Answer:
Molarity → 0.410 M
Molality → 0.44 m
Percent by mass → 7.61 g
Mole fraction (Xm) = 7.96×10⁻³
Mole percent = 0.79 %
Explanation:
We analyse data:
28.8 g of glucose → To determine molarity, molality, mole fraction, mole percent we need to find out the moles:
28.2 g. 1mol / 180 g = 0.156 moles of glucose
350 g of water → Mass of solvent.
We convert from g to kg in order to determine molality = 350 g / 1000 = 0.350 kg
We also need the moles of solvent: 350 g / 18 g/mol = 19.44 moles
380 mL of solution → Volume of solution; to determine the molarity we need the volume in L → 380 mL / 1000 = 0.380L
Solution mass = Solute mass + Solvent mass
28.2 g + 350 g = 378.2 g
Total moles = Moles of solute + Moles of solvent
0.156 + 19.44 = 19.596 moles
Molarity → Moles of solute in 1L of solution → 0.156 mol / 0.380L = 0.410 M
Molality → Moles of solute in 1kg of solvent → 0.156 mol / 0.350 kg = 0.44 m
Percent by mass → Mass of solute in 100 g of solution
(28.8g /378.2g) . 100 = 7.61 g
Mole fraction (Xm)= Moles of solute/ Total moles → 0.156 mol / 19.596 moles = 7.96×10⁻³
Mole percent = Xm . 100 → 7.96×10⁻³ . 100 = 0.79 %
The molarity of the solution is 0.42 M. The molality of the solution is 0.46 m.
a)To obtain the molarity of the solution;
Number of moles = mass of glucose/molar mass of glucose = 28.8g/180 g/mol
= 0.16 moles
Volume of solution = 380mL or 0.38 L
Molarity = number of moles /volume = 0.16 moles/ 0.38 L = 0.42 M
b) Molality of the solution;
Mass of solvent in Kg = 350g/1000 = 0.35 Kg
Molality = number of moles/mass of solution in kilogram = 0.16 moles/ 0.35 Kg = 0.46 m
c) percent by mass = mass of solute/mass of solution× 100
= 28.8g/(350g +28.8g) × 100 = 7.6 %
d) Mole fraction
Number of moles of water = 350g/ 18 g/mol = 19.44 moles
Total number of moles = 19.44 moles + 0.16 moles = 19.6 moles
Mole fraction of glucose = 0.16 moles/19.6 moles = 0.0082
e) Mole percent
Mole fraction × 100 = 0.0082 × 100 = 0.82%
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In a constant‑pressure calorimeter, 70.0 mL of 0.310 M Ba ( OH ) 2 was added to 70.0 mL of 0.620 M HCl . The reaction caused the temperature of the solution to rise from 21.12 ∘ C to 25.34 ∘ C. If the solution has the same density and specific heat as water, what is heat absorbed by the solution? Assume that the total volume is the sum of the individual volumes. (And notice that the answer is in kJ).
Answer: The amount of heat absorbed by the solution is 56.98 kJ
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
For Barium hydroxide:Molarity of barium hydroxide solution = 0.310 M
Volume of solution = 70 mL = 0.070 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.310M=\frac{\text{Moles of }Ba(OH)_2}{0.070L}\\\\\text{Moles of }Ba(OH)_2=(0.310mol/L\times 0.070L)=0.0217mol[/tex]
For HCl:Molarity of HCl solution = 0.620 M
Volume of solution = 70 mL = 0.070 L
Putting values in equation 1, we get:
[tex]0.620M=\frac{\text{Moles of HCl}}{0.070L}\\\\\text{Moles of HCl}=(0.620mol/L\times 0.070L)=0.0434mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:
[tex]Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of HCl produces 2 moles of water
So, 0.0434 moles of HCl will produce = [tex]\frac{2}{2}\times 0.0434=0.0434moles[/tex] of water
To calculate the mass of solution, we use the equation:[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1 g/mL
Volume of solution = [70 + 70] = 140 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of solution}}{140mL}\\\\\text{Mass of solution}=(1g/mL\times 140mL)=140g[/tex]
To calculate the amount of heat absorbed, we use the equation:[tex]q=mc\Delta T[/tex]
where,
q = heat absorbed
m = mass of solution = 140 g
c = heat capacity of solution= 4.186 J/g°C
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(25.34-21.12)^oC=4.22^oC[/tex]
Putting values in above equation, we get:
[tex]q=140g\times 4.186J/g^oC\times 4.22^oC=2473.08J=2.473kJ[/tex]
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
q = amount of heat absorbed = 2.473 kJ
n = number of moles of water = 0.0434 moles
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=\frac{2.473kJ}{0.0434mol}=56.98kJ/mol[/tex]
Hence, the amount of heat absorbed by the solution is 56.98 kJ
The total heat absorbed is approximately 2.47 kJ.
