The mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.
Given:
Volume of solution (V) = 375 mL = 0.375 L
Molarity of sodium ions (c) = 0.900 M
Using the formula for molarity: c = n/V
Since each mole of sodium phosphate produces 3 moles of sodium ions:
Effective molarity of sodium phosphate solution = c / 3 = 0.300 M
Number of moles of sodium phosphate (n) = c * V = 0.300 M * 0.375 L = 0.1125 mol
Molar mass of sodium phosphate (M) = 164 g/mol
Mass of sodium phosphate (m) = n * M = 0.1125 mol * 164 g/mol = 18.45 g
Therefore, the mass of sodium phosphate needed to make 375 mL of a solution with a molarity of sodium ions of 0.900 M is indeed 18.45 grams.
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To make a solution of tribasic sodium phosphate with a concentration of 1.50 M Na+ ions in 375 mL, you will need 276.38 grams of Na3PO4.
Explanation:To calculate the grams of Na3PO4 needed, we need to use the equation relating moles, volume, and concentration:
Moles of Na+ ions = Molarity of Na+ ions x Volume of solution (in L)
Moles of Na3PO4 = 3 x Moles of Na+ ions
First, convert the volume in mL to L (375 mL = 0.375 L). Then, substitute the values into the equation:
Moles of Na+ ions = 1.50 M x 0.375 L = 0.5625 mol
Moles of Na3PO4 = 3 x 0.5625 mol = 1.6875 mol
Finally, convert moles to grams using the molar mass of Na3PO4 (163.94 g/mol):
Mass of Na3PO4 = 1.6875 mol x 163.94 g/mol = 276.38 g
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The limiting reactant is completely consumed in a chemical reaction. (T/F)
Answer: Yes
Explanation:
Limiting reagent is the reagent which limits the formation of product as it gets completely consumed in the reaction.
Excess reagent is the reagent which is left unreacted in the reaction.
For example: [tex]2HCl+Ca\rightarrow CaCl_2+H_2[/tex]
If there are 2 moles of [tex]HCl[/tex] and 2 moles of [tex]Ca[/tex]
As can be seen from the chemical equation,
2 moles of hydrochloric acid react with 1 mole of calcium.
Thus 2 moles of [tex]HCl[/tex] will completely react with 1 mole of calcium and (2-1)=1 mole of calcium will remain as such.
Thus HCl is the limiting reagent as it limits the formation of product and calcium is the excess reagent as it is left unreacted.
Final answer:
The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction. It is completely consumed in the reaction.
Explanation:
The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. The other reactant or reactants are considered to be in excess. To determine the limiting reactant, you need to compare the amount of each reactant present in the reaction to the stoichiometric ratios in the balanced chemical equation. The reactant with the smallest amount is the limiting reactant.
If a mixture of total pressure 50 psia obeys Raoult's law and a species has a vapour presse of 20psia, what is the DePriester K-value for the species in the mixture?
Answer : The DePriester K-value for the species in the mixture is, 0.4
Explanation :
According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.
[tex]p_i=X_i\times p[/tex] ........(1)
According to the Raoult's law, the partial pressure exerted in gas phase by a component is equal to the product of the vapor pressure of that component and its mole fraction for an ideal liquid solution.
[tex]p_i=Y_i\times p_v[/tex] ........(2)
When the gas and the liquid are in equilibrium then these partial pressures must be the same.
[tex]X_i\times p=Y_i\times p_v[/tex]
or,
[tex]\frac{X_i}{Y_i}=\frac{p_v}{p}[/tex]
This ration is called as equilibrium ratio [tex]K_i[/tex] of the i-th component.
[tex]K_i=\frac{X_i}{Y_i}=\frac{p_v}{p}[/tex] .......(3)
As we are given that,
Total pressure = [tex]p=50psia[/tex]
The vapor pressure = [tex]p_v=20psia[/tex]
According to the relation (3), we get
[tex]K_i=\frac{X_i}{Y_i}=\frac{p_v}{p}=\frac{20}{50}=0.4[/tex]
Therefore, the DePriester K-value for the species in the mixture is, 0.4
The weak base ammonia, NH3, and the strong acid hydrochloric acid react to form the salt ammonium chloride, NH4Cl. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.289 M solution of ammonium chloride at 25∘C
Final answer:
To calculate the pH of a 0.289 M ammonium chloride solution, first determine the Ka of the ammonium ion from the Kb of ammonia and set up an ICE table to find the hydronium ion concentration. Then calculate the pH using the negative logarithm of the hydronium ion concentration.
Explanation:
The pH of a solution of ammonium chloride can be calculated by first determining the Ka (acid dissociation constant) of the ammonium ion (NH4+), which is the conjugate acid of ammonia (NH3). Given that the Kb for ammonia is 1.8×10−5, the Ka for ammonium can be calculated using the relationship Ka = Kw/Kb, where Kw is the ion-product constant for water (1.0×10−14 at 25°C). In this case, Ka = 1.0×10−14 / 1.8×10−5 = 5.6×10−10.
Knowing the Ka, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentration of hydronium ions, H3O+, produced. Since ammonium chloride is a strong electrolyte, it completely dissociates in water, thus initial [NH4+] is 0.289 M, and initial [H3O+] is 0. After the equilibrium is established, we calculate the concentration of H3O+ and subsequently find the pH of the solution. The pH is determined using the formula pH = -log[H3O+]. For a solution of ammonium chloride, this results in an acidic pH due to the formation of H3O+ ions.
