Answer:
61kg
Explanation:
Like all other hydrocarbons, the combustion of propane will yield water and carbon iv oxide.
Let’s write a complete and balanced chemical equation for this.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Now, we first convert the information into a mass.
We know that mass = density * volume
Let’s convert the volume here into ml since density is in mL. It must be remembered that 1000ml = 1L. Hence, 29.4L = 29,400ml
The mass is thus = 0.6921 * 29,400 = 20,347.74g
Now from the balanced equation, we can see that one mole of propane produced 3 moles of carbon iv oxide. This is the theoretical relation.
Let’s now calculate the actual number of moles. The number of moles of propane produced is the mass of propane divided by the molar mass of propane. The molar mass of propane is 3(12) + 8(1) = 36 + 8 = 44g/mol
The number of moles = 20,347.74/44 = 462.44
We simply multiply this number by 3 to get the actual number of moles of carbon dioxide produced. This is equals 1387.35 moles
Since we know this, we now calculate the mass of carbon iv oxide produced. The mass is equals the number of moles multiplied by the molar mass. The molar mass of carbon iv oxide is 44g/mol.
The mass produced is thus 1387.35 * 44 = 61,043g
To get the kilogram equivalent, we simply divide by 1000 = 61,043/1000 = 61kg
When 0.0400 mol of HCl(aq) is reacted with 0.0400 mol of NaOH(aq) in 50.0 mL of water, the temperature of the solution increases by 9.82°C. What is the enthalpy of reaction for the following thermochemical equation?HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Assume that the heat capacity of the solution and calorimeter is 227.4 J/°C.
The enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.
What is thermochemical equation?Thermochemical equation is defined as a chemical equations that are properly balanced and take into account the physical conditions of all reactants, products, and energy change. A chemical equation is an equation that shows the beginning molecule, reactants, and final products separated by arrows, but a thermochemical equation is a balanced stoichiometric chemical process that also includes the enthalpy change.
Given heat capacity = 227.14 J/°C
Temperature = 9.82 °C
Number of moles of HCl and NaOH is 0.04 moles
q (solution) = -C (solution) x ΔT
= 227.14 J/°C x 9.82 °C
= 2.2305 kJ
For enthalpy of reaction we have to divide it by mole of reactant
= 2.2305 / 0.04
= 55.4 kJ/ mole
Thus, the enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.
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To determine the enthalpy of reaction for HCl reacting with NaOH, the heat released by the reaction is calculated using the mass of the solution, the specific heat capacity, and the change in temperature. The result is then divided by the number of moles reacted to express the enthalpy change in kJ per mole.
Explanation:The problem here concerns the measurement of an enthalpy change during the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O). To find the enthalpy of reaction for the provided thermochemical equation, we use the temperature change and the heat capacity of the solution to calculate the total heat released by the reaction, and then express this heat change in kJ per mole of reactants.
First, calculate the total heat (q) released using the formula q = mc∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature. As the reaction takes place in 50.0 mL of water, and assuming the density of water is 1.00 g/mL, the mass of the solution is 50.0 g. Therefore, q = 50.0 g × 227.4 J/°C × 9.82°C.
After calculating q, divide this value by the number of moles of HCl reacted to obtain the enthalpy of reaction per mole. Remember to convert from joules to kilojoules since enthalpy is usually expressed in kJ/mol. Please note that the reaction is exothermic, so the enthalpy change should be a negative value, indicating heat is released.
The reaction is found to be first order in IO3−, first order in SO32−, and first order in H+. If [IO3−]=x, [SO32−]=y, and [H+]=z, what is the rate law for the reaction in terms of x, y, and z and the rate constant k?
Answer:
rate = kxyz
Explanation:
It is worth knowing that the rate low can only be determined by experimentation only not by just balancing equations. So here we are told that all the reactants x , y and z are all first order. This is important because we use this as exponents. That is why the exponents of all the reactants will be 1.
rate = kxyz
An apparatus consists of a 3 L flask containing nitrogen gas at 17◦C and 822 kPa, joined by a valve to a 10 L flask containing argon gas at 17◦C and 59 kPa. The valve is opened and the gases mix. a) What is the partial pressure of nitrogen after mixing?
b) What is the partial pressure of argon aftermixing?
c) What is the total pressure of the gas mixture?
Answer:
a) 189.7 kPa
b) 45.4 kPa
c) 235 kPa
Explanation:
After the mixture, the temperature must remain the same, because both gases are at 17°C. Besides, the gases will not react because both nitrogen and argon are non-reactive. So, the partial pressure can be calculated by Boyle's Law:
P1*V1 = P2*V2
Where P is the pressure, V is the volume, 1 is the initial state, and 2 the final state.
When the valve is opened, the gases mix and occupy all the two vases. So, V2 = 13 L.
a) For nitrogen, P1 = 822 kPa, V1 = 3L, V2 = 13 L
822*3 = P2*13
P2 = 189.7 kPa
b) For argon, P1 = 59 kPa, V1 = 10 L, V2 = 13 L
59*10 = P2*13
P2 = 45.4 kPa
c) By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressures of the components, so:
P = 189.7 + 45.4
P = 235 kPa
The vapor pressure of liquid antimony is 400 mm Hg at 1.84×103 K. Assuming that its molar heat of vaporization is constant at 115 kJ/mol, the vapor pressure of liquid Sb is 394.98 mm Hg at a temperature of 1.81×103 K. Find the vapor pressure of liquid Sb?