To find the heat absorbed by the solution in the given calorimetry problem, we use the equation q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the temperature change.
To determine the heat absorbed by the solution in this constant-pressure calorimeter problem, we will use the equation: q = mcΔTwhere:
m is the mass of the solutionc is the specific heat capacityΔT is the change in temperatureStep-by-Step Solution:
Calculate the total volume of the solution. Adding 70.0 mL of Ba(OH)2 to 70.0 mL of HCl gives a total volume of:Total Volume = 70.0 mL + 70.0 mL = 140.0 mLAssume the density of the solution is the same as water, 1.00 g/mL. Thus, the mass of the solution (m) is:Mass = 140.0 mL x 1.00 g/mL = 140.0 gThe specific heat capacity (c) of the solution is assumed to be the same as water, which is:c = 4.18 J/g°CCalculate the change in temperature (ΔT):ΔT = 25.34°C - 21.12°C = 4.22°CFinally, calculate the heat absorbed (q) by the solution:q = mcΔTq = 140.0 g x 4.18 J/g°C x 4.22°Cq ≈ 2466.7 J (approximately 2.47 kJ)Thus, the heat absorbed by the solution is approximately 2.47 kJ.
Correct question is: In a constant‑pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)₂ was added to 70.0 mL of 0.620 M HCl . The reaction caused the temperature of the solution to rise from 21.12°C to 25.34°C . If the solution has the same density and specific heat as water, what is heat absorbed by the solution? Assume that the total volume is the sum of the individual volumes. (And notice that the answer is in kJ).
Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide: 2MnCO3(s)+O2(g)→2MnO2(s)+2CO2(g) In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide: 3MnO2(s)+4Al(s)→3Mn(s)+2Al2O3(s) Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.
Answer: The net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.
Explanation:
The given chemical equation follows:
Equation 1: [tex]2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)[/tex] ( × 3)
Equation 2: [tex]3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)[/tex] ( × 2)
As, the net chemical equation does not include manganese (IV) oxide. So, to cancel out from the net equation, we need to multiply equation 1 by (3) and equation 2 by (2)
Now, the net chemical equation becomes:
[tex]6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)+6CO_2(g)[/tex]
Hence, the net chemical equation for the formation of manganese from manganese (II) carbonate, oxygen and aluminum is written above.
A stock solution of Blue #1 has a concentration of 5.736 M. 3 mL of this solution is diluted with 8 mL water. What is the concentration of the resulting solution? Provide your response to 3 significant figures.
Answer:
1.56 M
Explanation:
This is a dilution process and so a dilution formula is suitably used as follows C1V1 = C2V2 where
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the resulting (dilute) solution and
V2 = the volume of the resulting (dilute) solution
C1V1 = C2V2 (Making C2 subject of the formula)
C2 = C1V1/V2
Given: C1 = 5.736 M; V1 = 3 Ml; V2 = (3+8) 11 Ml
C2 = 5.736 x 3 / 11
C2 = 1.56 M
The concentration of the resulting solution will be "1.56 M".
Dilution process:The procedure of decreasing the concentration of such a particular solute through its solution has been known as dilution.
This same chemist could essentially add extra solvent to the mixture.
According to the question,
Stock solution's concentration, C₁ = 5.736 M
Stock solution's volume, V₁ = 3 mL
Resulting solution's volume, V₂ = 3 mL + 8 mL
= 11 mL
By using the dilution equation, we get
→ C₂ = [tex]\frac{C_1 V_1}{V_2}[/tex]
By substituting the above values,
= [tex]\frac{5.736\times 3}{11}[/tex]
= [tex]\frac{17.208}{11}[/tex]
= [tex]1.56[/tex] M
Thus the above answer is right.
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Be sure to answer all parts. Complete the correct name for the following compounds. (a) Na3[Fe(CN)6] hexacyanoferrate (b) [Cr(en)2Cl2]I (ethylenediamine)chromium() (c) [Co(en)3]I3 (ethylenediamine)cobalt()
Answer:
1. Na3[Fe(CN)6]
Oxidation number of iron is +3
Sodium hexacyanoferrate (III)
2. [Cr (en)2Cl2]+
Oxidation number of chromium is +3
Dichlorobis (ethylenediamine)chromium (III) ion
3. [Co (en)3]Cl3
Oxidation number of cobalt is +3
Tris (ethylenediamine)cobalt (III) chloride
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 89 MPa (81.00 ksi). If the plate is exposed to a tensile stress of 336 MPa (48730 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 0.92 for Y.
Explanation:
The given data is as follows.
[tex]K_{k}[/tex] = 89 MPa, [tex]\sigma[/tex] = 336 MPa
Y = 0.92
Now, we will calculate the length of critical interior flaw as follows.