The pH of a 0.289 M solution of ammonium chloride is approximately 4.89
This is calculated by determining the dissociation constant (Ka) for NH₄⁺ and using the concentration of H₃O⁺ ions to find the pH.
The Ka is found using the relation to the base dissociation constant (Kb) of ammonia.To determine the pH of a 0.289 M solution of ammonium chloride (NH₄Cl), we need to consider the dissociation of the ammonium ion (NH₄⁺) in water:
NH₄⁺ (aq) + H₂O (l) ⇔ H₃O⁺ (aq) + NH₃ (aq)The equilibrium constant for this reaction is the acid dissociation constant (Ka) of the ammonium ion.
We can calculate Ka using the relation Ka = Kw / Kb.
Given Kw = 1.0 × 10⁻¹⁴ and Kb for ammonia (NH₃) as 1.8 × 10⁻⁵, we find:
Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰Assuming the dissociation of NH₄⁺ is small, the concentration of H₃O⁺ is 'x', and using the initial concentration of NH₄⁺ (0.289 M) in the expression for Ka:
Ka = [tex]\frac{[H_{3}O^{+}][NH_{3} ] }{NH_{4} ^{+} }[/tex]5.6 × 10⁻¹⁰ = [tex]\frac{(x)(x)}{(0.289 - x) }[/tex] ≈ [tex]\frac{x^{2} }{0.289}[/tex]Simplifying for 'x', we get:
x² = 5.6 × 10⁻¹⁰ * 0.289x² = 1.62 × 10⁻¹⁰x (which is [H₃O⁺]) = [tex]\sqrt{(1.62 \times 10^{-10} )}[/tex] ≈ 1.27 × 10⁻⁵ MThe pH is then calculated as:
pH = -log[H₃O⁺] ≈ -log(1.27 × 10⁻⁵) ≈ 4.89Hence, the pH of a 0.289 M solution of ammonium chloride is approximately 4.89
An achiral hydrocarbon A of molecular formula C7H12 reacts with two equivalents of H2 in the presence of Pd-C to form CH3CH2CH2CH2CH(CH3)2. One oxidative cleavage product formed by the treatment of A with O3 is CH3COOH. Reaction of A with H2 and Lindlar catalyst forms B, and reaction of A with Na, NH3 forms C. Identify compounds A, B, and C. Be sure to answer all parts.
Answer:
A) 5-methylhex-2-yne
B) (2Z)-5-methylhex-2-ene
C) (2E)-5-methylhex-2-ene
Explanation:
The given compound must be alkyne as it is undergoing reduction with two equivalents of hydrogen molecule.
Also as it is giving acetic acid on oxidative ozonolysis, it must have triple bond after two carbons in the chain.
The structure of hydrocarbon formed after reduction will give us the structure of alkyne by these information.
Reaction with hydrogen molecule in presence of Lindlar's catalyst gives cis alkene.
Reaction with hydrogen molecule in presence of Na, ammonia gives trans alkene.
The structure of compound is shown in the figure
The achiral hydrocarbon (A) with a molecular formula C7H12 is identified as Hept-1-yne. Upon reacting with a Lindlar catalyst and H2, it forms Hept-1-ene (B). When Hept-1-yne is treated with Sodium in ammonia (Na, NH3), it forms 1-heptyne (C).
Explanation:The hydrocarbon A that has a molecular formula of C7H12 should be Hept-1-yne in light of the fact that Hept-1-yne upon hydrogenation, utilizing Pd-C as a catalyst and two equivalents of H2, forms 4-methyl hexane, which is exact to what was stated in the question. When Hept-1-yne is treated with ozone (O3), it gives two oxidative cleavage products, one being CH3COOH (acetic acid). The reaction gives us a clue about the presence of a triple bond at the end of the heptane chain. Hence, the structure of compound A (Hept-1-yne) is identified as CH3-(CH2)4-C≡CH.
Compound B can be identified as Hept-1-ene. This is due to the fact that Hept-1-yne, upon reacting with a Lindlar catalyst and H2, forms Hept-1-ene. This conversion is a result of the partial reduction of the triple bond to a double bond.Lastly, compound C can be identified as 1-heptyne. This is because when we treat Hept-1-yne with Sodium in ammonia (Na, NH3, it selectively reduces the triple bond to a trans double bond, a process known as dissolving metal reduction.
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Enzymes function most efficiently at the temperature of a typical cell, which is 37 degrees Celsius. What happens to enzyme function when the temperature rises? What happens to enzyme function when the temperature drops?
Answer:
At high temperature the enzyme becomes denatured.
Lower temperature the enzymes become inactive.
Explanation:
Enzymes do not work at high temperature since the temperature kills the cells. And at LOW temperature enzymes have no energy to perform their work hence becomes inactive.
vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compare this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?
Answer:
q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.
Explanation:
1. Heat absorbed
q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ
2. Change in volume
V(water) = 0.018 L
pV = nRT
1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K
V = 30.62 L
ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L
3. Work done
w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm
w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ
4. Why the difference?
Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.
The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.20. If the molar absorptivity of yellow dye at 427 nm is 27400 M–1cm–1, what is the concentration of the solution?