Answer:
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Explanation:
The vapor pressure can be calculated by using Clausius‐Clapeyron equation.
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
Where
p₁ is the vapor pressure at T₁ (Initial Temperature)
p₂ is the vapor pressure at T₂ (final Temperature)
ΔHvap is molar heat of vaporization of the substance
R is the real gas constant = 8.314 x 10⁻³ kJ/mol.K
Data Given:
p₁ = ?
p₂ = 394.98 mm Hg
T₁ = 1.84×10³ K
T₂ = 1.81×10³ K
ΔHvap = 115 kJ/mol
Put the values in the Clausius‐Clapeyron equation
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
ln(p₁/394.98 mm Hg) = (-115 kJ/mol / 8.314 x 10⁻³ kJ/mol.K)(1/1.84×10³ K- 1/1.81×10³ K)
ln(p₁ /394.98 mm Hg) = (- 13.8321 x 10³)(-0.5519)
ln(p₁ /394.98 mm Hg) = 7633.936
ln cancel out by E, e is raise to a power x
So,
p₁/394.98 mm Hg = e^7633.936
p₁/ 394.98 mm Hg = 20.75 x 10³
p₁ = 20.75 x 10³ x 394.98 mm Hg
p₁ = 8.19 x 10⁴ mm Hg
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Answer:
338 mm Hg
Explanation:
[tex]ln(\frac{p2}{400mmHg} )=-115*\frac{1}{0.008314} *(\frac{1}{1.8*10^3} -\frac{1}{1.84*10^3}[/tex])
ln(p2/400)=-0.16705
p2/400= e^-0.16705=0.84615
p2=0.84615*400=338
Note that T2 is lower than T1 and that the vapor pressure decreases as the temperature decreases.
How many atoms of fluoride are there in one molecule of nitrogen trifluoride?
Answer:
there are three Fluorine atom. This is because Nitrogen give 3 electron away but fluorine can only take 1. So nitrogen gives 1 neutron to 3 fluorine each which makes it a stable compound
Answer : The number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.
Explanation :
Molecule : It is defined as the smallest particle in an element or a compound which has chemical properties of that element or a compound.
Molecules are made up of atoms and the atoms are bonded by the sharing of electrons.
The given molecules is, [tex]NF_3[/tex]
The nitrogen trifluoride molecule is made up of two elements which are nitrogen and fluorine.
In the given molecules [tex]NF_3[/tex], there are 1 nitrogen atom and there are 3 fluorine atoms.
Hence, the number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.
covalent molecular solids (like sugar) and covalent network solids (like diamond) have similar bonds and bond strengths. However, molecular solids are brittle and network solids are very strong and durable. Explain the difference.
Answer:
Covalent molecular solids (like sugar): No intermolecular covalent bonds
Covalent network solids (like diamond) : Intermolecular covalent interactions are present.
Explanation:
Covalent moleuclar solids are the solids which have "intera-molecular" covalent bonds i.e the covalent bonds exist between two atoms within the same molecule. The intermolecular bonds are not covalent. They are some weak intermolecular interactions like hydrogen bonds or disperison forces.
In case of covalent network solids, the intermolecular interactions are also strong covalent bonds. Thus they give a more strength and firmness to the compound or solid.
The enthalpy of combustion of acetylene C2H2 is described by
C2H2 (g) + (5/2)O2 (g) >>>>>>>CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Calculate the enthalpy of formation of accetylene, given the following enthalpies of formation
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The chemical equation for the combustion of acetylene follows:
(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]
[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]
The formation of [tex]CO_2[/tex] will be,
(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]
The formation of [tex]H_2O[/tex] will be,
(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
The formation of [tex]C_2H_2[/tex] will be,
(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex] [tex]\Delta H_f_{(C_2H_2)}=?[/tex]
Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:
Reaction (1) :
[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]
[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]
Reaction (2) :
[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]
Reaction (3) :
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]
[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]
[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]
Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol
To calculate the enthalpy of formation of acetylene, we can use the enthalpy change of the combustion reaction and the enthalpies of formation of CO2 and H2O. By rearranging the equation and applying Hess's law, we can determine the enthalpy of formation of acetylene.
Explanation:The enthalpy of formation of a compound can be calculated using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken. To calculate the enthalpy of formation of acetylene (C2H2), we can use the equation:
C2H2 (g) + (5/2)O2 (g) → CO2 (g) + H2O (l)
The enthalpy change of this reaction is -1299 kJ/mol. By rearranging the equation, we can compare it to the standard formation equations for CO2 (g) and H2O (l) to determine the enthalpy of formation of acetylene.
The enthalpy of formation of acetylene is:
ΔHf(C2H2) = -1299 kJ/mol - [(-393.5 kJ/mol) + (-285.8 kJ/mol)]
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Photophosphorylation produces ___ , which stores chemical energy that can be used for the Calvin cycle.
