Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
(1) Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
(2) Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J
(3) H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
-802.3kJ
A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)
Answer:
D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)
E: CH3-CH2-(CH3)-CH2-CH3
(Please see the figures enclosed )
Explanation:
D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically inactive. The reason is that two enantiomers are present in an equal amount).
E is optically inactive, so its structure has to be symmetric.
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2 and I2 are 0.453 M, what will the concentration of HI be at equilibrium
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
uses of sodium chloride in daily life
Answer:
sodium chloride can be used as salt
extraction sodium metal by electrolysis
a common chemical in laboratory experiments
Answer:
sodium chloride can be used as preservatives,
in preserving foods.
A sample of chloroform, CHCl 3 , , was determined to have a molecular mass of 112.3g / (mol) . Its molecular mass is known to be 119.5g / (mol) . Calculate the absolute error and the percent error
Answer:
Explanation:
in your case ,
Meaured value = 112.3
actual value = 119.5
Absolute error= measured value - actual value
Percent error = [measured value - actual value / actual value ] x 100
Hope this help you to find the answer
a binary ionic compound is made of two components name one of them
Answer:
CATION
Explanation:
It's one is the action and the mother is a cation.
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
Indicate whether the following represents a Chemical or Physical change: Milk sours
Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
Which of the following reactions would be predicted by the activity series list
A. A metal ion reacts with another ion to form a precipitate.
B. A metal replaces a metallic ion below it on the list.
C. A metal replaces a metallic ion above it on the list.
D. A metal reacts with oxygen in a combustion reaction.
Answer:
The answer is B) A metal replaces a metallic ion below it on the list.
Explanation:
I just did it and got it correct, luckily I didn't use the other answer posted for this question.
A metal replaces a metallic ion below it on the list give reaction which would be predicted by the activity series list.
So, option B is correct one.
What is Electrochemical series?The list in which elements arranged in the increasing order of their electrode potential values is called Electrochemical series.
The Electrochemical series is also called activity series.
To learn more about Electrochemical series here.
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Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.
Answer:
1. Non-target
2. Bycatch
Explanation:
In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.
But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.
The eel fish which gets captured is known as bycatch fishes in the fishing operation.
Thus, Non-target and Bycatch are the correct answer.
Answer:
Its just bycatch
Explanation:
The eel species is called a(n) bycatch of the fishing operation.
Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, write only NR. LiNO₃
Answer:
It is neutral (NR)
Explanation:
Salts are formed when the ionizable hydrogens in an acid is replaced by metallic or ammonium ions from bases. The reaction is known as a neutralization reaction.
The nature of a salt formed from this reaction depends on the nature of the reacting acid and base.
If the reaction is between a strong acid and strong base, the salt produced is a neutral salt.
If the reaction occurs between a strong acid and a weak base, the salt produced is acidic.
If the reaction occurs between a strong base and a weak acid, the salt produced is a basic salt.
Considering the salt above, LiNO3.
On hydrolysis, addition of water, the following products are obtained:
LiNO3 + H2O ----> LiOH + HNO3
The products obtained, LiOH and HNO3 are a strong base and a strong acid respectively. Therefore, the salt, LiNO3, is a neutral salt.
The salt, LiNO₃ is a neutral, NR salt as it's a salt formed from the reaction of a strong acid and a strong base.
In neutralisation reactions, acids and bases react to form salt and water.
However, the salt formed may be acidic, basic or neutral. This is dependent on the type of acid and base which form the salt.
A strong acid and a strong base react to yield a neutral salt like, LiNO₃.
The equilibrium equation when LiNO₃ is dissolved in aqeous solution is;
LiNO₃ + H2O ==>. LiOH + HNO₃Evidently, LiOH and HNO₃ are an example strong base and acid respectively.
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One proposed mechanism of the reaction of HBr with O2 is given here. HBr + O2 → HOOBr (slow) HOOBr + HBr → 2HOBr (fast) HOBr + HBr → H2O + Br2 (fast) What is the equation for the overall reaction?