Answer:
The concentration of the solution, [tex]C=7.2992\times 10^{-6} M[/tex]
Explanation:
The absorbance of a solution can be calculated by Beer-Lambert's law as:
[tex]A=\varepsilon Cl[/tex]
Where,
A is the absorbance of the solution
ɛ is the molar absorption coefficient ([tex]L.mol^{-1}.cm^{-1}[/tex])
C is the concentration ([tex]mol^{-1}.L^{-1}[/tex])
l is the path length of the cell in which sample is taken (cm)
Given,
A = 0.20
ɛ = 27400 [tex]M^{-1}.cm^{-1}[/tex]
l = 1 cm
Applying in the above formula for the calculation of concentration as:
[tex]A=\varepsilon Cl[/tex]
[tex]0.20= 27400\times C\times 1[/tex]
[tex]C = \frac{0.20}{27400\times 1} M[/tex]
So , concentration is:
[tex]C=7.2992\times 10^{-6} M[/tex]
The concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
BEER-LAMBERT EQUATION:
The concentration of a solution/sample measured using a spectrophotometer can be calculated using beer-lambert's equation as follows:A = εbc
Where;
ε = molar absorptivity of the yellow dye solution b = the path length of cuvettec = the concentration of the yellow dye solutionA = absorbance of the yellow dyec = A ÷ εb
According to this question, the absorbance of the yellow dye solution at 427 nm is 0.20, its molar absorptivity at 427 nm is 27400 M-¹cm-¹ and the cuvette length is 1.0cm. Hence, the concentration can be calculated as follows:c = A ÷ εb
c = 0.20 ÷ (27400 × 1)
c = 0.20 ÷ 27400
c = 7.29 × 10-⁶M
Therefore, the concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 + 2HCl ⟶CaCl2 + H2O + CO2
A) How many grams of calcium chloride will be produced when 26.0g of calcium carbonate are combined whith 12.0g of hydrochloric acid?
B) Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
Answer: a) 18.3 grams
b) [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
Explanation:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of calcium carbonate}=\frac{26g}{100g/mol}=0.26moles[/tex]
[tex]\text{Number of moles of hydrochloric acid}=\frac{12g}{36.5g/mol}=0.33moles[/tex]
According to stoichiometry:
2 mole of [tex]HCl[/tex] react with 1 mole of [tex]CaCO_3[/tex]
0.33 moles of [tex]HCl[/tex] will react with=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCO_3[/tex]
Thus [tex]HCl[/tex] is the limiting reagent as it limits the formation of product and [tex]CaCO_3[/tex] is the excess reagent.
2 moles of [tex]HCl[/tex] produce = 1 mole of [tex]CaCl_2[/tex]
0.33 moles of [tex]HCl[/tex] produce=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCl_2[/tex]
Mass of [tex]CaCl_2=moles\times {\text{Molar Mass}}=0.165\times 111=18.3g[/tex]
As 0.165 moles of [tex]CaCO_3[/tex] are used and (0.33-0.165)=0.165 moles of [tex]CaCO_3[/tex] are left unused.
Mass of [tex]CaCO_3[/tex] left unreacted =[tex]moles\times {\text {Molar mass}}=0.165\times 100=16.5g[/tex]
Thus 18.3 g of [tex]CaCl_2[/tex] are produced. [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
An initially evacuated 1.5 m tank is fed (adiabatically) with steam from a line available at a constant 15 MPa and 400 °C until the tank pressure reaches 15 MPa. What is the final mass of water in the tank in kg?
Answer:
95.8 kg
Explanation:
At the end of the feeding process, there is steam in the tank at 15 MPa and 400ºC because the process is adiabatic. So, use the steam tables (In this case I use data from van Wylen Six Edition, table B.13) in order to get the specific volume of superheated steam.
The specific volume data reported is [tex]v=0.01565\frac{m^{3}}{kg}[/tex]
The mass can be calculated with the definition of specific volume:
[tex]v=\frac{V}{m}\\m=\frac{V}{v}=\frac{1.5m^{3}}{0.01565\frac{m^{3}}{kg}} =95.8kg[/tex]
Excited lithium ions emit radiation at a wavelength of 670.8 nm in the visible range spectrum. Calculate the frequency of a photon of this radiation
Answer : The frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]
Explanation : Given,
Wavelength of the radiation = 670.8 nm
First we have to convert wavelength form 'nm' to 'm'.
Conversion used : [tex](1nm=10^{-9}m)[/tex]
So, the wavelength of the radiation = 670.8 nm = [tex]670.8\times 10^{-9}m[/tex]
Now we have to calculate the frequency of a photon of radiation.
Formula used : [tex]\nu =\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of a photon of radiation
[tex]\lambda[/tex] = wavelength of the radiation
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get the frequency of a photon of radiation.
[tex]\nu =\frac{3\times 10^8m/s}{670.8\times 10^{-9}m}[/tex]
[tex]\nu =4.47\times 10^9s^{-1}[/tex]
Therefore, the frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]
Answer:
[tex]f=4.47x10^5GHz[/tex]
Explanation:
Hello,
In this case, we relate the speed of light, wavelength and frequency via the shown below equation expressed in the proper SI system of units:
[tex]f=\frac{c}{\lambda } =\frac{3x10^8m/s}{670.8nm*\frac{1x10^{-9}m}{1nm} } =4.47x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ f=4.47x10^5GHz[/tex]
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The vapor pressure of water is 23.76 mm Hg at 25°C. How many grams of urea, CH4N2O, a nonvolatile, nonelectrolyte (MW = 60.10 g/mol), must be added to 238.2 grams of water to reduce the vapor pressure to 23.22 mm Hg ? water = H2O = 18.02 g/mol.
Answer:
18.700 g
Explanation:
As the urea is a nonvolatile and nonelectrolyte solute, it will reduce the vapor pressure of the solution according to:
[tex]P_{vs} =P_{w} *x_{w}[/tex]
Where [tex]P_{vs}[/tex] is the vapor pressure of the solution, [tex]P_{w}[/tex] is the vapor pressure of the pure water, and [tex]x_{w}[/tex] is the molar fraction of water. This equation applies just for that kind of solutes and at low pressures (23.76 mmHg is a low pressure).