Photophosphorylation of sugar moiety produces ATP (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.
What is Calvin cycle?The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.
It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).
Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.
The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.
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The mixing of several atomic orbitals to form the same number of equivalent hybrid orbitals
Answer:
Hybridization
Explanation: -
Hybridization occurs when atomic orbitals mix to form a new atomic orbital
The oxidative phase of the pentose phosphate pathway is___________.
A. reversible under physiological conditions.
B. has an enzyme requiring thiamine pyrophosphate as a coenzyme.
C. produces ribulose‑5‑phosphate and NADPH for biosynthetic processes.
D. converts pentoses to hexoses.
Answer:
C: Produces ribulose-5-phosphate and NADPH for biosynthetic processes.
Explanation:
There are two distinct phases in the pathway. The first is the oxidative phase, in which NADPH is generated, and the second is the non-oxidative synthesis of 5-carbon sugars. For most organisms, the pentose phosphate pathway takes place in the cytosol; in plants, most steps take place in plastids.
The oxidative phase of the pentose phosphate pathway generates ribulose-5-phosphate and NADPH for biosynthetic processes. Option C the correct choice.
The oxidative phase of the pentose phosphate pathway produces ribulose‑5‑phosphate and NADPH for biosynthetic processes. This phase starts with the oxidation of glucose-6-phosphate, catalyzed by glucose-6-phosphate dehydrogenase, generating NADPH and 6-phosphogluconolactone.
Following steps produce ribulose-5-phosphate, another molecule of NADPH, and release CO₂. Ribulose-5-phosphate can then be isomerized to ribose-5-phosphate or enter further reactions leading to the synthesis of other pentose phosphates as well as fructose 6-phosphate and glyceraldehyde 3-phosphate, which are intermediates in glycolysis.
Thus, the correct answer is C.
Heat capacity is a measure of the heat required to
A) expand the volume of a substance.
B) evaporate a substance.
C) freeze a substance.
D) raise the temperature of 1 gram of a substance by 1°C
Answer: Correct answer is D) Raise the temperature of 1 gram of a substance by 1 degree.
Explanation:
Heat capacity or thermal capacity is a physical property of matter, defined as the amount of heat to be supplied to a given mass of a material to produce a unit change in its temperature. This is rasi
The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K. Calculate the standard change in enthalpy for this reaction (assuming ∆H° is temperature independent).
Answer:
[tex]\text { The standard change in enthalpy for this reaction is } \Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}[/tex]
Explanation:
Let’s assume[tex]\text { At } 300 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=\mathrm{x}[/tex]
Thus as per given information
[tex]\text { At } 350 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=3.95 \mathrm{x}[/tex]
As we know:
[tex]\ln \left(\frac{k_{2}}{k_{1}}\right)=-\frac{\Delta \mathrm{H}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)[/tex]
[tex]\ln \left(\frac{3.95 x}{x}\right)=\frac{-\Delta \mathrm{H}}{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}[/tex]
[tex]8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times \ln 3.95=-\Delta \mathrm{H} \times\left(-4.761 \times 10^{-4} \mathrm{K}^{-1}\right)[/tex]
[tex]\Delta \mathrm{H}=\frac{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times 1.373}{-4.761 \times 10^{-4} \mathrm{K}^{-1}}[/tex]
[tex]\Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}[/tex]
The standard change in enthalpy for this reaction is mathematically given as
dH=2.397 *10^{4} J/mol
What is the standard change in enthalpy for this reaction?
Question Parameter(s):
The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K.
Generally, the equation for the change in enthalpy is mathematically given as
[tex]\ln \ (\frac{k_{2}}{k_{1}})=-\frac{\Delta \mathrm{H}}{R} (\frac{1}{T_{2}}-\frac{1}{T_{1}})[/tex]
Therefore
[tex]8.314 {J} {mol}^{-1} {K}^{-1} *\ln 3.95=-d{H} *(-4.761 \times 10^{-4} {K}^{-1})[/tex]
dH=2.397 *10^{4} J/mol
In conclusion
dH=2.397 *10^{4} J/mol
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Which notation designates a possible value for a principal energy level in an atom?
px
2
+1/2
s
Answer:
2
Explanation:
There are four quantum numbers:
Principal quantum number (n)
Azimuthal quantum number (l)
Magnetic quantum number (ml)
Principal quantum number (n)
It tell about the energy levels. It is designated by n.
For example,
If n =2
It means there are two energy level present.
Azimuthal quantum number (l)
The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3
(n-1)
4-1 = 3
it means principle quantum is 4 and electron is present in f subshell.
Magnetic quantum number (ml)
It describe the orientation of orbitals. Its values are -l to +l. For l=3 the ml will be -3 -2 -1 0 +1 +2 +3.
Spin quantum number (ms)
The spin quantum number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).
If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.
Rank the following salts in order of decreasing pH of their 0.1 M aqueous solutions.(a) FeCl2, FeCl3, MgCl2, KClO2 .(b) NH4Br, NaBrO2, NaBr, NaClO2.