Answer:
4 HBr + O2 → + 2H2O + 2Br2
Explanation:
Based on the following reaction mechanism:
HBr + O2 → HOOBr (slow)
HOOBr + HBr → 2HOBr (fast)
HOBr + HBr → H2O + Br2 (fast)
The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):
HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2
4 HBr + O2 → + 2H2O + 2Br2Beeing this reaction the equation of the overall reaction.
Of the following three atoms, one tends to lose three electrons, another tends to lose two electrons, and another tends to lose one electron. Rank these atoms in order of the number of electrons they tend to lose, from most to fewest: potassium, K; calcium, Ca; gallium, Ga. Rank from most to fewest.
Answer: Ga > Ca > K
Explanation:
Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.
The electrons are filled according to Afbau's rule in order of increasing energies. The metals tend to get stable by losing electrons to attain noble gas configuration.
[tex]K:19:1s^22s^22p^63s^23p^64s^1[/tex]
[tex]K^+:18:1s^22s^22p^63s^23p^6[/tex]
[tex]Ca:20:1s^22s^22p^63s^23p^64s^2[/tex]
[tex]Ca^{2+}:18:1s^22s^22p^63s^23p^6[/tex]
[tex]Ga:31:1s^22s^22p^63s^23p^64s^23d^{10}4s^24p^1[/tex]
[tex]Ga^{3+}:28:1s^22s^22p^63s^23p^64s^23d^{10}[/tex]
Thus gallium (Ga) loses three electrons, Calcium (Ca) loses 2 electrons and Potassium (K) loses one electron.
Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.
Answer:
The percentage yield is 50%
For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à
Answer:
Explanation:
a. Determine the number of valence electrons for each atom in the molecule
In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.
b. Draw the Lewis Dot structure
In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1
c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)
In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).
d. Show the polarity of each bond and for the molecule by drawing in the dipole +d
The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".
I hope it helps!
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
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Real images can be projected onto a screen.
A. True
B. False
Answer:
True
Explanation:
A real image can be projected or seen on a screen but a virtual image cannot because a real image is formed when light rays coming from an object actually meet at a point after refraction through a lens while a virtual image is formed when light rays coming from an object only appear to meet at a point when produced ...
Hope this helps you, and Good luck!
Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of 12.5 in long, 3.50 in wide, and 2.50 in high. (1in = 2.54 cm).
Answer:
[tex]m=14,105.71 g Fe[/tex]
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]
Then, we compute the volume in cubic centimetres:
[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]
Finally, as the density is given by:
[tex]\rho =\frac{m}{V}[/tex]
We solve for the mass:
[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]
Best regards.
Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is
Explanation:
A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.
On the other hand;
Bronsted-Lowry acid is the substance that donates the proton.
HF (aq) + SO32- ⇌ F- + HSO3-
In the forward reaction;
Bronsted-Lowry acid : HF
Bronsted-Lowry base: SO32-
In the backward reaction;
Bronsted-Lowry acid : HSO3-
Bronsted-Lowry base: F-
The conjugate base of HF is F-
The conjugate acid of SO32- is HSO3-
How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?
Answer:
[tex]m_{CaF_2}0.375gCaF_2[/tex]
Explanation:
Hello,
In this case, for the studied reaction:
[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]
Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:
[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]
Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:
[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]
Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:
[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]
Best regards.
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K. If the final temperature of the system is 31.10°C, what is the specific heat capacity of the alloy? J g·°C
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43
Answer:
Less than 0.033 M:
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Explanation:
Hello,
In this case, the described equilibrium is:
[tex]2A+B\rightarrow 2Z[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]
Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]
But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
7.Which one of the following statements is not true?
1 point
O The molecules in a solid vibrate about a fixed position
O The molecules in a liquid are arranged in a regular pattern
The molecules in a gas exert negligibly small forces on each other, except during
collisions
The molecules of a gas occupy all the space available
Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of water produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
1.7 moles of ethanol would be needed.
Explanation:
* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
First off, we have to state the equation for the reaction.
So we know that;
ethanol + oxygen → acetic acid + water
This leads us to;
C2H5OH + O2 → CH3COOH + H2O
1 1 1 1
To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.
