Answer:
the concentration remains = 0.138 M
Explanation:
Let us calculate the charge passed:
The charge passed during the electrolysis = current X time(s)
charge passed = 2.40X30X60= 2160 C
we know that
96485 C = 1 F
Therefore 2160 C = 0.0224 F
according to Faraday's law of electrolysis if we pass one Faraday of charge through an electrolytic solution, 1 gram equivalent of metal will be deposited.
the gram equivalent of nickel deposited = 0.0224
The moles of Nickel deposited = 0.0224/2 = 0.0112 mol
the initial moles of Nickel ions present in solution is
[tex]moles=molarityXvolume(L)=0.250X0.1=0.0250mol[/tex]
The moles of nickel ion gets consumed = 0.0112
So moles of nickel ion left after electrolysis = 0.0138
[tex][Ni^{+2}]=\frac{moles}{volume(L)}\frac{0.0138}{0.1}=0.138M[/tex]
The concentration of Ni2+ ions remaining in the solution can be found by applying Faraday's first law of electrolysis and subtracting the moles of Ni2+ discharged from the initial moles present in the solution.
Explanation:The electrolysis process involves a chemical reaction in which electricity is used to break down a substance into its constituent elements. In this scenario, we are determining the concentration of Ni2+ in the solution after the electrolysis process of NiSO4.
As per Faraday's first law of electrolysis, the mass of any element discharged during electrolysis is directly proportional to the quantity of electricity (current) passed. Thus, using Faraday's constant (96485 C/mol e-) and the given current and time, we can calculate how many moles of Ni2+ are plated out.
The number of moles of Ni2+ discharged would be equal to the number of moles initially present in the 100.0 mL of 0.250 M NiSO4 solution. By subtracting the moles of Ni2+ that were discharged from the total initial moles of Ni2+, we can obtain the number of moles of Ni2+ remaining in the solution. This value can be converted back into molarity by dividing the moles of Ni2+ remaining by the volume of the solution in liters.
In this way, we can find the concentration of Ni2+ ions that remain in the solution after electrolysis.Learn more about Electrolysis here:https://brainly.com/question/33301115
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A container holds 0.490 m3 of oxygen at an absolute pressure of 5.00 atm. A valve is opened, allowing the gas to drive a piston, increasing the volume of the gas until the pressure drops to 1.20 atm. If the temperature remains constant, what new volume (in m3) does the gas occupy? HINT m3
Answer: The new volume will be [tex]2.04m^3[/tex]
Explanation:
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex] (at constant temperature)
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=5.00atm\\V_1=0.490m^3\\P_2=1.20atm\\V_2=?m^3[/tex]
Putting values in above equation, we get:
[tex]5atm\times 0.490m^3=1.20\times V_2\\\\V_2=2.04m^3[/tex]
Hence, the new volume will be [tex]2.04m^3[/tex]
Suppose 23.8 g of oxygen (O2) is heated at constant atmospheric pressure from 27.4°C to 149°C. (a) How many moles of oxygen are present? (Take the molar mass of oxygen to be 32.0 g/mol) (b) How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate.) (c) What fraction of the heat is used to raise the internal energy of the oxygen?
Answer:
For a: The number of moles of oxygen gas is 0.74375 moles.
For b: The energy transferred to oxygen as heat is [tex]2.631\times 10^3[/tex]
For c: The fraction of heat used to raise the internal energy of oxygen is 0.714.
Explanation:
For a:To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of oxygen gas = 23.8 g
Molar mass of oxygen gas = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxygen gas}=\frac{23.8g}{32g/mol}=0.74375mol[/tex]
Hence, the number of moles of oxygen gas is 0.74375 moles.
For b:Oxygen is a diatomic gas.
