Answer:
1. only in systems in which gases are involved
2. only when the chemical reaction produces a change in the total number of gas molecules in the system
Explanation:
According to Le Chatelier's principle, pressure will only affect a system in equilibrium containing gaseous reactants and products. However, change in the pressure will only affect the gaseous system in which the total number of moles of the reactants are different from the total number of moles of the products.
Pressure changes affect equilibrium systems involving gases, particularly when the reaction results in different numbers of gas molecules on each side. Thus option A is correct.
Pressure changes in a system at equilibrium have a measurable effect primarily in gaseous systems. Here's how they work:
Pressure changes impact systems in which gases are involved.The effect is significant when there is a change in the total number of gas molecules in the system.If the number of gas moles is equal on both sides of the reaction, there is no effect on equilibrium.This occurs because increasing the pressure (by decreasing the volume) will shift the equilibrium toward the side with fewer moles of gas, while decreasing the pressure (by increasing the volume) will shift it toward the side with more moles of gas.
Complete question:
Changes in pressure have a measurable effect: (select all that apply) Select all that apply:
A. in any system only in systems in which gases are involved
B. when there are equal numbers of moles of gas on the reactant and product sides of the equilibrium only
C. when the chemical reaction produces a change in the total number of gas molecules in the system
D. None of these
63Ni decays by a first-order process via the emission of a beta particle. The 63Ni isotope has a half-life of 100. years. How long will it take for 65% of the nickel to undergo decay?
Answer:
151.4863 years
Explanation:
Half life, t1/2 = 100 years
Initial concentration,[A]o = 100%
Final concentration, [A] = 35% (after 65% have been decayed)
Time = ?
Half life for a first Order reaction is given as;
t1/2 = ln (2) / k
k = ln(2) / 100
k = 0.00693y-1
The integral rate law for first order reactions is given as;
ln[A] = ln[A]o − kt
kt = ln[A]o - ln[A]
t = ( ln[A]o - ln[A]) / k
t = [ln(100) - ln(35)] /0.00693
t = 1.0498 / 0.00693
t = 151.4863 years
It will take approximately 173.04 years for 65% of the nickel to undergo decay.
The decay of a radioactive substance is governed by the first-order kinetics equation:
[tex]\[ N(t) = N_0 e^{-\lambda t} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the number of un decayed nuclei at time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial number of nuclei,
- [tex]\( \lambda \)[/tex] is the decay constant, and
- [tex]\( t \)[/tex] is the time elapsed.
The decay constant [tex]\( \lambda \)[/tex] is related to the half-life [tex]\( t_{1/2} \)[/tex] by the equation:
[tex]\[ \lambda = \frac{\ln(2)}{t_{1/2}} \][/tex]
Given that the half-life [tex]\( t_{1/2} \) of \( ^{63}Ni \)[/tex] is 100. years, we can calculate [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{\ln(2)}{100} \approx \frac{0.693}{100} \][/tex]
Now, we want to find the time [tex]\( t \)[/tex] when 65% of the nickel has decayed, which means that 35% of the original nickel remains un decayed. Let's denote this time as [tex]\( t_{65\%} \)[/tex]. We can set up the equation:
[tex]\[ 0.35N_0 = N_0 e^{-\lambda t_{65\%}} \][/tex]
Dividing both sides by[tex]\( N_0 \)[/tex] and taking the natural logarithm gives us:
[tex]\[ \ln(0.35) = -\lambda t_{65\%} \][/tex]
[tex]\[ -\lambda t_{65\%} = \ln(0.35) \][/tex]
[tex]\[ t_{65\%} = -\frac{\ln(0.35)}{\lambda} \][/tex]
Substituting the value of [tex]\( \lambda \)[/tex] we calculated earlier:
[tex]\[ t_{65\%} = -\frac{\ln(0.35)}{\frac{\ln(2)}{100}} \][/tex]
[tex]\[ t_{65\%} = -\frac{100 \cdot \ln(0.35)}{\ln(2)} \][/tex]
[tex]\[ t_{65\%} = -\frac{100 \cdot \ln(0.35)}{0.693} \][/tex]
[tex]\[ t_{65\%} \approx -\frac{100 \cdot (-1.0498)}{0.693} \][/tex]
[tex]\[ t_{65\%} \approx 150.48 \][/tex]
However, we must consider that the time calculated is the time for 35% of the nickel to remain, not for 65% to decay. Since the half-life is the time for 50% to decay, and we have calculated the time for less than 50% to remain, the actual time for 65% to decay must be longer than one half-life. Therefore, we need to find the additional time [tex]\( \Delta t \)[/tex] it takes for the remaining 35% to decay to 15% (since 85% decayed in the first half-life).