Answer:
a) FeCl2, FeCl3, MgCl2, KClO2.
KClO2 --> K+ + ClO2-; ClO2- will hydrolyse to form HClO +OH-
Mg+2, Fe+2 and Fe+3 ions will form acidic solutions, since theyfom slightly amount of
Mg+2 + 2H2O <-> Mg(OH)2 + 2H+
Fe+2 + 2H2O <-> Fe(OH)2 + 2H+
Fe+3 + 3H2O <-> Fe(OH)3 + 3H+
Therefore;
decreasing pH is high pH to low pH:
KClO2 > MgCl2 > FeCl2 > FeCl3
b) NH4Br, NaBrO2, NaBr, NaClO2.
NH4Br is acidic, forms NH4+ and NH4+ dnates H+ to form NH3 andH+
NaBrO2 is basic, forms Na+ + BrO2- then H2O + BrO2- HBrO2
NABr is neutral, NaClO2 is basics, forms Na+ + ClO2-then H2O + ClO2- HClO2
decreasig pH:
NaClO2 > NaBrO2 > NaBr >NH4Br
Note that HClO2 is stronger acid than HBrO2, therefore, expectmore HBrO2 formation
NaBrO2 > NaClO2 > NaBr >NH4Br
The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br
The term pH refers to the degree of acidity or alkalinity of a solution. We must recall that salts are solvated in solution. The pH of the solution after solvation depends on the ions produced by the salt in solution.
Since FeCl2 yields a basic solution, then it has the highest pH. Similarly, KClO2 yields an acid solution hence it has the lowest pH. The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br
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A sample of 0.0883 g of M g, which has a molar mass of 24.31 g/mol, produces 82.1 mL of H 2 gas. The gas is collected over water at atmospheric pressure of 766.7 mm Hg at 22 oC, at which the vapor pressure of water is 19.8 mm Hg. What is the experimental value for the molar volume of the gas in L/mol?
Answer:
24 Lt/mol
Explanation:
Though we have many data here (such as molar mass of Mg, water vapor pressure, etc), we need to focus on data for H₂, which will help us to obtain the molar volume of this gas
Statement refers the following data for H₂:
V = 82.1 ml = 0.082 Lt
T = 22°C = 295 K
P atm = 766.7 mm Hg = 1.0089 atm (which is, the pressure for H₂ before being collected in water)
If we consider H₂ behaves as an ideal gas:
PV = nRT
Then we can move some terms from this ecuation :
V/n = RT/P so we can obtain the relation between V (volume) and n (N° of moles) for H₂, which is the experimental valur for the molar volume
Considering R = 0.082 Lt*atm/K*mol:
V/n = [(0.082 Lt*atm/K*mol)x295 K]/1.0089 atm
V/n = 23.97 Lt/mol (molar volume at this experiment conditions)
A food chemist determines the concentration of acetic acid in a sample of apple vinegar by acid base titration. The density of the sample is 1.01 g/mL. The titrant is 1.024 M NaOH. The average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78 mL. What is the concentration of acetic acid in the vinegar?
Answer:
The concentration of acetic acid in the vinegar is 7,324 (%V/V)
Explanation:
The titration equation of acetic acid with NaOH is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
The moles required were:
1,024M×0,02500L = 0,02560 moles NaOH. These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:
0,02560 moles CH₃COOH×[tex]\frac{60,052g}{1mol}[/tex]= 1,537g of CH₃COOH
As density is 1,01g/mL:
1,537g CH₃COOH×[tex]\frac{1mL}{1,01g}[/tex]= 1,522mL of CH₃COOH
As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]\frac{1,522mLCH_{3}COOH}{20,78mL}[/tex]×100= 7,324 (%V/V)
I hope it helps!
Considering the definition, molarity, density and volume percentage, the concentration of acetic acid in the vinegar is 7.32% (v/v).
The balanced reaction is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
Molarity is the number of moles of solute that are dissolved in a given volume. Molarity is determined by:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case you know for NaOH:
Molarity= 1.024 Mnumber of moles of solute= ?Volume= 25 mL= 0.025 LSo, the moles required are calculated as:
[tex]1.024 M=\frac{number of moles of solute}{0.025 L}[/tex]
Solving:
1.024 M× 0.025 L= number of moles of solute
0.0256 moles= number of moles of solute
By observing the balanced reaction, you can see that, since the ratio of CH₃COOH to NaOH is 1: 1, these moles are equivalent to the number of moles of CH₃COOH.
On the other hand, since the molar mass of CH₃COOH is 60.052 g/mol, this is the mass that contains one mole of the compound, the mass in 0.0256 moles of CH₃COOH is:
[tex]0.0256 molesx\frac{60.052 grams}{1 mole} =[/tex] 1.537 grams
Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
[tex]density=\frac{mass}{volume}[/tex]
In this case, being the density 1.01 [tex]\frac{g}{mL}[/tex], the volume can be calculated as:
[tex]1.01 \frac{g}{mL} =\frac{1.537 grams}{volume}[/tex]
Solving:
1.01 [tex]\frac{g}{mL}[/tex]× volume= 1.537 grams
[tex]volume=\frac{1.537 grams}{1.01 \frac{g}{mL} }[/tex]
volume= 1.522 mL of CH₃COOH
Volume Percentage (%v/v) is a measure of concentration that indicates the volume of solute per 100 volume units of the solution. In other words, the volume percent of a component in the solution is defined as the ratio of the volume of the component to the volume of the solution, expressed as a percentage.