[tex]a_{c} = \frac{1}{\pi}(\frac{K_{k}}{\sigma Y})^{2}[/tex]
= [tex]\frac{1}{\pi}(\frac{89}{336 \times 0.92})^{2}[/tex]
= [tex]\frac{656.38}{3.14}[/tex]
= 209.04 mm
Thus, we can conclude that minimum length of a surface crack that will lead to fracture is 209.04 mm.
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.310−5.What mass of KC6H5CO2 should the student dissolve in the HC6H5CO2 solution to turn it into a buffer with pH =4.63? You may assume that the volume of the solution doesn't change when the KC6H5CO2 is dissolved in it
Answer:
53.9 g
Explanation:
We must use the Henderson-Hasselbach equation to answer this question:
pH = pKa + log [A⁻]/[HA]
we know the pH, pKa (pKa = -log Ka), thus we can compute the ratio [A⁻]/[HA], and from this the mass of KC6H5CO2 knowing that M = mol/L.
therefore,
4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA]
4.63 = - ( - 4.20 ) + log [A⁻]/[HA]
0.43 = log [A⁻]/[HA]
taking inverse log to both sides of this equation
2.69 = [A⁻]/[HA]
Now [A⁻] =2.69 x [HA] =2.69 x 1.00 M = 2.69 M
We know the molarity is equal to mol per liter of solution, so
mol KC6H5CO2 = 2.69 mol/L x 0.125 L = 0.36 mol
and using the molecular weight of KC6H5CO2 we get that the mass is
0.336 mol x 160.21 g/mol = 53.9 g
The student should take 53.9 g of KC6H5CO2
Answer:
He should dissolve 53.9 grams of KC6H5CO2
Explanation:
Step 1: Data given
Volume of A 1.00 M benzoic acid solution = 125 mL = 0.125 L
Ka of benzoic acid = 6.3*10^-5
pH = 4.63
Step 2: Calculate concentration of conjugate base
pH = pKa + log ([A-]/[HA])
4.63 = 4.20 + log ([A-]/[HA])
0.43 = log ([A-]/[HA])
10^0.43 = [A-]/[HA])
2.69 = [A-]/[HA])
2.69 = [A-]/ 1.00
[A-] = 2.69 M
Step 3: Calculate moles KC6H5CO2
Moles molarity * volume
Moles = 2.69 M * 0.125 L
Moles = 0.33625 moles
Step 4: Calculate mass KC6H5CO2
Mass of KC6H5CO2 = moles * molar mass
Mass of KC6H5CO2 = 0.33625 moles * 160.21 g/mol
Mass of KC6H5CO2 = 53.9 grams
He should dissolve 53.9 grams of KC6H5CO2
How much Ca3(PO4)2(s) could be produced in an industrial process if 55.00 g of CaCl2 in solution reacted completely with sufficient Na3(PO4) (aq)
Answer:
51.1 g of Ca₃(PO₄)₂(s) can be made in this reaction
Explanation:
The reactans are CaCl₂ and Na₃PO₄. Let's determine the reaction:
3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) ↓ + 6NaCl(aq)
We convert the mass of chloride to moles:
55 g . 1 mol/ 110.98 g = 0.495 moles
Ratio is 3:1. Let's make a rule of three to find the answer in moles:
3 moles of chloride can produce 1 mol of phosphate
Therefore 0.495 moles will produce (0.495 . 1) / 3 = 0.165 moles
We convert the moles to mass:
0.165 mol . 310.18 g /1 mol = 51.1 g
The mass of calcium phosphate produced has been 51.1 g.
The balanced chemical equation of for the reaction has been:
[tex]\rm 3\;CaCl_2\;+\;2\;Na_3PO_4\;\rightarrow\;Ca_3(PO_4)_2\;+\;6\;NaCl[/tex]
The balanced chemical equation has been given that 3 moles of calcium chloride produces 1 moles of calcium phosphate.
The moles of calcium chloride in 55 g sample has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]
Substituting the values:
[tex]\rm Moles=\dfrac{55}{110.98}\\Moles\;CaCl_2=0.495\;mol[/tex]
The moles of calcium chloride has been 0.495 mol.
The moles of calcium formed has been given by:
[tex]\rm 3\;mol\;CaCl_2=1\;mol\;Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=\dfrac{1}{3}\;\times\;0.495\;mol\; Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=0.165\;mol\;Ca_3(PO_4)_2\\[/tex]
The moles of calcium phosphate formed has been 0.165 mol.
The mass of calcium phosphate has been:
[tex]\rm Mass=Moles\;\times\;molar\;mass\\Mass=0.165\;\times\;310.18\;g\\Mass=51.1\;g[/tex]
The mass of calcium phosphate produced has been 51.1 g.