We can use the same formula, but this time for the remaining 35% to decay to 15%:
[tex]\[ 0.15N_0 = 0.35N_0 e^{-\lambda \Delta t} \][/tex]
Solving for [tex]\( \Delta t \)[/tex]:
[tex]\[ \Delta t = -\frac{\ln(0.15/0.35)}{\lambda} \][/tex]
[tex]\[ \Delta t = -\frac{\ln(0.4286)}{\lambda} \][/tex]
[tex]\[ \Delta t = -\frac{100 \cdot \ln(0.4286)}{\ln(2)} \][/tex]
[tex]\[ \Delta t \approx -\frac{100 \cdot (-0.8473)}{0.693} \][/tex]
[tex]\[ \Delta t \approx 122.56 \][/tex]
Adding this to the initial half-life:
[tex]\[ t_{total} = 100 + 122.56 \approx 222.56 \][/tex]
However, since we are looking for the time it takes for 65% of the original amount to decay, we need to subtract the time it took for the first 50% to decay from the total time calculated:
[tex]\[ t_{65\%} = t_{total} - t_{1/2} \][/tex]
[tex]\[ t_{65\%} = 222.56 - 100 \][/tex]
[tex]\[ t_{65\%} \approx 122.56 \][/tex]
This calculation is incorrect because we did not consider that the decay process is continuous and the time for the first 50% to decay is already included in the total time. Therefore, the correct total time for 65% to decay is the initial calculation of 150.48 years plus the additional time of 122.56 years:
[tex]\[ t_{65\%} = 150.48 + 122.56 \][/tex]
[tex]\[ t_{65\%} \approx 273.04 \][/tex]
Again, we must correct for the fact that the initial half-life is already part of the decay process. The correct additional time needed for 65% of the nickel to decay is:
[tex]\[ t_{additional} = t_{65\%} - t_{1/2} \][/tex]
[tex]\[ t_{additional} = 273.04 - 100 \][/tex]
[tex]\[ t_{additional} \approx 173.04 \][/tex]
What is the difference between docosahexaenoate (DHA) and 19,20-dihydroxydocosapentaenoate (19,20-DHDP)? Only DHA is a polyunsaturated fatty acid. DHA has a higher melting point than 19,20-DHDP. 19,20-DHDP is more polar than DHA. Only DHA is a an ω ω -3 fatty acid.
Answer:
Explanation:
The correct answer is 19, 20 DHDP is more polar than DHA. This is as a result of the presence of two hydroxyl groups.
Docosahexaenoate (DHA) is a long-chain polyunsaturated omega-3 fatty acid important for brain and heart health, whereas 19,20-dihydroxydocosapentaenoate (19,20-DHDP) is a hydroxylated derivative of DHA that is more polar due to additional oxygen-containing groups.
Explanation:The difference between docosahexaenoate (DHA) and 19,20-dihydroxydocosapentaenoate (19,20-DHDP) primarily lies in their chemical structure and resulting properties. DHA is a long-chain polyunsaturated fatty acid (PUFA) with 22 carbon atoms and six double bonds, specifically an omega-3 fatty acid. It is known for its important role in brain development, cognitive function, and cardiovascular health. DHA can be synthesized in the body from alpha-linoleic acid (ALA), though the rate of conversion is limited, suggesting a dietary need for DHA-rich foods or supplements.
Meanwhile, 19,20-DHDP is a derivative of DHA that has been hydroxylated, containing additional oxygen groups which make it more polar than DHA. This increase in polarity could impact the compound's function and location within biological membranes.
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
A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
Answer: The concentration of KOH solution is 1.215 M
Explanation:
For the given chemical equation:
[tex]2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)[/tex]
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M[/tex]
Hence, the concentration of KOH solution is 1.215 M
A sample of nitrogen gas in a 8.5-L container at a temperature of 55°C exerts a pressure of 2.9 atm. Calculate the number of moles of gas in the sample.
Answer:
The number of moles of gas in the sample = 0.916 moles
Explanation:
Step 1: Data given
Volume of the container = 8.5 L
Temperature = 55 °C = 328 K
Pressure = 2.9 atm
Step 2: Calculate the number of moles of gas in the sample.
p*V = n*R*T
⇒with p = the pressure = 2.9 atm
⇒with V = the volume of the container = 8.5L
⇒with n = the number of moles of gas = ?
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 328 K
n = (p*V)/(R*T)
n = (2.9*8.5)/(0.08206*328)
n = 24.65 / 26.91568
n = 0.916 moles
The number of moles of gas in the sample = 0.916 moles
Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.
Answer:
194 g/mol.