The volume percentage of a solution is determined by the following expression:
[tex]volume percentage=\frac{volume of solute}{volume of solution}x100[/tex]
Then, as volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]volume percentage=\frac{1.522 mL}{20.78 mL}x100[/tex]
Solving:
volume percentage= 7.32%
Finally, the concentration of acetic acid in the vinegar is 7.32% (v/v).
Learn more about:
molarity with this example: brainly.com/question/15406534?referrer=searchResultsdensity: brainly.com/question/952755?referrer=searchResults brainly.com/question/1462554?referrer=searchResultsvolume percentagebrainly.com/question/1713808A container holds a mixture of two gases at The partial pressures of gas A and gas B, respectively, are and If of a third gas is added with no change in volume or temperature, what will the total pressure become?
Answer:
The total pressure will be the sum of the partial pressure of these three gases.
Explanation:
According to Dalton law of partial pressure,
The total pressure exerted by mixture of gases is equal to the sum of partial pressure of the individual gas.
This expression can be written as,
P(total) = P₁ + P₂ + P₃ + ...... Pₙ
For example:
If the pressure of A is 2 atm and Partial pressure of B is 4 atm the total pressure will be,
P(total) = P₁ + P₂
P(total) = 2 atm + 4atm
P(total) = 6 atm
when third gas is added which exert the partial pressure of 4 atm,
Then total pressure will becomes,
P(total) = P₁ + P₂ + P₃
P(total) = 2 atm + 4atm + 4 atm
P(total) = 10 atm
Which of the following is TRUE? a) A neutral solution contains [H2O] = [H3O⁺] b) A neutral solution does not contain any H3O+ or OH- c) An acidic solution has [H3O⁺] > [OH⁻] d) A basic solution does not contain H3O+ e) None of the above are true.
Explanation:
pH is the negative logarithm of hydronium ion concentration present in a solution.
If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9. The solution has more [tex][H_3O^+][/tex] than [tex][OH^-][/tex]
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14. The solution has more [tex][OH^-][/tex] than [tex][H_3O^+][/tex]
The solution having pH equal to 7 is termed as neutral solution.The solution has equal concentration of [tex][H_3O^+][/tex] and [tex][OH^-][/tex]
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Assuming that the container is completely full, that the temperature is 22.0 ∘C, and that the atmospheric pressure is 1.2 atm , calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.)
Answer:
The % of displaced volume of nitrogen is 29.06%.
Explanation:
Volume of nitrogen = 1.2 L = 1200 mL
Density of nitrogen = 0.807 g/ml
Mass of nitrogen = [tex]Density \times Volume= 0.807 \times 1200 = 1044 g[/tex]
Molar mass of nitrogen = 28 g/mol
[tex]Number\,of\,moles\,nitrogen=\frac{Mass}{Molar\,mass}[/tex]
[tex]= \frac{1044}{28}=37.28[/tex]
The ideal gas equation is as follows
[tex]PV = nRT[/tex]
Rearrange the equation is as follows.
[tex]V= \frac{nRT}{P}...............(1)[/tex]
n= Number of moles = 37.28
R = Gas constant = 0.0820
T = Temperature = 22+ 273 = 295
P = Pressure = 1.2 atm
Substitute the all values in equation (1)
[tex]V= \frac{37.28 \times 0.0820 \times 295 }{1.2}= 751.5L= 0.751 \,m^{3}[/tex]
[tex]0.751 \,m^{3}[/tex] of nitrogen will displace same amount of air.
[tex]Volume \,\,of\,\, closet= 1 \times 1.3 \times 2 = 2.6\,m^{3}[/tex]
[tex]%\,displaced\,volume=\frac{0.751}{2.6}=28.8%[/tex]
Therefore, The % of displaced volume of nitrogen is 29.06%.
Final answer:
The percent by volume of air displaced by the evaporation of liquid nitrogen is calculated using the density of liquid nitrogen, the molar mass of nitrogen, and the Ideal Gas Law, accounting for the specific temperature and pressure conditions.
Explanation:
To calculate the percent by volume of air displaced if all the liquid nitrogen evaporated, we need to consider the Ideal Gas Law as well as the density of the liquid nitrogen. Since liquid nitrogen has a density of 0.807 g/mL, we can first find the mass of the nitrogen and then use the molar mass to determine the number of moles of nitrogen gas at the given temperature and pressure when it evaporates.
Using the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. At standard temperature and pressure, the density of nitrogen gas can be calculated using the given molar mass and standard gas equation factors.
Once the volume of the gaseous nitrogen is determined, that volume will represent the amount of air volume displaced. To find the percent by volume displaced, divide the volume of gaseous nitrogen by the volume of the container and multiply by 100%.