To calculate the amount of heat transferred, we use the equation:
[tex]Q=nC_p\Delta T[/tex]
where,
Q = heat absorbed or released
n = number of moles of oxygen gas = 0.74375 moles
[tex]C_p[/tex] = specific heat capacity at constant pressure = [tex]\frac{7}{2}R[/tex] (For diatomic gas)
R = gas constant = 8.314 J/mol K
[tex]\Delta T[/tex] = change in temperature = [tex](149-27.4)^oC=121.6^oC=121.6K[/tex]
Putting values in above equation, we get:
[tex]Q=0.74375mol\times (\frac{7}{2})\times 8.314J/mol.K\times 121.6K\\\\Q=2631.71J=2.631\times 10^3J[/tex]
Hence, the energy transferred to oxygen as heat is [tex]2.631\times 10^3[/tex]
For c:To calculate the fraction of heat, we use the equation:
[tex]f=\frac{U}{Q}[/tex]
where,
U = internal energy = [tex]nC_v\Delta T[/tex]
Calculating the value of U:
n = number of moles of oxygen gas = 0.74375 moles
[tex]C_v[/tex] = specific heat capacity at constant pressure = [tex]\frac{5}{2}R[/tex] (For diatomic gas)
R = gas constant = 8.314 J/mol K
[tex]\Delta T[/tex] = change in temperature = [tex](149-27.4)^oC=121.6^oC=121.6K[/tex]
Putting values in above equation, we get:
[tex]U=0.74375mol\times (\frac{5}{2})\times 8.314J/mol.K\times 121.6K\\\\U=1879.79J=1.879\times 10^3J[/tex]
Taking the ratio of 'U' and 'Q', we get:
[tex]f=\frac{1.879\times 10^3}{2.631\times 10^3}\\\\f=0.714[/tex]
Hence, the fraction of heat used to raise the internal energy of oxygen is 0.714.
A 3.4 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.9 g. The two substances react, releasing carbon dioxide gas to the atmosphere. After the reaction, the contents of the reaction vessel weigh 11.6 g. What is the mass of carbon dioxide released during the reaction?
Answer: 2.7 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
[tex]NaHCO_3(aq)+CH_3COOH(aq)\rightarrow CH_3COONa(aq)+H_2O(l)+CO_2(g)[/tex]
Given: mass of sodium hydrogen carbonate = 3.4 g
mass of acetic acid = 10.9 g
Mass of reactants = mass of sodium hydrogen carbonate+ mass of acetic acid = 3.4 + 10.9= 14.3 g
Mass of reactants = Mass of products in reaction vessel + mass of carbon dioxide (as it escapes)
Mass of carbon dioxide = 14.3 - 11.6 =2.7 g
Thus the mass of carbon dioxide released during the reaction is 2.7 grams.
A solution is prepared by diluting 43.5 mL of 6.5×10?2M Ba(OH)2 to a volume of 270.5 mL .
Express the pH to two decimal places.
Answer:
The pH of the solution is 12.31 .
Explanation:
Initial molarity of barium hydroxide =[tex]M_1=6.5\times 10^{-2}M[/tex]
Initial volume of barium hydroxide =[tex]V_1=43.5 mL[/tex]
Final molarity of barium hydroxide =[tex]M_2[/tex]
Final volume of barium hydroxide =[tex]V_2=270.5 mL[/tex]
[tex]M_1V_1=M_2V_2[/tex]
[tex]M_2=\frac{6.5\times 10^{-2}M\times 43.5 mL}{270.5 mL}[/tex]
[tex]M_2=0.0104 M[/tex]
[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^-[/tex]
1 mol of barium gives 2 mol of hydroxide ions.
Then 0.0104 M of barium hydroxide will give:
[tex]2\times 0.0104 M=0.0208 M[/tex] of hydroxide ions
[tex][OH^-]=0.0208 M[/tex]
[tex]pH=14-pOH=14-(-\log[OH^-])[/tex]
[tex]pH=14-(-\log[0.0208 M])=12.31[/tex]
The pH of the solution is 12.31 .
The compound AX2 decomposes according to the equation, 2 AX2(g) => 2 AX(g) + X2(g). In one experiment, AX2 was measured at various times, and these data recorded: Time (s) {AX2} mol/L 0 0.0500 2.0 0.0448 6.0 0.0300 8.0 0.0249 10.0 0.0209 20.0 0.0088 What is the average rate from 8.0 to 20.0 seconds in mol / L s? Use 4 decimal places.
Answer:
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
Explanation:
[tex]2 AX_2(g)\rightarrow 2 AX(g) + X_2(g)[/tex]
Average rate of the reaction =[tex]R_a[/tex]
[tex]R_a=-\frac{\Delta [x]}{\Delta T}=-\frac{x_2-x_1}{t_2-t_1}[/tex]
[tex]R_a[/tex] =Average rate of the reaction during the given time interval.
[tex]\Delta [x][/tex] = Change in concentration of reactant with respect to time.
[tex]\Delta T[/tex] = Change in time.
[tex]x_1[/tex]=Concentration of reactant at time[tex]t_1[/tex]
[tex]x_2[/tex]=Concentration of reactant at time[tex]t_2[/tex]
So, at [tex]t_1=8.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_1=0.0249 mol/L[/tex]
And at [tex]t_2=2.0 sec[/tex] the concentration of [tex]AX_2[/tex] :
[tex]x_2=0.0088 mol/L[/tex]
The average rate of the reaction at given interval will be given as:
[tex]R_a=-\frac{x_2-x_1}{t_2-t_1}=-\frac{0.0088 mol/L-0.0249mol/L}{20.0s-8.0 s}=0.001341 mol/L s[/tex]
0.001341 mol/L s is the average rate from 8.0 to 20.0 seconds.