From the equation above lets calculate the water molar fraction:
[tex]23.22mmHg=23.76mmHg*x_{w}\\ x_{w}=\frac{23.22mmHg}{23.76mmHg}=0.977[/tex]
So, the molar fraction of the urea should be: [tex]x_{urea}=1-x_{w}=0.023[/tex]
Then, calculate the average molecular weight:
[tex]M=x_{w}*MW_{w}+x_{urea}*MW_{urea}\\ M=0.977*18.02+0.023*60.10=18.989[/tex]
The molar fraction of urea is:
[tex]0.023=\frac{X urea mol}{S solution moles}=\frac{x urea grams}{238.2+x (solution grams)}*\frac{1 urea mol}{60.10 g}*\frac{18.989 solution grams}{1 solution mol}[/tex]
Solving for x,
[tex]x=18.700g[/tex]
Part C Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm. What is the radius of a gallium atom? Express your answer numerically in picometers. View Available Hint(s) radius = nothing pm p m Submit
Final answer:
The radius of a gallium atom in a primitive cubic unit cell, with an edge length of 362 pm, is calculated to be 181 picometers by dividing the edge length by 2.
Explanation:
The question asks for the radius of a gallium atom given that gallium crystallizes in a primitive cubic unit cell with an edge length of 362 pm. In a primitive cubic unit cell, the atoms are located at the corners of the cube, and the length of the edge of the cube is equal to twice the atomic radius. Therefore, to find the radius of the gallium atom, we divide the edge length by 2.
Radius of gallium atom = edge length / 2 = 362 pm / 2 = 181 pm.
This calculation reveals that the radius of a gallium atom is 181 picometers in a primitive cubic unit cell structure.
A chef is making deluxe sandwiches for a special guest. They call for 3 slices of jalapeno cheddar cheese and 5 slices of honey ham. If he has 180 slices of each in the kitchen storage, how many sandwiches can he make before one of his ingredients run out.?
Answer:
The chef can be able to make 36 sandwiches.
Explanation:
3 slices of jalapeno cheddar cheese + 5 slices of honey ham → 1 sandwich
According to information, 3 slices of jalapeno cheddar cheese will combine with 5 slices of honey ham to give 1 sandwich.
180 slices of jalapeno cheddar cheese will combine with:
[tex]\frac{5}{3}\times 180=300 [/tex] slices of honey ham
But we are having only 180 slices of honey ham. tghe number of sandwiches will depend upon number of slices of honey ham.
180 slices of honey ham will combine with:
[tex]\frac{3}{5}\times 180=108 [/tex] slices of jalapeno cheddar cheese
From 5 slices of honey ham we can make 1 sandwich,then from 180 slices of hinry ham we will be able make:
[tex]\frac{1}{5}\times 180=36 sandwiches[/tex]
The chef can be able to make 36 sandwiches.
Calculate the pH of a 0.22 M ethylamine solution.
Answer:
answer is 12.18
Explanation:
(C2H5NH2, Kb = 5.6 x 10-4.)
Answer:
pH = 10.1
Explanation:
For weak base solutions [OH] = SqrRt([Base]·Kb
Then, from pH + pOH = 14 => pH = 14 - pOH
[OH} = SqrRt[(0.22)(5.6 x 10⁻⁴)] = 3.91
pH = 14 - 3.91 = 10.1
1. In an equilibrium experiment, acetic acid (which is a weak acid) is mixed with sodium acetate (a soluble salt), with methyl orange as an indicator. Explain this phenomenon by using the common ion effect. Include equations in your explanation.
Explanation:
It is known that acetic acid is a weak acid. It's equilibrium of dissociation will be represented as follows.
[tex]CH_{3}COOH(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)[/tex]
On the other hand, sodium acetate ([tex]CH_{3}COONa[/tex]) is a salt of weak acid, that is, [tex]CH_{3}COOH[/tex] and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.
Since, acetic acid is a weak acid but still it is an acid. So, when methyl orange is added in a solution of acetic acid then it given a reddish-orange color because of its acidity.
When sodium acetate is mixed into this solution then it will dissociate as follows.
[tex]CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)[/tex]
As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.
So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.
As a result, there will be decrease in the concentration of hydronium ions. As a result, there will be increase in the pH of the system.
Hence, color of methyl orange will change from reddish orange to yellow. This shift in equilibrium is due to the common ion which is [tex]CH_{3}COO^{-}[/tex] ion.
Ethylene diamine tetra-acetic acid (EDTA) is a water-soluble compound that readily combines with metals, such as calcium, magnesium, and iron. The molecular formula for EDTA is C10N2O8H16. One EDTA molecule complexes (associates with) one metal atom. A factory produces an aqueous waste that contains 20 mg/L calcium and collects the waste in 44-gallon drums. What mass (g) of EDTA would need to be added to each drum to completely complex all of the calcium in the barrel? (1 gal = 3.785 L)
Answer: The mass of EDTA that would be needed is 24.3 grams.
Explanation:
We are given:
Concentration of [tex]Ca^{2+}[/tex] ions = 20 mg/L
Converting this into grams/ Liter, we use the conversion factor:
1 g = 1000 mg
So, [tex]\Rightarrow \frac{20mg}{L}\times {1g}{1000mg}=0.02g/L[/tex]
Now, we need to calculate the mass of calcium present in 44 gallons of drum.