1 mol of ethanol produces 1 mole of water
x mol of thanol would produce 1.7 mol of water
Thus we have;
1 = 1
x = 1.7
x = 1.7 moles of ethanol would be needed.
Calcium chloride reacts with sodium hydroxide to form solid calcium hydroxide, Ca(OH)2. The balanced net ionic equation is
Group of answer choices
Ca+2(aq) + 2OH-(aq) + 2 Na+(aq) + 2 Cl-(aq) → Ca(OH)2(s).
Ca+2(aq) + 2OH-(aq) + Na+(aq) + Cl-(aq) → Ca(OH)2(s).
Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s).
CaCl2(aq) + 2NaOH(aq) → CaOH(s) + NaCl(aq).
CaCl2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaCl(aq).
Answer:
Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)
Explanation:
In chemistry, the net ionic equation is a way to write a chemical reaction whereas you write only the ions that are involved in the reaction.
When calcium chloride, CaCl₂ reacts with sodium hydroxide, NaOH to produce Ca(OH)₂ the only ions involved in the reaction are Ca²⁺ and OH⁻, thus, the balanced net ionic equation is:
Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)
Cl⁻ and Na⁺ are not involved in the reaction and you don't have to write them.
The balanced net ionic equation for the reaction between calcium chloride and sodium hydroxide is
Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)The ionic equation for the reaction between calcium chloride and sodium hydroxide can be written as follow:
Calcium chloride => CaCl₂
Sodium hydroxide => NaOH
In solution,
CaCl₂(aq) —> Ca²⁺(aq) + 2Cl⁻(aq)
NaOH(aq) —> Na⁺(aq) + OH⁻(aq)
CaCl₂(aq) + NaOH(aq) —>
Ca²⁺(aq) + 2Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) —> Ca(OH)₂(s) + 2Cl⁻(aq) + Na⁺(aq)
Cancel the spectator ions (i.e Cl⁻ and Na⁺) and write 2 before OH⁻ to obtain the net ionic equation as shown below:
Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)Learn more: https://brainly.com/question/9489859
A solution of pentane and ethanol (CH3CH2OH)that is 50.% pentane by mass is boiling at 57.2°C. The vapor is collected and cooled until it condenses to form a new solution.
Calculate the percent by mass of pentane in the new solution. Here's some data you may need:
normal boiling point density vapor pressure at
57.2°C
pentane 36.°C 0.63gmL 1439.torr
ethanol 78.°C 0.79gmL 326.torr
Be sure your answer has 2 significant digits.
dont round during math only for answer!
Note for advanced students: you may assume the solution and vapor above it are ideal.
Answer:
The correct answer is 81.52 percent.
Explanation:
Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.
In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.
The moles or n can be calculated by using the formula,
n = weight/molecular mass
The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.
The moles of pentane is,
= 50 g/72.15 g/mol = 0.6930 mol
The moles of ethanol is,
= 50 g/46.07 g/mol = 1.0853 mol
The mole fraction of pentane is,
= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897
The mole fraction of ethanol is,
= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103
Now the vapor pressure of solution will be,
= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol
= (1439 * 0.3897) + (326 * 0.6103)
= 759.736 torr
The vapor pressure of pentane within the solution,
= vapor pressure of pentane * mole fraction of pentane
= 1439 torr * 0.3897
= 560.778 torr
The fraction of pentane is,
= 560.778 / 759.736 = 0.738
Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262
The mass of pentane = 0.738 * 72.15 = 53.2467
The mass of ethanol = 0.262 * 46.07 = 12.07034
The percent by mass of pentane in new solution will be,
Mass% = mass of pentane/Total mass * 100%
= 53.2467/(53.2467 + 12.07034) * 100%
= 53.2467/65.31704 * 100 %
= 81.52 %
3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm. The density of the metal is 5.30 g/cm3. Assume that 68% of the unit cell is occupied by Ba atoms. The molar mass of barium is 137.3 g/mol. Using this information, calculate Avogadro’s number. Show your calculation procedure that allows you to derive Avogadro’s number. Your answer must show six digits after the decimal point (i.e., 6.pppx1023) that is not necessarily the same as the known value. By showing your calculation-result down to six digits after the decimal point, you showcase that you did calculate the number, instead of simply adopting the known Avogadro’s number available in open resources.