During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as CaCO3 . If a 11.3 kg bag of CaCO3 is used to treat the sidewalk, how many moles of CaCO3 does this bag contain?
Answer:
The number of moles of CaCO3 on the bag is 112.90 moles
Explanation:
number mole (n) = mass (m) divided by molecular mass (Mm)
Mm of CaCO3 = 100.0869 g/mole
mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams
number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.
The number of moles of CaCO3 in an 11.3 kg bag is calculated by dividing the mass of the CaCO3 by its molar mass.
First, we need to determine the molar mass of CaCO3. The molar mass of calcium (Ca) is approximately 40.08 g/mol, carbon (C) is approximately 12.01 g/mol, and oxygen (O) is approximately 16.00 g/mol. Since there is one atom of calcium, one atom of carbon, and three atoms of oxygen in CaCO3, the molar mass of CaCO3 is calculated as follows:
Molar mass of CaCO3 = [tex](40.08 g/mol) + (12.01 g/mol) + (3 -16.00 g/mol)[/tex]
Molar mass of CaCO3 = [tex]40.08 g/mol + 12.01 g/mol + 48.00 g/mol[/tex]
Molar mass of CaCO3 = [tex]100.09 g/mol[/tex]
Now, we convert the mass of the bag from kilograms to grams because the molar mass is given in grams per mole:
[tex]11.3 kg - 1000 g/kg = 11300 g[/tex]
Finally, we calculate the number of moles of CaCO3 in the bag:
Number of moles = mass of CaCO3 / molar mass of CaCO3
[tex]Number of moles[/tex] =[tex]11300 g / 100.09 g/mol[/tex]
[tex]Number of moles = 113 moles[/tex]
Therefore, the bag contains approximately 113 moles of CaCO3.
The correct answer is [tex]\boxed{113 \text{ moles}}.[/tex]
The answer is: [tex]113 \text{ moles}.[/tex]
An ideal gas is brought through an isothermal compression process. The 3.00 mol of gas goes from an initial volume of 261.6×10−6 m3 to a final volume of 138.2×10−6 m3 . If 9340 J is released by the gas during this process, what are the temperature ???? and the final pressure ???????? of the gas?
Answer : The temperature and the final pressure of the gas is, 586.83 K and [tex]1.046\times 10^{9}atm[/tex] respectively.
Explanation : Given,
Initial volume of gas = [tex]261.6\times 10^{-6}m^3[/tex]
Final volume of the gas = [tex]138.2\times 10^{-6}m^3[/tex]
Heat released = -9340 J
First we have to calculate the temperature of the gas.
According to the question, this is the case of isothermal reversible compression of gas.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
[tex]q=-w[/tex]
Thus, w = -q = 9340 J
The expression used for work done will be,
[tex]w=nRT\ln (\frac{V_2}{V_1})[/tex]
where,
w = work done = 9340 J
n = number of moles of gas = 3 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = ?
[tex]V_1[/tex] = initial volume of gas
[tex]V_2[/tex] = final volume of gas
Now put all the given values in the above formula, we get the temperature of the gas.
[tex]9340J=3mole\times 8.314J/moleK\times T\times \ln (\frac{261.6\times 10^{-6}m^3}{138.2\times 10^{-6}m^3})[/tex]
[tex]T=586.83K[/tex]
Now we have to calculate the final pressure of the gas by using ideal gas equation.
[tex]PV=nRT[/tex]
where,
P = final pressure of gas = ?
V = final volume of gas = [tex]138.2\times 10^{-6}m^3=138.2\times 10^{-9}L[/tex]
T = temperature of gas = 586.83 K
n = number of moles of gas = 3 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get:
[tex]P\times (138.2\times 10^{-9}L)=3mole\times (0.0821L.atm/mole.K)\times (586.83K)[/tex]
[tex]P=1.046\times 10^{9}atm[/tex]
Therefore, the temperature and the final pressure of the gas is, 586.83 K and [tex]1.046\times 10^{9}atm[/tex] respectively.
g On the basis of their respective pKa values, determine which is the stronger acid, acetic acid (pKa 5 4.7) or benzoic acid (margin; pKa 5 4.2). By what factor do their acidities differ?
Answer:
Benzoic acid is the stronger acid
Explanation:
Weak acids do not dissociate completely in the solution. They exists in equilibrium with their respective ions in the solution.