Conversion factor used: 1 gallon = 3.785 L
So, 44 gallons = (44 × 3.785)L = 166.54 L
Calculating the mass of calcium ions in given amount of volume, we get:
In 1L of volume, the mass of calcium ions present are 0.02 g.
Thus, in 166.54 L of volume, the mass of calcium ions present will be = [tex]\frac{0.02g}{1L}\times 166.54L=3.3308g[/tex]
The chemical equation for the reaction of calcium ion with EDTA to form Ca[EDTA] complex follows:
[tex]EDTA+Ca^{2+}\rightarrow Ca[EDTA][/tex]
Molar mass of EDTA = 292.24 g/mol
Molar mass of [tex]Ca^{2+}[/tex] ion = 40 g/mol
By Stoichiometry of the reaction:
40 grams of calcium ions reacts with 292.24 grams of EDTA.
So, 3.3308 grams of calcium ions will react with = [tex]\frac{292.24g}{40g}\times 3.3308g=24.33g[/tex] of EDTA.
Hence, the mass of EDTA that would be needed is 24.3 grams.
A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from the nitrogen atoms and compound Y is produced from the metal atoms in the reactant. What are the compounds X and Y? X is nitrogen dioxide, and Y is a metal halide. X is nitrogen dioxide, and Y is a metal oxide. X is nitrogen gas, and Y is a metal sulfate. X is nitrogen gas, and Y is a metal oxide.
Answer:
The correct answer is: X is nitrogen dioxide, and Y is a metal oxide
Explanation:
Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.
The general equation is given as:
[tex]4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x[/tex]
Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.
In a combustion reaction, a compound containing nitrogen and a metal typically forms nitrogen dioxide (compound X) from the nitrogen atoms and a metal oxide (compound Y) from the metal atoms.
Explanation:In the context of your question about how a compound containing nitrogen and a metal reacts in a combustion reaction, the outcome depends on the specific reactant. Usually, compounds containing nitrogen atoms tend to form nitrogen oxides under high heat or combustion conditions, with nitrogen dioxide (NO2) being a common example. This would be compound X.
Regarding the metal component, in a combustion reaction, metals commonly react with oxygen in the environment to produce metal oxides. This would make compound Y a metal oxide. Therefore, the correct pair of products according to your question tends to be: X is nitrogen dioxide, and Y is a metal oxide.
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Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800˚C (1073 K) is 3.6 × 1023 m-3 . The atomic weight and density (at 800˚C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3 .
Answer:
The energy for vacancy formation in silver is 1.1 ev/atom
Explanation:
The total number of sites is equal to:
[tex]N=\frac{N_{A} \rho }{A}[/tex]
Where
NA = Avogadro´s number = 6.023x10²³atom/mol
A = atomic weight of silver = 107.9 g/mol
ρ = density of silver = 9.5 g/cm³
Replacing:
[tex]N=\frac{6.023x10^{23}*9.5 }{107.9} =5.3x10^{22} atom/cm^{3} =5.3x10^{28} m^{-3}[/tex]
The energy for vacancy is equal:
[tex]Q=-RTln(\frac{N_{v} }{N} )[/tex]
Where
R = 8.314 J/mol K = 8.614x10⁻⁵ev/atom K
T = 800°C = 1073 K
Nv = number of vacancy = 3.6x10²³m⁻³
Replacing:
[tex]Q=-8.614x10^{-5} *1073*ln(\frac{3.6x10^{23} }{5.3x10^{28} } )=1.1ev/atom[/tex]
Final answer:
The energy for vacancy formation in silver is 2.87 x 10^-19 J.
Explanation:
To calculate the energy for vacancy formation in silver, we can use the equation: E = kTln(N/V), where E is the energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, N is the number of vacancies, and V is the volume.
Given that the equilibrium number of vacancies is 3.6 x 10^23 m^-3, we can use the density of silver to calculate the volume. The density of silver at 800˚C is 9.5 g/cm^3, which is equivalent to 9.5 x 10^6 kg/m^3.
Therefore, the energy for vacancy formation in silver is:
E = (1.38 x 10^-23 J/K) x (1073 K) x ln(3.6 x 10^23 / (9.5 x 10^6)) = 2.87 x 10^-19 J.
C2H4(g) + H2(g) → C2H6(g) ΔH = –137.5 kJ; ΔS = –120.5 J/K Calculate ΔG at 25 °C and determine whether the reaction is spontaneous. Does ΔG become more negative or more positive as the temperature increases?
Answer:-ΔG=-101.5KJ
Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.
We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin
Also we need to convert the temperature in Kelvin as it is given in degree celsius.
[tex]\Delta H=-137.5\\ \Delta S=-120J/K\\ \Delta S=-0.120KJ/K\\ T=25^{.C}\\ T=273+25=298 K\\ \Delta G=?\\ \Delta G=\Delta H-T\Delta S\\\Delta G=-137.5KJ-(278\times -0.120)\\ \Delta G=-137.5+35.76\\\Delta G=-101.74\\\Delta G=Negative[/tex]
After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ
For a reaction to be spontaneous the value of \Delta G \ must be negative .
As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.
In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .
Hence the value of ΔG would become more positive with the increase in temperature.
So we found the value of ΔG to be -101.74KJ
Answer:
ΔG = -101.591 KJ
Explanation:
Gibbs free energy -
It is a thermodynamic quantity , which is given by the change in enthalpy minus the product of the change in entropy and absolute temperature.
i.e.,
ΔG is given as the change in gibbs free energy ( KJ )
ΔS is given as the change in entropy ( KJ /K )
ΔH is given as the change in ethalphy ( KJ )
T = temperature ( Kelvin ( K ))
ΔG = ΔH - TΔS
The sign of ΔG determines the reaction spontaneity , as
ΔG = negative , the reaction is spontaneous and
If ΔG = positive , the reaction is non spontaneous .