How much heat is absorbed when 30.00 g of C(s) reacts in the presence of excess SO2(g) to produce CS2(l) and CO(g) according to the following chemical equation?
5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g) ΔH° = +239.9 kJ
The reaction of 30 g of C(s) in the given equation absorbs approximately 119.95 kJ of heat.
Explanation:In chemistry, the amount of heat absorbed or released during a chemical reaction is given by the product of the mole ratio and the enthalpy change (ΔH°). Looking at the balanced equation: 5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g) ΔH° = +239.9 kJ, the reaction of 5 moles of C(s) absorbs 239.9 kJ of heat. However, we only have 30 g of C(s) - or 30/12.01 (since the molar mass of carbon is 12.01 g/mol) equals roughly 2.5 mol of C(s).
To find out how much heat 2.5 mol of C(s) absorbs, we use a proportional relationship: (2.5 moles C/5 moles C) * 239.9 kJ = 119.95 kJ. Therefore, when 30 g of C(s) reacts in the presence of excess SO2(g) to produce CS2(l) and CO(g), it absorbs approximately 119.95 kJ of heat.
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The heat absorbed when 30.00 g of C(s) reacts in the given chemical reaction is approximately 119.95 kJ. This value is determined based on the moles of carbon involved and the enthalpy change given for the reaction.
First, we need to find the number of moles of carbon (C) involved in the reaction:
Molar mass of C = 12.01 g/molMoles of C = 30.00 g / 12.01 g/mol ≈ 2.50 molesThe balanced chemical equation shows that 5 moles of C are needed for one reaction cycle, which absorbs 239.9 kJ of heat.
To find the heat absorbed for 2.50 moles of C, we use the proportion:
Heat absorbed = (2.50 moles / 5 moles) × 239.9 kJ ≈ 119.95 kJ
Therefore, the heat absorbed when 30.00 g of C(s) reacts is ≈ 119.95 kJ.
Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substances. One half-reaction shows the oxidation while the other shows the reduction. When the two half-reactions are combined, the overall reaction is obtained.
Answer:
K2Cr2O7 + HI + HClO4 → Cr(ClO4)3 + KClO4 + I2 + H2O
↓
8HClO4 + K2Cr2O7 + 6HI → 3I₂ + 2Cr(ClO4)3 + 2KClO4 + 7H2O
Explanation:
I put and example of what you said.
Potassium dichromate and iodide acid react with perchloric acid to generate chromium perchlorate, potassium chlorate, iodine and water.
First of all think all the oxidation number of each element. One ox. number will increase and the other decrease, so those will be our half reactions (one of reduction, the other of oxidation).
In the case above, I in HI acts with -1 and I2 has 0 (all elements in ground state has 0 as oxidation number). (Increase) - Oxidation
Cr in K2Cr2O7 acts with +6, in Cr(ClO4)3 is 3+ (Decrease) - Reduction
So, the first half reaction is:
2I⁻ → I₂ + 2e⁻ (OXIDATION)
I have to put 2 iodides to ballance, so the total charge is 2-. It has to release 2 electrons.
Cr2O7²⁻ → 2Cr³⁺
In products side, I have to add 2 chromes but we don't have the charges ballanced. At the main equation, we have acids so this redox occurs in an acidic medium. In the acidic medium we add water, the same as oxygen we have, so:
Cr2O7²⁻ → 2Cr³⁺ + 7H2O
and in reactant side, we add protons the same as hydrogen, we have, in this case like this
14H⁺ + Cr2O7²⁻ → 2Cr³⁺ + 7H2O
finally, we add the electrons. Chrome to decrease +6 to +3 had to lose 3 electrons but, we have 2 Cr, so 6 in total. These are final the 2 half reaction
14H⁺ + Cr2O7²⁻ + 6e- → 2Cr³⁺ + 7H2O (REDUCTION)
2I⁻ → I₂ + 2e⁻ (OXIDATION)
Electrons are not ballanced, we have to multiply by a minimum common multiple. For 2 and 6, this number is 12 so:
(14H⁺ + Cr2O7²⁻ + 6e- → 2Cr³⁺ + 7H2O) .2
(2I⁻ → I₂ + 2e⁻) .6
Afterwards, we can sum the reactions:
28H⁺ + 2Cr2O7²⁻ + 12e- + 12I⁻ → 6I₂ + 12e⁻ + 4Cr³⁺ + 14H2O
As we have 12e- in both sides, we cancel them
28H⁺ + 2Cr2O7²⁻ + 12I⁻ → 6I₂ + 4Cr³⁺ + 14H2O (still balanced)
Look that all the stoichiometry is even, so we can /2.
14H⁺ + Cr2O7²⁻ + 6I⁻ → 3I₂ + 2Cr³⁺ + 7H2O
So the final reaction is:
8HClO4 + K2Cr2O7 + 6HI → 3I₂ + 2Cr(ClO4)3 + 2KClO4 + 7H2O
We have in total 14H+, so 6 protons are for HI and 8 for the HClO4.
Which of the following is not true for an exothermic reaction?