The extent of dissociation of the acid furnising hydrogen ions can be determined by using dissociation constant of acid ([tex]K_a[/tex]).
Thus for a weak acid, HA
[tex]HA \rightleftharpoons A^- + H^+[/tex]
The [tex]K_a[/tex] is:
[tex]K_a= \frac{[A^-][H^+]}{[HA]}[/tex]
The more the [tex]K_a[/tex], the more the acid dissociates, the more the stronger is the acid.
Also,
[tex]pK_a[/tex] is defined as the negative logarithm of [tex]K_a[/tex].
So, more the [tex]pK_a[/tex], less is the [tex]K_a[/tex] and vice versa
All can be summed up as:
The less the value of [tex]pK_a[/tex], the more the [tex]K_a[/tex] is and the more the acid dissociates and the more the stronger is the acid.
Given,
[tex]pK_a[/tex] of acetic acid = 54.7
[tex]pK_a[/tex] of benzoic acid = 54.2
[tex]pK_a[/tex] of benzoic acid < [tex]pK_a[/tex] of acetic acid
So, benzoic acid is the stronger acid.
If 1.00 mol of argon is placed in a 0.500-L container at 27.0 degree C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol. Express your answer to two significant figures and include the appropriate units.
Answer:
2.0 atm is the difference between the ideal pressure and the real pressure.
Explanation:
If 1.00 mole of argon is placed in a 0.500-L container at 27.0 °C
Moles of argon = n = 1.00 mol
Volume of the container,V = 0.500 L
Ideal pressure of the gas = P
Temperature of the gas,T = 27 °C = 300.15 K[/tex]
Using ideal gas equation:
[tex]PV=nRT[/tex]
[tex]P=\frac{1.00 mol\times 0.0821 L atm/mol K\times 300.15 K}{0.500 L}=49.28 atm[/tex]
Vander wall's of equation of gases:
The real pressure of the gas= [tex]p_v[/tex]
For argon:
[tex]a=1.345 L^2 atm/mol^2[/tex]
b=0.03219 L/mol.
[tex](p_v+(\frac{an^2}{V^2})(V-nb)=nRT[/tex]
[tex](p_v+(\frac{(1.345 L^2 atm/mol^2)\times (1.00 mol)^2}{(0.500 L)^2})(0.500 L-1.00 mol\times 0.03219L/mol)=1.00 mol\times 0.0821 L atm/mol K\times 300.15 K[/tex]
[tex]p_v = 47.29 atm[/tex]
Difference :[tex] p - p_v= 49.28 atm - 47.29 atm = 1.99 atm\approx 2.0 atm[/tex]
2.0 atm is the difference between the ideal pressure and the real pressure.
Write the formula for the complex ion formed by the metal ion Cr3+ and six NO2- ions as ligands. Decide whether the complex could be isolated as a chloride salt or a potassium salt, and write the formula for the appropriate salt
Answer:The metal complex formed would have the following formula [Cr(NO₂)₆]³⁻. The complex has a net negative charge and hence it can only be isolated as a salt with a positive cation so the formed complex could be isolated as potassium salt. The formula for salt would be K₃[Cr(NO₂)₆].
Explanation:
The metal ion given to us is Cr³⁺ (Chromium) in +3 oxidation state.
The electronic configuration for the metal ion is [Ar]3d³ so there are vacant 3d metal orbitals which are available and hence 6 NO₂⁻ ligands can easily attack the metal center and form a metal complex.
The charge on the overall complex can be calculated using the oxidation states of metal and ligand which is provided.
The (chromium ) Cr³⁺ metal has +3 charge and 6 NO₂⁻ (nitro) ligands have -6 charge and since the ligands will be providing a total of 6 - (negative) charge and hence only 3- (negative ) charge can be neutralized so a net 3- negative charge would be present on the overall complex which is basically present at the metal center :
charge on the complex=+3-6=-3
Let X be the Oxidation state of Cr in complex =[Cr(NO₂)₆]³⁻
X-6=-3
X=-3+6
X=+3
so our calculated oxidation state of Cr is +3 which matches with the provided in question.
As we can see that the overall metal complex has a net negative charge and hence and only positively charged cations can form a salt with this metal complex and hence only potassium K⁺ ions can form salt with the metal complex.
since overall charge present on the metal complex is -3 so 3 K⁺ ion would be needed to neutralize it and hence the formula of the metal salt would be K₃[Cr(NO₂)₆].
With organolithium and organomagnesium compounds, approach to the carbonyl carbon from the less hindered direction is generally preferred. Assuming that this is the case, draw the structure of the major product formed when 4-tert-butylcyclohexanone reacts with phenylmagnesium bromide, followed by treatment with aqueous acid.
