Answer:
W= -11KJ
Explanation:
Given:
volume expands from 28 L to 51 L
pressure =4.9 atm.
We will need to Convert the pressure to Pascal SI
But 1 atm = 101,325 Pa.
Then,
Pressure= (4.9*101323)/1atm = 5*10^5 pa
Then we need to Convert the volumes to cubic meters
But we know that1 m³ = 1,000 L.
V1= 28L * 1m^3/1000L = 0.028m^3
V2=51L × 1m^3 /1000L =0.051m^3
The work done during the expansion of a gas can be calculated as
W= -P(V2-V1)
W= - 5*10^5(0.051m^3 - 0.028m^3)
W= -1.1× 10^4J
Then we can Convert the work to kiloJoule
But1 kJ = 1,000 J.
W= -1.1× 10^4J× 1kj/1000J
= -11KJ
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33 g of lead. Determine simplest formula for the compund
Answer: The empirical formula is [tex]PbO_2[/tex]
Explanation:
Mass of Pb = 4.33 g
Mass of O = (5.00-4.33) g = 0.67 g
Step 1 : convert given masses into moles
Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Pb = [tex]\frac{0.021}{0.021}=1[/tex]
For O = [tex]\frac{0.042}{0.021}=2[/tex]
The ratio of Pb O= 1: 2
Hence the empirical formula is [tex]PbO_2[/tex]
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.
What does the empirical formula tell you about the compound?
The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?
Can you see what type of functional group this compound could have?
Answer:
See explanation.
Explanation:
Hello,
In this case, we can show how the empirical formula is found by following the shown below procedure:
1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:
[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]
2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:
[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]
3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:
[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]
4. Compute the moles of oxygen by using its molar mass:
[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]
5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:
[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]
6. Search for the closest whole number (in this case multiply by 2):
[tex]C_3H_6O_2[/tex]
Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:
[tex]M=12*3+1*6+16*2=74g/mol[/tex]
Which is about three times in the molecular formula, for that reason, the actual formula is:
[tex]C_9H_{18}O_6[/tex]
It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.
Best regards.
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
Name the compound Ga S3
Answer:
Gallium(III) sulfideA 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)
a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.
Answer and Explanation:
The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.
If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.
Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol
Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol
moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-
According to this, the following are the answers to the sentences:
a. The number of moles of CH₃COOH will remain the same ⇒ FALSE
The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions
b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE
Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)
c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE
The equilibrium concentration of OH⁻ is increased
d. The pH will decrease⇒ FALSE
pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:
[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]
pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15
Thus, the pH will increase.
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings
Answer:
[tex]V=27992L=28.00m^3[/tex]
Explanation:
Hello,
In this case, the combustion of methane is shown below:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]
Best regards.
Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
Learn more about chemical bonds:
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https://brainly.com/question/13178368
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How fast are the atoms moving if the temperature of a gas is cold?
A. very, very slowly
B. they are stagnant
C. very, very quickly
Answer:
i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.
Answer:
A. very, very slowly
Explanation:
A is the answer because atoms will move faster in hot gas than in cold gas.
Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene
The question is incomplete as the options are missing, however, the correct complete question is attached.
Answer:
The correct answer is option A. ( check image)
Explanation:
The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.
Thus, the correct answer is option A. ( check image)
The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.
Answer:
1. stability factor
2. steric hindrance factor
Explanation:
stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.
ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive
You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3
Complete Question
The complete question is shown on the first uploaded image
Answer:
The initial concentration is [tex]C_f = 0.0022 \ M[/tex]
Explanation:
From the question we are told that
The volume of solution A is [tex]V_i = 10.0 mL[/tex]
The concentration of A is [tex]C_i = 0.0200 \ M[/tex]
The volume of solution B is [tex]V_B = 10.0mL[/tex]
The volume of water is [tex]V_{w } = 70.0 mL[/tex]
Generally the law of dilution is mathematically represented as
[tex]C_i * V_i = C_f * V_f[/tex]
Where [tex]C_f[/tex] is the concentration of the mixture
[tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as
[tex]V_f = 10 + 10 + 70[/tex]
[tex]V_f = 90mL[/tex]
So
[tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]
substituting values
[tex]C_f = \frac{0.0200 * 10 }{90}[/tex]
[tex]C_f = 0.0022 \ M[/tex]
Note the mixture obtained is [tex]KIO_3[/tex]
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
Chemistry question. Image attached.
Answer:
The answer to your question is given below
Explanation:
The balanced equation for the reaction is given below:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
1. Writing an expression for the equilibrium constant, K.
The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Thus, we can write the equilibrium constant, K for the reaction as follow:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²
2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis
If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state?
gaseous
liquid
solid
Answer:
Gaseous
Explanation:
Gasses can move freely and do not form the shape of their containers
Liquids are more free than solids, but they conform to the shape of their container
Solids are not free
According to Le Châtelier's principle, how would a change in pressure affect a
gaseous system in equilibrium?
Answer:
Le Châtelier's principle states that when a chemical system at equilibrium is distributed by a change in conditions, the equilibrium position will shift in a direction that tends to counteract the change.
Therefore, when there is a change in pressure, the equilibrium will counteract the change by reducing/increasing the pressure through adjusting the no. of moles of gas.
Note: At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.
For example, when there's an increase in total pressure, the equilibrium position will shift to the side with a smaller no. of moles of gas so as to reduce the pressure.
Answer:
The equilibrium would shift to reduce the pressure change
Explanation:
The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.
Answer:
- N2 does not exist as a liquid at pressures below 0.127 atm.
- N2 is a solid at 16.7 atm and 56.5 K.
- N2 is a liquid at 1.00 atm and 73.9 K
- N2 is a gas at 0.127 atm and 84.0 K.
Explanation:
Hello,
At first, we organize the information:
- Normal melting point: 63.2 K.
- Normal boiling point: 77.4 K.
- Triple point: 0.127 atm and 63.1 K.
- Critical point: 33.5 atm and 126.0 K.
In such a way:
- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).
- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.
- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.
- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.
Best regards.