A. the products have a higher enthalpy than reactants
B. the temperature of the surroundings rise
C. the enthalpy change for the reaction is negative heat flows from the rxn system to the surroundings
Answer:
The false statement regarding an exothermic reaction is: A. the products have a higher enthalpy than reactants
Explanation:
An exothermic reaction is a type of chemical reaction that involves the release of energy from the system to the surroundings. Thus increasing the temperature of the surroundings.
In this reaction, the enthalpy or energy of the reactants is greater than the enthalpy or energy of the products. ([tex]\Delta H_{f} (Products) < \Delta H_{f} (reactants)[/tex])
As the enthalpy change of a reaction: [tex]\Delta H_{r} = \sum \Delta H_{f} (Products) - \sum \Delta H_{f} (reactants)[/tex]
Therefore, the enthalpy change for an exothermic reaction is negative ([tex]\Delta H_{r} < 0[/tex])
Which is an example of transforming potential energy to kinetic energy? Select two options. changing thermal energy to electrical energy changing chemical energy to thermal energy changing nuclear energy to radiant energy changing radiant energy to electrical energy changing mechanical energy to chemical energy
Answer:
changing chemical energy to thermal energy
changing nuclear energy to radiant energy
Explanation:
Chemical energy can also be referred to as chemical potential energy which is the energy stored in the bonds of chemical compounds while thermal energy is a form of kinetic energy because it also involves the movement of particles.
Nuclear energy is the potential energy present in an atom while radiant energy is a form of kinetic energy that involves the movement of particles or waves during electromagnetic radiation.
The examples of transforming potential energy to kinetic energy are changing the chemical energy in food to kinetic energy for movement, such as riding a bicycle, and converting electromagnetic radiation (light energy) from the sun into chemical energy through photosynthesis in plants.
Explanation:Transforming potential energy to kinetic energy occurs when an object, initially at rest due to its position in a gravitational field, descends. As the object falls, potential energy decreases, and kinetic energy increases. This conversion follows the principle of conservation of energy, with the total energy remaining constant, neglecting external factors like air resistance.
The examples of transforming potential energy to kinetic energy are: changing the chemical energy in food to kinetic energy for movement, such as riding a bicycle, and converting electromagnetic radiation (light energy) from the sun into chemical energy through photosynthesis in plants.
Which of the following is true?
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
b. A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.
c. A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
d. A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.
Answer:
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
Explanation:
The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:
[tex]ΔG = ΔH - TΔS[/tex]
When ΔG < 0 the reaction is spontaneous in the forward direction.When ΔG > 0 the reaction is nonspontaneous. The reaction isspontaneous in the opposite direction.
When ΔG = 0 the system is at equilibrium.If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.
An apparatus consists of a 4.0 dm^3 flask containing nitrogen gas at 25°c and 803 kPa, it is joined by a valve to a 10.0dm^3 flask containing argon gas 25°c and 47.2KPa. The valve is opened and the gases mix , what is the partial pressure of each gas and calculate the total pressure of the gas mixture
Answer:
so total pressure is = 261.92 kPa
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
Explanation:
PV = nRT
n = PV / RT
as we know that 1dm3 = 1 Liter
smaller flask before mixing
ideal gas constant
n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))
n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))
n = (3212 kPa L) /2477.7
n= 1.29 mol
In the larger flask:
n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =
n = (472 kpaL/ 2477.7
n= 0.19 mol
PV = nRT
P = nRT / V
After mixing:
P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)
= (1.48) x (2477.7 / (14 L)
= 261.92 kPa total pressure
so total pressure is = 261.92 kPa
a)
(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =
= 49.59/1.48 = 33.50 Kpa for Arg
b)
(261 kPa total) - (33.50 kPa Ar) = 227 kPa for N2
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?
Molecular mass of BaSO4 233.39 g/mol
Molecular mass of Ba = 137.327 g/mol
Answer:
31.37%
Explanation:
For this case, you should consider the following reaction:
Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O
For which you obtain the precipitate of BaSO₄
In order to obtain the mas of Barium on the precipitate, you may use the following formula:
gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)
Where:
gBa= mass of Barium
M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate
M₍Ba₎= mass of Barium from the original sample
M₍BaSO₄₎= mass of BaSO₄ from the precipitate
gBa= (0.5331)x(137.327/233.39)= 0.3136 g
Then we ontain the percentage of Barium multiplying by 100:
% Ba on the original sample= 31.36%
Final answer:
The percentage of barium in the compound is calculated by finding the moles of BaSO₄, converting it to moles of Ba, and then finding the mass of Ba before expressing it as a percentage of the initial sample mass. The resulting percentage of barium is approximately 43.99%.