Answer:
trans-4-tert-butyl-1-phenylcyclohexanol
Explanation:
The bulky tert-butyl group will occupy the equatorial position in the cyclohexanone ring.
It will hinder approach of the Grignard reagent from the bottom side of the molecule.
The reagent will attack from the top, so the product alcohol will be
trans-4-tert-butyl-1-phenylcyclohexanol.
A sample of gas contains 0.1300 mol of N2(g) and 0.2600 mol of O2(g) and occupies a volume of 23.9 L. The following reaction takes place: N2(g) + 2O2(g)2NO2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
Answer : The volume of the sample after the reaction takes place is, 15.93 liters.
Explanation : Given,
Moles of [tex]N_2[/tex] = 0.13 mole
Moles of [tex]O_2[/tex] = 0.26 mole
Initial volume of gas = 23.9 L
First we have to calculate the moles of [tex]NO_2[/tex] gas.
The balanced chemical reaction is :
[tex]N_2(g)+2O_2(g)\rightarrow 2NO_2(g)[/tex]
From the balanced reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react with 2 moles of [tex]O_2[/tex] to give 2 moles of [tex]NO_2[/tex].
So, 0.13 mole of [tex]N_2[/tex] react with [tex]2\times 0.13=0.26[/tex] moles of [tex]O_2[/tex] to give [tex]2\times 0.13=0.26[/tex] moles of [tex]NO_2[/tex].
According to the Avogadro's Law, the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.
[tex]V\propto n[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 23.9 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]n_1[/tex] = initial moles of gas = 0.13 + 0.26 = 0.39 mole
[tex]n_2[/tex] = final moles of gas = 0.26 mole
Now put all the given values in the above formula, we get the final temperature of the gas.
[tex]\frac{23.9L}{V_2}=\frac{0.39mole}{0.26mole}[/tex]
[tex]V_2=15.93L[/tex]
Therefore, the volume of the sample after the reaction takes place is, 15.93 liters.
What two stereoisomeric alkanes are formed in the catalytic hydrogenation of (Z)-3-methyl-2-hexene? Draw and label the alkanes. In this reaction, what are the relative amounts of each of the two alkanes generated?
Resolution of the attached issue .
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The catalytic hydrogenation of (Z)-3-methyl-2-hexene generates two stereoisomeric alkanes: 3-ethylheptane and 2,2,4-trimethylpentane. The alkane with less steric hindrance, 2,2,4-trimethylpentane, is formed in greater quantities than 3-ethylheptane.
Explanation:The catalytic hydrogenation of (Z)-3-methyl-2-hexene leads to the formation of two stereoisomeric alkanes: 3-ethylheptane and 2,2,4-trimethylpentane. These alkanes are formed due to the addition of hydrogen to the double bond in the alkene under the influence of a catalyst, usually a metal such as platinum or palladium. The reaction involves the breaking of the weaker pi bond in the alkene and the formation of stronger sigma bonds, with the hybridization of the carbon atoms changing from sp2 to sp3.
Regarding the relative amounts of each alkane generated, the alkane formed depends on the steric hindrance around the double bond. Because 2,2,4-trimethylpentane poses less steric hindrance compared to 3-ethylheptane, it's formed in greater quantities.
The reaction looks like this for 2,2,4-trimethylpentane (the major product):
(Z)-3-methyl-2-hexene + H2 ---> 2,2,4-trimethylpentane
And like this for 3-ethylheptane (the minor product):
(Z)-3-methyl-2-hexene + H2 ---> 3-ethylheptane
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Atmospheric pressure at sea level is 760 mm Hg, and oxygen makes up 20.9% of this air when it is dry. Scientists at the Mt. Washington Observatory in New Hampshire measured the atmospheric pressure at the summit of Mt. Washington (6,289 feet above sea level) as 609 mm Hg. When the air is dry, the partial pressure of oxygen at the summit is approximately _____ mm Hg.
Answer:
127.28 mmHg
Explanation:
The molar fraction of oxygen in dry air at 760 mmHg is 20.9%. This molar fraction is not affected too much by the height, so it may be taken as a constant. The partial pressure of oxygen may be calculated as:
[tex]P_{O_{2}}=y_{O_{2}}*P[/tex]
So, if the total pressure is 609 mmHg,
[tex]P_{O_{2}}=0.209*609=127.28[/tex] mmHg.
How many grams of NaHCO3 (FM 84.01 g/mol) should be mixed with Na2CO3 to produce a 1.00 L buffer solution with pH 9.50. The final concentration of Na2CO3 in this solution is 0.10 M. pKa1 = 6.37 and pKa2 = 10.33 for H2CO3.