Given -
For the reaction ,
C₂H₄ (g) + H₂(g) ---> C₂H₆(g)
ΔH = - 137.5 KJ
ΔS = - 120.5 J /K
Since ,
1 KJ = 1000 J
1 J = 1 / 1000KJ
ΔS = - 120.5 / 1000 KJ /K
ΔS = -0.1205 KJ /K
T = 25°C
(adding 273 To °C to convert it to K)
T = 25 + 273 = 298 K
Putting the values on the above equation ,
ΔG = ΔH - TΔS
ΔG = -137.5 KJ - 298 * (-0.1205 KJ / K)
ΔG = -137.5 KJ + 35.909 KJ
ΔG = -101.591 KJ
Since,
the value of ΔG is negative ,
hence, the reaction is spontaneous.
For the above reaction ,
If the temperature is increased ,
ΔG = ΔH - TΔS
From the above equation ,
the value of TΔS will increase ,
As a result the value of ΔG will be more positive , by increasing the temperature.
A rigid tank contains 20 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psia and 90°F, respectively. Determine the amount of air added to the tank. The gas constant of air is R
Answer : The amount of air added to the tank will be, 1.2062 Kg.
Explanation :
First we have to calculate the volume of air by using ideal gas equation.
[tex]P_1V_1=\frac{m_1RT_1}{M}[/tex]
where,
[tex]P_1[/tex] = initial pressure of air = [tex]20psia=1.36atm[/tex]
conversion used : [tex]1psia=0.068046atm[/tex]
[tex]T_1[/tex] = initial temperature of air = [tex]70^oF=294.261K[/tex]
conversion used : [tex](70^oF-32)\frac{5}{9}+273.15=294.261K[/tex]
[tex]V_1[/tex] = initial volume of air = ?
[tex]m_1[/tex] = initial mass of air = [tex]20Ibm=9071.85g[/tex]
conversion used : [tex]1lbm=453.592g[/tex]
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.36atm)\times V_1=\frac{(9071.85g)\times (0.0821L.atm/mole.K)\times (294.261K)}{M}[/tex]
[tex]V_1=\frac{161150.9299}{M}L[/tex]
Now we have to calculate the final amount of air by using ideal gas equation.
[tex]P_2V_2=\frac{m_2RT_2}{M}[/tex]
where,
[tex]P_2[/tex] = final pressure of air = [tex]23.5psia=1.599atm[/tex]
[tex]T_2[/tex] = final temperature of air = [tex]90^oF=305.37K[/tex]
[tex]V_2[/tex] = final volume of air = [tex]V_1=\frac{161150.9299}{M}L[/tex]
[tex]m_2[/tex] = final mass of air = ?
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.599atm)\times (\frac{161150.9299}{M}L)=\frac{m_2\times (0.0821L.atm/mole.K)\times (305.37K)}{M}[/tex]
[tex]m_2=10278.074g[/tex]
Now we have to calculate the amount of air added to the tank.
[tex]m_2-m_1=10278.074g-9071.85g=1206.224g=1.2062Kg[/tex]
conversion used : (1 Kg = 1000 g)
Hence, the amount of air added to the tank will be, 1.2062 Kg.
The relation between the volume, the pressure, and the temperature is PV = mRT. Then the amount of air added to the tank is 1.2062 kg.
What is thermodynamics?
It is a branch of science that deals with heat and work transfer.
A rigid tank contains 20 lbm of air at 20 psi and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psi and 90°F, respectively.
The ideal gas equation is
[tex]PV = \dfrac{mRT}{M}[/tex]
[tex]P_1 = 20 \ psi = 1.36 \ atm\\\\T_1 = 70 ^oF = 294.261\ K\\\\V_1 = \ \ ? \\\\m_1 = 20 \ lbm = 9071.85 \ kg[/tex]
R = 0.0821 L atm/mole K
M = molar mass of air
Now put all the given values in the ideal gas equation, we have
[tex]\rm V_1 = \dfrac{9071.85*0.0821*294.261}{1.36M}\\\\\\V_1 = \dfrac{161150.8299}{M}[/tex]
Now we have to calculate the final amount of air by using an ideal gas equation, we have
[tex]P_2 = 23.5 \ psi = 1.599\ atm\\\\T_2 = 90^oF = 305.37\ K\\\\V_2 = V_1\\\\m_2 = \ \ ?[/tex]
Then we have
[tex]m_2 = \dfrac{1.599M* \dfrac{161150.9299}{M}}{0.0821*305.37}\\\\m_2 = 10278.074 \ g[/tex]
Now we have to calculate the amount of air added to the tank will be
⇒ m₂ - m₁ = 10278.074 - 9071.85 = 1206.244 g = 1.12062 kg
Hence, the amount of air added to the tank is 1.2062 kg.
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The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K. CH4 (g) + CCl4 (g) 2 CH2Cl2 (g) Calculate the equilibrium concentrations of reactants and product when 0.377 moles of CH4 and 0.377 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.
Final answer:
To calculate the equilibrium concentrations of reactants and product, we need to use the given equilibrium constant, Kc, and the initial concentrations of the reactants.
Explanation:
To calculate the equilibrium concentrations of reactants and product, we need to use the given equilibrium constant, Kc, and the initial concentrations of the reactants. The balanced equation for the reaction is CH4 (g) + CCl4 (g) ⇌ 2 CH2Cl2 (g).