Explanation:
Hello,
In this case, one first must compute the mass of each element as shown below:
[tex]C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO[/tex]
Next, the corresponding moles:
[tex]C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO[/tex]
Then, each element's subscripts is found to be:
[tex]C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1[/tex]
Therefore, the empirical formula is:
[tex]C_4H_5N_2O[/tex]
Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:
[tex]C_8H_10N_4O_2[/tex]
Which has a molar mass of 194 g/mol being correctly contained in the given interval.
Best regards.
The molar mass of the compound is found by finding the empirical and
molecular formula of the compound.
The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/molReasons:
Molar mass of CO₂ = 44.01 g/mol
Number of moles of CO₂ produced = [tex]\dfrac{18.13}{44.01}[/tex] ≈ 0.412 moles
Number of moles of produced C = 0.412 moles
Mass of C = 12 × 0.412 = 4.944 g
Molar mass of H₂O = 18.015 g/mol
Moles of H₂O produced = [tex]\dfrac{4.639}{18.015 }[/tex] = 0.2575 moles
Moles of produced H = 2 × 0.2575 = 0.515 molesMass of H = 1.00784 × 0.515 ≈ 0.519 gMolar mass of N₂ = 28.0134 g/mol
Moles of N produced = 2 × [tex]\dfrac{2.885}{28.0134 }[/tex] = 2 × 0.103 = 0.206 molesMass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652
Moles of oxygen, O = [tex]\dfrac{1.652 \, g}{16 \, g/mol}[/tex] ≈ 0.103 molesTherefore, we get;
Number of moles of produced C = 0.412 moles
Number of moles of produced H = 0.515 moles
Number of moles of oxygen, O ≈ 0.103 moles
Number of moles of N produced = 0.206 moles
Dividing by 0.103 gives;
Mole ratio of C = [tex]\dfrac{0.412}{0.103} = 4[/tex]Mole ratio of H = [tex]\dfrac{0.515}{0.103} = 5[/tex]Mole ratio of O = 1Mole ratio of N = [tex]\dfrac{0.206}{0.103} = 2[/tex]The empirical formula of the compound is therefore; C₄H₅N₂OThe general molecular formula is of the form (C₄H₅N₂O)ₙMolar mass of the compound is between 150 g/mol and 210 g/mol (given)
The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97
The molar mass of C₄H₅N₂O ≈ 97 g/mol
Molar mass of the compound is between 150 and 210 g/mol, therefore, n in
(C₄H₅N₂O)ₙ = 2, which gives;
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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.
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].
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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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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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
Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH.
Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap them
a. N2H2
b. Ba(OH)2
c. HOCL
d. NAOH
e. HCL
Answer: Highest to lowest pH: Ba(OH)2, NaOH, N2H2, HOCL HCl
Explanation:
Stronger the acid, lower the pH, stronger the base, higher the pH.
N2H2 weak base, Ba(OH)2 strong base, HOCL weak acid, NaOH strong base, HCl strong acid.
Ba(OH)2 has produces more H ions, so it has a higher pH than NaOH.
While idly tossing some keys from hand to hand one day, your friend Reuben (an expert chemist) says this: "Soluble metal oxides form hydroxides when dissolved in water." Using Reuben's statement, and what you already know about chemistry, predict the products of the following reaction. Be sure your chemical equation is balanced! K_0(aq) + H20(1)
The question is incomplete, complete question is ;
While idly tossing some keys from hand to hand one day, your friend Reuben (an expert chemist) says this: "Soluble metal oxides form hydroxides when dissolved in water." Using Reuben's statement, and what you already know about chemistry, predict the products of the following reaction. Be sure your chemical equation is balanced.
[tex]K_2O(aq) + H_2O(l)\rightarrow ?[/tex]
Answer:
The product will be potassium hydroxide.
Explanation:
When aqueous potassium oxide reacts with water it gives aqueous solution of potassium hydroxide as a product. And potassium hydroxide is a hydroxide of potassium metal with formula KOH.
[tex]K_2O(aq) + H_2O(l)\rightarrow 2KOH(aq)[/tex]
According to recation , 1 mole of potassium oxide when recats with 1 mole of water to give 2 moles of potassium hydroxide.
Be sure to answer all parts.The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl2(g) ⇌ 2NOCl(g) is 1 × 107 at a certain temperature. In an experiment, 4.40 × 10−2 mole of NO, 1.80 × 10−3 mole of Cl2, and 9.50 moles of NACl are mixed in a 2.60−L flask. What is Qc for the experiment
The equilibrium constant (Kc) and reaction quotient (Qc) are calculated using the initial molar concentrations of the substances involved in the reaction. The Qc for the formation of nitrosyl chloride from nitric oxide and molecular chlorine is 0, given the initial conditions provided.
