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.
Explanation:
A line-angle formula shows a ring with six vertices and alternating single and double bonds. An OCH3 group is attached to the first vertex. A CH2CH3 group is attached to the third (clockwise) vertex. Spell out the full name of the compound.Part A. A line-angle formula shows a ring with six vertices and alternating single and double bonds. A COOH group is attached to the first vertex. A Br atom is attached to the second (clockwise) and the third vertices. Spell out the full name of the compound.Part B. A line-angle formula shows a ring with six vertices and alternating single and double bonds. A CH3 group is attached to the first vertex. An F atom is attached to the third (clockwise) vertex. Spell out the full name of the compound.
Answer:
1) 3-Ethylanisole
2) 2,3-Dibromobenzoic acid
3) 3-Fluorotoluene
Explanation:
Let us try to look at the structures of each compound one after the other as described in the question.
1) A ring with six vertices and alternating double and single bonds must refer to a benzene ring. A benzene ring having -OCH3 attached to the first vertex is called anisole. If a -CH2CH3 group is now attached at position 3, we now name the compound 3-Ethylanisole.
2) A ring with six vertices and alternating single and double bonds is a benzene ring. If the ring has -COOH attached to the first vertex, we call it benzoic acid. If bromine atoms are attached to the second and third vertices respectively, the compound is now named 2,3-Dibromobenzoic acid.
3) A ring with alternating single and double bonds is a benzene ring. If a -CH3 group is attached to the first vertex, we call the compound toluene. If a fluorine atom is now attached to position 3, the compound can now be named 3-Fluorotoluene
during the electrolysis of an aqueous solution of sodium nitrate, a gas forms at the anode, what gas is it?
Answer: The answer is B
Explanation:
Answer:
oxygen
Explanation:
A general chemistry student found a chunk of metal in the basement of a friend's house. To figure out what it was, she tried the following experiment. First she measured the mass of the metal to be 385.8 grams. Then she dropped the metal into a measuring cup and found that it displaced 17.8 mL of water. Calculate the density of the metal. Density = _______ g/mL Use the table below to decide the identity of the metal. This metal is most likely _________.
substances density g/cm3
water 1.00
aluminium 2.72
chromium 7.25
nickel 8.91
silver 10.50
lead 11.34
1. 21.67g/ml
2. aluminium
Explanation:
1. density = mass/volume
385.8/17.8= 21.67ml
2. 1g/ml=0.1g/cm^3
21.67g/ml = 2.167g/cm^3
..... substance is probably aluminium
1. the density of the metal is 21.67g/ml
2. This metal is most likely aluminum
The calculation is as follows;
1.
[tex]density = mass \div volume[/tex]
[tex]385.8\div 17.8= 21.67ml[/tex]
2.
1g/ml=0.1g/cm^3
So,
21.67g/ml = 2.167g/cm^3
Therefore, substance is probably aluminum
Learn more: https://brainly.com/question/17429689?referrer=searchResults
Draw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.