First, calculate the initial moles of each reactant by multiplying the initial concentration by the volume of the vessel. Then, we can set up an ICE (Initial-Change-Equilibrium) table to find the change in concentration of each species and the equilibrium concentrations.
Using the ICE table and the equilibrium constant expression, we can solve for the equilibrium concentrations of CH4, CCl4, and CH2Cl2.
If the tip of the syringe, "The Titrator", was not filled with NaOH before the initial volume reading was recorded, would the concentration of acetic acid in vinegar of that trial be greater than or less than the actual concentration? Please explain your answer.
Answer:
The concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
If the syringe tip was not filled with NaOH before recording the initial volume, the concentration of acetic acid calculated would be less than the actual concentration.
If the tip of the syringe was not filled with NaOH before the initial volume reading was recorded, the concentration of acetic acid in vinegar calculated from that trial would be less than the actual concentration. This is because the actual volume of NaOH dispensed during titration would be over-reported.
When the titration is performed, the volume of NaOH required to reach the equivalence point appears larger than it truly is, leading to a miscalculation of the moles of NaOH used. Consequently, this will result in the calculated concentration of acetic acid being lower than its true value.
In titration, ensuring that the titrant (in this case, NaOH) is ready and properly measured is crucial for accurate results. The initial volume reading must be correct to avoid errors in determining the volume of NaOH added, which directly affects the accuracy of the acetic acid concentration estimation.
True or False Titanium's corrosion resistance is so strong that even titanium with oxygen impurities does no reduction in corrosion resistance.
Answer :
true
Explanation:
when titanium react with oxygen it form TiO₂ Titanium oxide which is passive in nature it forms layer of TiO₂ when large amount of oxygen is passed through titanium at very high temperature it does not react with oxygen impurities for this reason titanium mostly used in aerospace and chemical industries
Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the following information: 1. 2CO2(g)+2H2O(l)⇌CH3COOH(l)+2O2(g), K1 = 5.40×10−16 2. 2H2(g)+O2(g)⇌2H2O(l), K2 = 1.06×1010 3. CH3COOH(l)⇌2C(s)+2H2(g)+O2(g), K3 = 2.68×10−9
Answer : The value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]
Explanation :
The following equilibrium reactions are :
(1) [tex]2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2[/tex] [tex]K_1=5.40\times 10^{-16}[/tex]
(2) [tex]2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)[/tex] [tex]K_2=1.06\times 10^{10}[/tex]
(3) [tex]CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)[/tex] [tex]K_3=2.68\times 10^{-9}[/tex]
The final equilibrium reaction is :
[tex]CO_2(g)\rightleftharpoons C(s)+O_2(g)[/tex] [tex]K_{goal}=?[/tex]
Now we have to calculate the value of [tex]K_{goal}[/tex] for the final reaction.
First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:
[tex]K_{goal}=\sqrt{K_1\times K_2\times K_3}[/tex]
Now put all the given values in this expression, we get :
[tex]K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}[/tex]
[tex]K_{goal}=1.238\times 10^{-7}[/tex]
Therefore, the value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]
The value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g) can be calculated by multiplying the equilibrium constants of the individual reactions involved.
Explanation:The value of the equilibrium constant, Kgoal, for the reaction CO2(g) ⇌ C(s) + O2(g) can be determined using the given information:
2CO2(g) + 2H2O(l) ⇌ CH3COOH(l) + 2O2(g), K1 = 5.40×10-162H2(g) + O2(g) ⇌ 2H2O(l), K2 = 1.06×1010CH3COOH(l) ⇌ 2C(s) + 2H2(g) + O2(g), K3 = 2.68×10-9Since reaction 3 is the sum of reactions 1 and 2, we can use the equations to calculate the value of Kgoal:
Kgoal = K1 × K2 × K3
Substituting the values:
Kgoal = (5.40×10-16) × (1.06×1010) × (2.68×10-9) = 1.47×10-14
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In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL + 10.00 mL of the acid have been added?
Answer: The pH of the solution is 1.136
Explanation:
To calculate the moles from molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
For ammonia:Molarity of ammonia = 0.3764 M
Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol[/tex]
For nitric acid:Molarity of nitric acid = 0.3838 M
Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L
Putting values in above equation, we get:
[tex]0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol[/tex]
After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol
For the reaction of ammonia with nitric acid, the equation follows:
[tex]NH_3+HNO_3\rightarrow NH_4NO_3[/tex]
At [tex]t=0[/tex] 0.0178 0.022
Completion 0 0.0042 0.0178
As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.
To calculate the pH of the reaction, we use the equation:
[tex]pH=-\log[H^+][/tex]
where,
[tex][H^+]=\frac{0.0042mol}{0.05741L}=0.0731M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(0.0731)\\\\pH=1.136[/tex]
Hence, the pH of the solution is 1.136
A glycosidic bond can join two monosaccharide molecules to form a disaccharide.(T/F)
Answer:
True
Explanation:
A disaccharide is a sugar and the general molecular formula of a disaccharide is C₁₂H₂₂O₁₁.
A disaccharide is formed when two monosachharide units are joined by a covalent bond called the glycosidic bond.
The glycosidic bond in a disaccharide is formed by dehydration reaction between the two monosachharide units. The removal of the water molecule results in the formation of the glycosidic linkage.
For example: maltose a disaccharide, is formed when two molecules of glucose are joined by a (1→4) glycosidic bond. As, the glycosidic bond is formed between the carbon 1 of one glucose unit and carbon 4 of another glucose unit.