Answer: 58.0 grams of sodium bicarbonate will be required.
Explanation: pH of buffer solutions is calculated using Handerson equation:
[tex]pH=pKa+log(\frac{base}{acid})[/tex]
sodium carbonate acts as a base since it has carbonate ion where as sodium bicarbonate acts as an acid since it has bicarbonate ion that has a proton.
pKa2 value will be used here since we have sodium bicarbonate and not carbonic acid. Concentration of sodium carbonate is given as 0.10 M, pH is given as 9.50 and pKa2 is given as 10.33.
Let's plug in the values in Handerson equation and calculate the concentration of sodium bicarbonate.
[tex]9.50=10.33+log(\frac{0.10}{x})[/tex]
(where [tex]x[/tex] is the concentration of sodium bicarbonate)
[tex]9.50-10.33=log(\frac{0.10}{x})[/tex]
[tex]-0.83=log(\frac{0.10}{x})[/tex]
taking antilog to both sides:
[tex]10^-^0^.^8^3=\frac{0.10}{x}[/tex]
[tex]0.145=\frac{0.10}{x}[/tex]
[tex]x=\frac{0.10}{0.145}[/tex]
[tex]x=0.69[/tex]
Concentration of sodium bicarbonate is 0.69 M. Volume of solution is given as 1.00 L. So, the moles of sodium bicarbonate will be 0.69 moles.
Molar mass of sodium bicarbonate is 84.0 gram per mol. Multiply the moles by molar mass to calculate the required grams of it.
[tex]0.69mol(\frac{84.0g}{1mol})[/tex]
= 57.96 g
So, 57.96 g which is almost 58.0 g of [tex]NaHCO_3[/tex] will be required.
The required amount of NaHCO3 required to obtain a buffer solution with pH of 9.50 will be 56.80 g
pH relationship for a buffer solution :
[tex]pH = pKa + log(\frac{base}{acid}[/tex]
pH = 9.50pKa = 10.33Concentration of base = 0.10Concentration of acid = a[tex]9.50 = 10.33 + log(\frac{0.10}{a}[/tex]
[tex]9.50 - 10.33 = log(\frac{0.10}{a}[/tex]
[tex]-0.83 = log(\frac{0.10}{a}[/tex]
Take the inverse log of both sides
[tex] 10^{-0.83} = \frac{0.10}{a}[/tex]
[tex] 0.1479 = \frac{0.10}{a}[/tex]
[tex] a = \frac{0.10}{0.1479}[/tex]
[tex] a = 0.676 [/tex]
Concentration = mole × Molar mass Molar mass of NaHCO3 = 84.01 g/molConcentration of NaHCO3 = 84.01 × 0.676 = 56.80 g
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Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium amide is polar and methanol is nonpolar. Sodium amide undergoes an acid-base reaction with methanol. There would be no ion-dipole attractive forces between the two compounds.
Answer: sodium amide undergoes an acid -base reaction
Explanation:
sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.
This leads to formation of ammonia and sodium methoxide.
Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.
Hence the 3rd statement is a corrects statement.
So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.
The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.
The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.
The following reaction occurs:
NaNH₂+CH₃OH→NH₃+CH₃ONa
Sodium amide undergoes an acid -base reaction
Why methanol is not used as a solvent for sodium amide?Methanol is acidic in nature whereas sodium amide is a strong base so when we add or combine these two chemicals, they undergoes an acid -base reaction so that's why we can't use methanol as a solvent for sodium amide.
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You have performed a serial dilution of an unknown sample and counted 73 CFU on a countable plate that was marked as a 10^-4 dilution and you used a 0-.1 mL to inoculate the plate. What is the population density of the original sample?
Hey there!:
To find the original sample population density, divide the conted colonies by the dilution factor and the inoculate:
= 73 / (10⁻⁴ * 0.1)
= 73*10⁵
Thus, the original sample's population density was 7.3*10⁶ or 73* 10⁵.
Hope this helps!
The population density of the original sample will be 73 × [tex]10^{5}[/tex] .
What is population density?Population density would be a quantity of the type integer density; it would be a measuring of the population per unit area, even particularly per unit volume.
Calculation of population density
It is given that, counted colonies = 73
The dilution factor = 0.1 × [tex]10^{-4}[/tex]
population density = counted colonies / dilution factor = 73 / 0.1 × [tex]10^{-4}[/tex].
= 73 × [tex]10^{5}[/tex] .
Therefore, the population density will be 73 × [tex]10^{5}[/tex] .