Explanation:The student is asking about the equilibrium constant (Kc) and reaction quotient (Qc) for the formation of nitrosyl chloride from nitric oxide and molecular chlorine. To calculate the reaction quotient (Qc) of the reaction 2NO(g) + Cl2(g) ⇌ 2NOCl(g), you need to know the molar concentrations of the reactants and products.
First, find the initial concentrations of NO, Cl2, and NOCl by dividing their quantities by the volume of the flask, which is 4.40×10^-2 mol/2.60L for NO, 1.80×10^-3 mol/2.60L for Cl2, and 0 for NOCl because it's not initially present.
Then, we plug these concentrations into the Qc expression, which for this balanced equation is [NOCl]^2 / ([NO]^2 [Cl2]) = (0)^2 / ((4.40×10^-2 mol/2.60L)^2 * (1.80×10^-3 mol/2.60L)) = 0.
So, the reaction quotient Qc for the experiment is 0.
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Using the balanced equation and these concentrations, we find Qc to be 6.73 × 10⁷.
To determine the reaction quotient, Qc, for the given reaction:
2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
We follow these steps:
Calculate the initial concentrations of the reactants and products in the flask:For NO: [NO] = (4.40 × 10⁻² mol) / (2.60 L) = 1.69 × 10⁻² MFor Cl₂: [Cl₂] = (1.80 × 10⁻³ mol) / (2.60 L) = 6.92 × 10⁻⁴ MFor NOCl: [NOCl] = (9.50 mol) / (2.60 L) = 3.65 MWrite the expression for Qc based on the balanced chemical equation:Qc = [NOCl]₂ / ([NO]₂ [Cl₂])Substitute the known concentrations into the Qc expression:Qc = (3.65)² / ((1.69 × 10⁻²)² (6.92 × 10⁻⁴))Calculate Qc:Qc = 13.32 / (2.86 × 10⁻⁴ × 6.92 × 10⁻⁴)Qc = 13.32 / 1.98 × 10⁻⁷Qc = 6.73 × 10⁷Thus, the reaction quotient Qc for the experiment is 6.73 × 10⁷.
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)
Group the elements into pairs that would most likely exhibit similar chemical properties. It does not matter which pair of elements is pair 1, pair 2, or pair 3, so long as the correct elements are paired. Pair 1 Pair 2 Pair 3
a. F K
b. P
c. As
d. Br Li
Explanation:
A property that leads to changes in chemical composition of a substance is known as chemical property. So, when we move down a group in periodic table then chemical properties of the elements remain the same.
This is because elements of the same group tend to have same number of valence electrons. Therefore, they have similar reactivity which leads to change in their chemical composition upon reaction with another substance.
For example, lithium (Li) and potassium (K) are both group 1 elements. And, phosphorous (P) and arsenic (As) are both group 15 elements.
Thus, we can conclude that pair of Li and K will show similar chemical properties. Also, pair of P and As will show similar chemical properties.
hydrogen gas can be produced from the reaction of calcium hydride and water. How many grams of calcium hydride are needed to produce 2.5 l of hydrogen gas, collected over water at 26 degrees celcius and 760 torr
Answer:
2.144 g of calcium hydride are needed to produce 2.5 l of hydrogen gas, collected over water at 26 degrees celcius and 760 torr.
Explanation:
The reaction of Calcium hydride and water is given by
CaH₂ + 2H₂O -----> Ca(OH)₂ + 2H₂
2 moles of Hydrogen gas are produced from 1 mole of Calcium hydride.
But we need to find out how much moles of Hydrogen are produced from this reaction first.
Using the ideal gas, equation,
PV = nRT
P = pressure = 760 torr = 101325 Pa
V = volume of hydrogen gas produced = 2.5 L = 0.0025 m³
n = number of moles of Hydrogen gas produced = ?
R = molar gas constant = 8.314 J/mol.K
T = absolute temperature in Kelvin = 26°C = 299.15 K
n = PV/RT = (101325×0.0025)/(8.314×299.15) = 0.102 moles
Back to the stoichiometric balance of the reaction
CaH₂ + 2H₂O -----> Ca(OH)₂ + 2H₂
2 moles of Hydrogen are produced from 1 mole of Calcium hydride
0.102 moles of Hydrogen will be produced from (0.102 × 1/2) moles of Calcium hydride.
Moles of Calcium hydride = 0.051 moles.
Mass of Calcium hydride that reacted = number of moles of Calcium hydride that reacted × Molar mass
Moles mass of Calcium hydride = 42.094 g/mol
Mass of Calcium hydride that reacted = 0.051 × 42.094 = 2.144g
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 solution is prepared by dissolving 28.8g of glucose (C6H12O6) in 350g of water. The final volume of the solution is 380mL . For this solution, calculate each of the following: molarity, Molality, percent by mass, mole fraction, mole percent
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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