Answer:
The Avogadro's number is [tex]N_A = 6.02289 *10^{23}[/tex]
Explanation:
From the question we are told that
The edge length is [tex]L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}[/tex]
The density of the metal is [tex]\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3[/tex]
The molar mass of Ba is [tex]Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol[/tex]
Generally the volume of a unit cell is
[tex]V = L^3[/tex]
substituting value
[tex]V = [5.02 *10^{-10}]^3[/tex]
[tex]V = 1.265*10^{-28}\ m^3[/tex]
From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),
The volume of barium atom is
[tex]V_a = \frac{V}{2} * 0.68[/tex]
substituting value
[tex]V_a = \frac{ 1.265*10^{-28}}{2} * 0.68[/tex]
[tex]V_a = 4.301 *10^{-29} \ m^3[/tex]
The Molar mass of barium is mathematically represented as
[tex]Z = N_A V_a * \rho[/tex]
Where [tex]N_A[/tex] is the Avogadro's number
So
[tex]N_A = \frac{ Z}{ V_a * \rho}[/tex]
substituting value
[tex]N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}[/tex]
[tex]N_A = 6.02289 *10^{23}[/tex]
Which of the following best describe an atomic number? An element identity is defined by its atomic number this means it represents the number of
a)Atomic number is the number of protons in the nucleus of an atom.
b)atomic number represents the number of protons
An element's identity is defined by its atomic number; this means it represents the number of protons in its nucleus.
The question is incomplete, the complete question is;
Which of the following best describes an atomic number? An element's identity is defined by its atomic number; this means it represents the number of
A) protons plus neutrons in its nucleus.
B) electrons in the element.
C) protons in its nucleus.
D) neutrons in its nucleus.
The subatomic particles in the atom are;
ElectronsProtonsNeutronsNeutrons and protons are contained in the nucleus hence they are collectively called nucleons.
Electrons are found in the orbits. The number of protons must be equal to the number of electrons for the atom to be electrically neutral.
The number of protons in an atom or atomic number serve as a means of identifying an atom.
Learn more; https://brainly.com/question/18460120
Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.
a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253
Answer:
The answer is "Option b"
Explanation:
In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:
Formula:
[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]
[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]
[tex]=\ 0.0025 \ moles[/tex]
[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]
[tex]= 0.00315 \ moles[/tex]
[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]
[tex]= 0.00005 \ moles[/tex]
[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]
[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]
[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]
[tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]
How has human action affected the
population?
Answer:
Human action has affected the population in quite a negative way. Technological innovations and urbanization gave rise to a high degree of pollution on the land, air and water.
Emission of hydrocarbons from automobiles and factories are known to pollute and cause sicknesses related to the respiratory system thereby shortening the lifespan of humans. Pollution of water bodies also cause death of sea animals and thereby reducing food availability of humans.
A chemistry student is given 600. mL of a clear aqueous solution at 37° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg. Using only the information from above, can you calculate the solubility of X at 21° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.
Answer:
The solubility is [tex]S = 0.0014 \ g[/tex]
Explanation:
From the question we are told that
The volume of the solution is [tex]V = 600 mL[/tex]
The initial temperature is [tex]T_i = 37 ^oC[/tex]
The final temperature is [tex]T_f = 21^oC[/tex]
The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]
Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent (for solubility of a solute to be calculated the solute must be able to saturate the solvent)
now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away
The solubility at 21 ° C is mathematically represented as
[tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]
Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as
[tex]m_w = V * \rho_w[/tex]
Where [tex]\rho = 1 \frac{g}{mL}[/tex]
So
[tex]m_w =600 * 1[/tex]
[tex]m_w =600g[/tex]
So
[tex]S = \frac{84}{600 * 100 g \ of water }[/tex]
[tex]S = 0.0014 \ g[/tex]