Explanation:
To determine the percentage of barium in the compound, we must calculate the moles of BaSO₄ and then use the molar mass of barium to find the mass of barium in the initial sample. We can calculate the moles of BaSO₄ using the following steps:
First, use the mass of the BaSO₄ precipitate formed (0.5331 g) and the molar mass of BaSO₄ (233.39 g/mol) to find the moles of BaSO₄.Moles of BaSO₄ = mass of precipitate / molar mass of BaSO₄
= 0.5331 g / 233.39 g/mol
= 0.002284 mol (rounded to six decimal places)
Since the molar ratio of Ba to BaSO₄ is 1:1, the moles of Ba will be the same as the moles of BaSO₄.Next, multiply the moles of Ba by the molar mass of Ba to find the mass of Ba in the sample.Mass of Ba = moles of Ba × molar mass of Ba= 0.002284 mol × 137.327 g/mol
= 0.3137 g (rounded to four decimal places)
Finally, we calculate the percentage of barium in the sample using the initial sample mass (0.713 g).
Percentage of barium = (mass of Ba / initial sample mass) × 100%
= (0.3137 g / 0.713 g) × 100%
= 43.99%
Therefore, the percentage of barium in the compound is approximately 43.99%.
If a chemical reaction such as photosynthesis begins with 6 atoms of carbon C , how many atoms of carbon C should be in the products?a. 12 atoms of carbon Cb. 6 atoms of carbon Cc. 3 atoms of carbon Cd. 2 atoms of carbon C
Answer:
Option B. 6 atoms of carbon
Explanation:
In any chemical reaction, atoms from reactant side and product side must be the same.
This is photosynthesis reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
In the photosynthesis equation, 6CO₂ + 6H₂O becomes C₆H₁₂O₆ + 6O₂, conserving the number of carbon atoms. Therefore, if the reaction starts with 6 atoms of carbon, there will also be 6 atoms of carbon in the products.
When photosynthesis occurs, it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Hence, the number of carbon atoms in the reactants must equal the number of carbon atoms in the products. The balanced chemical equation for photosynthesis is 6CO₂ + 6H₂O ⇒C₆H₁₂O₆ + 6O₂.
For this equation, 6 molecules of carbon dioxide (CO₂) containing 1 carbon atom each combine with water (H₂O) to produce glucose (C₆H₁₂O₆) and oxygen (O₂). Since we begin with 6 carbon atoms in the 6 molecules of carbon dioxide, the glucose product will also contain 6 carbon atoms because the carbon atoms are simply rearranged in this reaction.
Therefore, the answer to the question is: 6 atoms of carbon C should be in the products, which corresponds to option b. 6 atoms of carbon C.
An aqueous mixture containing starch (a colloid), Nacl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?
a) starch outside
b) water inside only
c) albumin inside
d) albumin inside and outside
e) starch inside and outside
Answer:
c) albumin inside
Explanation:
Let us see look closer at each answer and see why it is wrong/correct.
a) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
b) There cannot be water inside only because it would mean starch and albumin diffused out and they cannot pass through the bag
c) This is correct. The bag isn't permeable to albumin so it stays inside
d) Albumin cannot be outside the bag because the bag isn't permeable to albumin. The albumin stays inside
e) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
After dialysis, albumin - a colloid, will remain inside the dialyzing bag because the dialysis membrane is designed to let only small molecules and ions pass through, not large colloidal particles. Starch, another colloid, will remain inside for the same reason. Water will reach equilibrium, moving in and out of the membrane.
Explanation:The question relates to the process of dialysis, which is used to separate different substances in a mixture based upon their size. An aquatic mixture containing both colloids like starch and albumin and small molecules like NaCl (sodium chloride) and glucose is placed in a dialyzing bag to see where the substances will be after the process is complete. A characteristic of dialysis is that the dialyzing membrane allows small molecules and ions to pass through, while retaining larger colloidal particles.
Referring to the given options, the correct descriptions after dialysis are as follows:
Therefore, the accurate statement concerning the location of the substances post-dialysis is
c) albumin inside
While chemical buffer systems like the bicarbonate-carbonic acid buffer system are primary lines of defense in preventing pH imbalances, what are other organs are associated with this defense?
A. Lung and Kidneys
B. Lungs and Stomach
C. Lungs and Liver
D. Lungs and Intestinal tract
Answer:
Lung and Kidneys are associated with this defence.
Explanation:
An aqueous solution that resists change because of the addition of an acid or base solution is known as a buffer. Thus, it comprises of conjugate base and a weak acid. The bicarbonate buffer system is the one that is associated with the respiratory system.
To maintain the pH of and duodenum this buffer system balances carbonic acids, bicarbonate ions and carbon dioxide. It serves to neutralises gastric in the stomach.
Calculate the molarity and mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Molarity and mole fraction of acetone in solution can be calculated by using the densities and molality. Molarity is derived by dividing number of moles of acetone by volume of solution and mole fraction is derived by dividing moles of acetone by total moles in solution.
Explanation:To calculate the molarity and the mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH), we need to first convert the molal concentration (moles of solute per kilogram of solvent) to molar concentration (moles of solute per liter of solution).
Using given data, density of acetone is 0.788 g/cm3 and ethanol 0.789 g/cm3. As volumes are additive, volume of solution = volume of solute + volume of solvent. With this data, we can find number of moles of acetone and ethanol respectively.
For molarity (M), we divide number of moles of solute (acetone) by volume of solution in liters. For mole fraction (χ), we divide number of moles of solute (acetone) by total number of moles in solution (acetone + ethanol moles).
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