Therefore, in a disaccharide the two monosaccharide units are joined by a glycosidic bond or linkage.
Therefore, the given statement is TRUE.
Which of the following acids is the STRONGEST? The acid is followed by its Ka value. Which of the following acids is the STRONGEST? The acid is followed by its Ka value. HF, 3.5 × 10-4 HCOOH, 1.8 × 10-4 HClO2, 1.1 × 10-2 HCN, 4.9 × 10-10 HNO2, 4.6 × 10-4
Answer:
chlorous acid HClO₂
Explanation:
The Ka is the acidity constant and it tells you about the acidity strength of a compound. If the value of Ka is high the compound is a strong acid. If the value of Ka is low the compound is a weak acid.
The problem gives us the following compounds with the Ka values:
HF, 3.5 × 10⁻⁴
HCOOH, 1.8 × 10⁻⁴
HClO₂, 1.1 × 10⁻²
HCN, 4.9 × 10⁻¹⁰
HNO₂, 4.6 × 10⁻4
The chlorous acid HClO₂ have the highest Ka, 1.1 × 10⁻², so this one is the strongest acid.
The strength of an acid is determined by its Ka value. Among the given options: HF, HCOOH, HClO2, HCN, and HNO2, the strongest acid is HClO2 as it has the highest Ka value of 1.1 × 10-2.
Explanation:The acidity of a substance is determined by its Ka value, which is the acid dissociation constant. A higher Ka value indicates a stronger acid because it shows the acid dissociates more completely in solution. The substances you've listed are all acids and their respective Ka values are given. Here, the strongest acid would have the highest Ka value.
Looking at the given options: HF (Ka = 3.5 × 10-4), HCOOH (Ka = 1.8 × 10-4), HClO2 (Ka = 1.1 × 10-2), HCN (Ka = 4.9 × 10-10), and HNO2 (Ka = 4.6 × 10-4), the strongest acid based on their Ka values would be HClO2 as it has the highest Ka value of 1.1 × 10-2.
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Calculate the equilibrium constants K’eq for each of the following reactions at pH 7.0 and 25oC, using the ∆Go’ values given: (a) Glucose-6-phosphate + H2O → glucose + PI ∆Go’= -13.8 kJ/mol (b) Lactose + H2O → glucose + galactose ∆Go’= -15.9 kJ/mol (c) Malate → fumarate + H2O ∆Go’= +3.1 kJ/mol
The equilibrium constants K'eq for the given reactions can be calculated using the formula K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin.
Explanation:For the given reactions the equilibrium constants K'eq can be calculated using the formula: K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol K at 25°C or 298.15 K) and T is the temperature in Kelvin.
For the reaction of Glucose-6-phosphate + H2O → glucose + PI with ∆Go'= -13.8 kJ/mol, K'eq = exp([(-13.8 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])For the reaction of Lactose + H2O → glucose + galactose with ∆Go'= -15.9 kJ/mol, K'eq = exp([(-15.9 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])For the reaction of Malate → fumarate + H2O with ∆Go'= +3.1 kJ/mol, K'eq = exp([(3.1 * 10^3 J/mol) / (8.314 J/mol K *298.15 K)])
Note: In the formulas above, the Gibbs free energy change is converted to J/mol (from kJ/mol) by multiplying by 10^3.
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Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X2O3 and write a balanced equation for the reaction.
Answer:The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Explanation:
Let the molecular formula of the oxide of metal be [tex]X_2O_y[/tex]
[tex]X_2O_y+yCO\rightarrrow yCO_2+2X[/tex]
Mass of metal product = 1.68 g
Moles of metal X =[tex]\frac{1.68 g}{55.9 g/mol}=0.03005 mol[/tex]
1 mol of metal oxide produces 2 moles of metal X.
Then 0.03005 moles of metal X will be produced by:
[tex]\frac{1}{2}\times 0.03005 mol=0.01502 mol[/tex] of metal oxide
Mass of 0.01502 mol of metal oxide = 2.40 g (given)
[tex]0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g[/tex]
y = 2.999 ≈ 3
The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Final answer:
To show the simplest formula of the oxide is X2O3, we calculate the moles of metal (X) and oxygen from given masses, find their ratio, and deduce the empirical formula. The balanced equation for the reaction with carbon monoxide is X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g).
Explanation:
To prove that the simplest formula of the oxide is X2O3, first we need to calculate the moles of metal X produced. Since the molar mass of X is given as 55.9 g/mol, we divide the mass of metal product (1.68 g) by the molar mass of X to obtain the number of moles:
moles of X = 1.68 g / 55.9 g/mol = 0.03005 mol
We know that the initial mass of the oxide is 2.40 g and the product (X) is 1.68 g, so the mass of oxygen in the oxide is:
mass of O = 2.40 g - 1.68 g = 0.72 g
rationalizing the ratio, we get approximately 2:3
Thus, the empirical formula of the oxide is X2O3.
Balanced Equation for the Reaction
The balanced equation for the reaction of metal X's oxide with carbon monoxide to obtain metal X and carbon dioxide is:
X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g)
This equation shows that the oxide of metal X reacts with carbon monoxide in a 1:3 mole ratio to produce the pure metal and carbon dioxide in a 2:3 mole ratio.
Consider the following reactions: A: Uranium-238 emits an alpha particle B: Plutonium- 239 emits an alpha particle C: thorium-239 emits a beta particle
a. Rank the resulting nucleus by atomic number, from highest to lowest
b. Rank the resulting nucleus by the number of neutrons, from most to least
Answer:
Rank the resulting by neutrons, from most to least