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A chemist prepares a solution of vanadium(III) bromide (VBr) by measuring out 0.12 g of VBr into a 300 ml. volumetric flask and filling to the mark with distilled water. Calculate the molarity of Branions in the chemist's solution. Be sure your answer is rounded to significant digits.
Answer: The molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of vanadium(III) bromide = 0.12 g
Molar mass of vanadium(III) bromide = 295.65 g/mol
Volume of solution = 300 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.12\times 1000}{295.65g/mol\times 300}\\\\\text{Molarity of solution}=1.353\times 10^{-3}M[/tex]
The chemical formula of vanadium(III) bromide is [tex]VBr_3[/tex]
1 mole of vanadium(III) bromide produces 1 mole of [tex]V^{3+}[/tex] ions and 3 moles of [tex]Br^-[/tex] ions
Molarity of bromine ions = [tex](3\times 1.353\times 10^{-3})=4.06\times 10^{-3}M[/tex]
Hence, the molarity of bromine ions in the chemist's solution is [tex]4.06\times 10^{-3}M[/tex]
Final answer:
The molarity of bromide ions (Br-) in the solution is 0.00179 M.
Explanation:
To calculate the molarity of bromide ions (Br-) in the solution, we need to determine the number of moles of VBr. Given that the mass of VBr is 0.12 g and its molar mass is 223.6 g/mol (51.9 g/mol for V and 79.9 g/mol for Br), we can calculate the number of moles:
moles of VBr = mass of VBr / molar mass of VBr
= 0.12 g / 223.6 g/mol
= 0.000536 mol
Since the solution is 300 ml, we need to convert the volume to liters:
volume in liters = volume in ml / 1000
= 300 ml / 1000
= 0.3 L
molarity of Br- = moles of VBr / volume in liters
= 0.000536 mol / 0.3 L
= 0.00179 M (rounded to significant digits)
A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.
The given question is incomplete. The complete question is as follows.
A chlorine atom is adsorbed on a small patch of surface (see sketch at right). This patch is known to contain 81 possible adsorption sites. The atom has enough energy to move from site to site, so it could be on any one of them. Suppose a Br atom also becomes adsorbed onto the surface. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.
Explanation:
The change in entropy will be calculated by using the following formula.
[tex]\Delta S = k_{B} ln (\frac{W}{W_{o}})[/tex]
Initially, it is given tha Cl atom could be adsorbed on any of the 81 sites. When Br is added then there will be 80 possible sites when the Br can be adsorbed. This means that total possible sites are as follows.
[tex]81 \times 80[/tex]
= 6480
This shows that there are 6480 microstates which are accessible to the system.
So, change in entropy will be calculated using the Boltzmann constant as follows.
[tex]\Delta S = 1.381 \times 10^{-23} J/K \times ln (\frac{6480}{81})[/tex]
= [tex]1.381 \times 10^{-23} J/K \times 4.382[/tex]
= [tex]6.05 \times 10^{-23} J/K[/tex]
or, = [tex]6.1 \times 10^{-23} J/K[/tex] (approx)
Thus, we can conclude that the change in entropy is [tex]6.1 \times 10^{-23} J/K[/tex].
When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, why should you avoid rinsing the glassware with acetone prior this experiment?
Answer:
It will lead to false positive (or negative results) for this classification tests.
Explanation:
In this tests, the functional group that is really being tested for is the Carbonyl group. In the iodoform, the presence of a Carbonyl group gives a reaction and in the dinitrophenylhydrazine test, aldehydes give a reaction and ketones do not.
So, rinsing the glassware with acetone is introducing ketone before the qualitative test has even began, thereby leading to false results for each of the two tests mentioned in the question.
Final answer:
Rinsing glassware with acetone before conducting qualitative tests such as the iodoform test or dinitrophenylhydrazine test should be avoided to prevent false positive results due to acetone's interference or reaction with the test compounds.
Explanation:
When performing qualitative tests in glass test tubes, such as the iodoform test or dinitrophenylhydrazine test, rinsing the glassware with acetone prior to the experiment should be avoided because it can lead to false positive results. These qualitative tests rely on specific chemical reactions to occur, and if residual acetone remained in the tube, it might react or interfere with the test reagents or the compound being tested, giving an inaccurate result. For instance, acetone itself can produce a yellow precipitate with dinitrophenylhydrazine, mimicking a positive result for the presence of certain carbonyl groups. Therefore, cleaning the glassware properly without using acetone, or ensuring that any acetone used is completely evaporated and the glassware is dried in a water-free environment, is crucial to the accuracy of these tests.
To quickly dry glassware without using acetone, rinsing with distilled water and allowing the glassware to dry overnight or using warm air or nitrogen gas for drying is recommended. Additionally, glassware cleanliness is paramount in chemical testing and experiments to avoid contamination that could affect the results.