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Question 5 A solution is prepared at 25°C that is initially 0.35M in chlorous acid HClO2, a weak acid with =Ka×1.110−2, and 0.29M in sodium chlorite NaClO2. Calculate the pH of the solution. Round your answer to 2 decimal places.
Answer : The pH of the solution is, 1.88
Explanation : Given,
[tex]K_a=1.1\times 10^{-2}[/tex]
Concentration of [tex]HClO_2[/tex] = 0.35 M
Concentration of [tex]NaClO_2[/tex] = 0.29 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log [K_a][/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (1.1\times 1-^{-2})[/tex]
[tex]pK_a=2-\log (1.1)[/tex]
[tex]pK_a=1.96[/tex]
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=1.96+\log (\frac{0.29}{0.35})[/tex]
[tex]pH=1.88[/tex]
Therefore, the pH of the solution is, 1.88
The pH of a buffer solution consisting of a weak acid and its conjugate base can be calculated using the Henderson–Hasselbalch equation, with the concentrations of the weak acid and base, and the Ka or pKa of the weak acid.
Explanation:In this scenario, we are dealing with a weak acid, chlorous acid (HClO2), and the salt of its conjugate base, sodium chlorite (NaClO2). When a weak acid is in solution with the salt of its conjugate base, a buffer solution is formed. Buffer solutions resist significant changes in pH upon the addition of small quantities of acid or base. The pH of the buffer solution can be calculated using the Henderson–Hasselbalch equation:
pH = pKa + log([A-]/[HA]).
Where,
pKa is the negative log of the acid dissociation constant Ka,[A-] is the concentration of the base form, here it is sodium chlorite NaClO2,[HA] is the concentration of the weak acid, here it is chlorous acid HClO2.Using the given concentrations and Ka or pKa, substitute them into the formula to calculate the pH of the solution.
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Why are the concentrations of [H3O+] and [OH−] equal in pure water?
A)[H3O+]=[OH−] because one of each is produced every time an [H+] transfers from one water molecule to another.
B)[H3O+]=[OH−] because any solution should contain an equal concentration of acidic and basic particles. C)[H3O+]=[OH−] because water is amphoteric substance.
D)[H3O+]=[OH−] because [H3O+] and [OH−] form a conjugate acid-base pair.
Answer : The correct option is, (A) [tex][H_3O^+]=[OH^-][/tex] because one of each is produced every time an [tex][H^+][/tex] transfers from one water molecule to another.
Explanation :
As we know that, when the two water molecule combine to produced hydronium ion and hydroxide ion.
The balanced reaction will be:
[tex]HOH+HOH\rightarrow H_3O^++OH^-[/tex]
Acid : It is a substance that donates hydrogen ion when dissolved in water.
Base : It is a substance that accepts hydrogen ion when dissolved in water.
From this we conclude that, the hydrogen ion are transferred from one water molecule to the another water molecule to form hydronium ion and hydroxide ion. In this reaction, one water molecule will act as a base and another water molecule will act as an acid.
Hence, the correct option is, (A)
In pure water, the concentrations of [H3O+] and [OH−] are equal because water undergoes self-ionization. The process results in equal formation of [H3O+] and [OH−] ions for every one water molecule that donates a proton.
Explanation:In pure water, the concentrations of hydronium ions [H3O+] and hydroxide ions [OH−] are equal because water undergoes a process called self-ionization or auto-ionization. In this process, one water molecule donates a proton (H+) to another water molecule, forming a hydronium ion [H3O+] and a hydroxide ion [OH−]. Therefore, for every hydronium ion formed, there is a hydroxide ion formed as well, leading to equal concentrations of [H3O+] and [OH−] in pure water. This is represented by the equation 2H2O(l) ⇌ H3O+(aq) + OH−(aq).
This does not mean that all solutions should contain equal concentrations of acidic and basic particles, nor does it imply that [H3O+] and [OH−] form a conjugate acid-base pair. Additionally, although water is an amphoteric substance (meaning it can act as both an acid and a base), that is not why the concentrations of [H3O+] and [OH−] are equal in pure water.
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The Bohr model gives a simple, but reasonably accurate, formula for the energy levels of an electron in an isolated hydrogen atom. When an electron moves from one energy level to another lower level, the difference in the energies between the two levels is emitted in a photon. What is the wavelength of the photon emitted when an electron falls from the fourth level to the first level?
Answer: [tex]0.98\times 10^{-7}m[/tex]
Explanation:
For calculating wavelength, when the electron will jump from n=4 to n= 1
Using Rydberg's Equation: for hydrogen atom
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation = ?
[tex]R_H[/tex] = Rydberg's Constant = [tex]1.097\times 1067m[/tex]
[tex]n_f[/tex] = Higher energy level = 4
[tex]n_i[/tex]= Lower energy level = 1
Z= atomic number = 1 (for hydrogen)
Putting the values, in above equation, we get
[tex]\frac{1}{\lambda}=1.097\times 10^7\left(\frac{1}{1^2}-\frac{1}{4^2} \right )\times 1^2[/tex]
[tex]\lambda=0.98\times 10^{-7}m[/tex]
Thus the wavelength of the photon emitted will be [tex]0.98\times 10^{-7}m[/tex]
Enter your answer in the provided box. One of the half-reactions for the electrolysis of water is 2H2O(l) → O2(g) + 4H+(aq) + 4e− If 3.696 L of O2 is collected at 25°C and 755 mmHg, how many faradays of electricity had to pass through the solution?
Answer : The number of faradays of electricity had to pass through the solution will be, 0.596 F
Explanation :
First we have to calculate the moles of oxygen gas.
using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 755 mm Hg = 0.99 atm
conversion used : (1 atm = 760 mmHg)
V = volume of gas = 3.696 L
T = temperature of gas = [tex]25^oC=273+25=298K[/tex]
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get the number of moles of oxygen gas.
[tex](0.99atm)\times (3.696L)=n\times (0.0821L.atm/mole.K)\times (298K)[/tex]
[tex]n=0.149mole[/tex]
Now we have to calculate the number of faradays of electricity had to pass through the solution.
The balanced half-reactions for the electrolysis of water is,
[tex]2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-[/tex]
From this we conclude that,
As, 1 mole of oxygen gas require 4 mole of electrons that means 4 F (faraday) of electricity.
So, 0.149 mole of oxygen gas require [tex]0.149\times 4=0.596F[/tex] of electricity.
Therefore, the number of faradays of electricity had to pass through the solution will be, 0.596 F
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in kPa
Answer:
Pressure exerted by the needle = 78.037 kPa
Explanation:
Pressure is defined as the force exerted on a surface per unit area. The SI unit of pressure is Pa (N.m⁻²).
The force exerted by the phonograph needle is equal to the gravitational force.
Thus,
F=m×g
Where,
F is the force
m is the mass of the body
g is the acceleration due to gravity
Also, g = 9.81 ms⁻²
Given, mass of the needle = 1.00 g
The conversion of g into kg is shown below:
1 g = 10⁻³ kg
Thus, mass of the needle = 1×10⁻³ kg
Thus, F= 1×10⁻³×9.81 N = 9.81×10⁻³ N
Ares of the circle = πr²
Given : Radius of tip of the needle = 0.200 mm
The conversion of mm into m is shown below:
1 mm = 10⁻³ m
Thus, mass of the needle = 0.2×10⁻³ m
Thus, Area of the tip of the needle = π×(0.2×10⁻³)² m² = 1.2571×10⁻⁷ m²
So, Pressure =F/A
P = (9.81×10⁻³)/1.2571×10⁻⁷ Pa = 7.8037×10⁴ Pa
The conversion of Pa into kPa is shown below:
1 Pa = 10⁻³ kPa
Thus, pressure exerted by the needle = 7.8037×10⁴×10⁻³ kPa = 78.037 kPa
Calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm (a typical value at high altitude). Atmospheric Gas Mole Fraction kH mol/(L*atm) N2 7.81 x 10-1 6.70 x 10-4 O2 2.10 x 10-1 1.30 x 10-3 Ar 9.34 x 10-3 1.40 x 10-3 CO2 3.33 x 10-4 3.50 x 10-2 CH4 2.00 x 10-6 1.40 x 10-3 H2 5.00 x 10-7 7.80 x 10-4
Answer : The solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]
Explanation :
First we have to calculate the partial pressure of nitrogen.
Formula used :
[tex]p_{N_2}=X_{N_2}\times P_{atm}[/tex]
where,
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = ?
[tex]X_{N_2}[/tex] = mole fraction of nitrogen = [tex]7.81\times 10^{-1}[/tex]
[tex]p_{atm}[/tex] = atmospheric pressure = 0.480 atm
Now put all the given values in the above formula, we get :
[tex]p_{N_2}=7.81\times 10^{-1}\times 0.480 atm[/tex]
[tex]p_{N_2}=0.375atm[/tex]
Now we have to calculate the solubility of nitrogen in water.
Formula used :
[tex]s_{N_2}=p_{N_2}\times K_H[/tex]
where,
[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 0.375 atm
[tex]s_{N_2}[/tex] = solubility of nitrogen in water = ?
[tex]K_H[/tex] = Henry's constant = [tex]6.70\times 10^{-4}mole/L.atm[/tex]
Now put all the given values in the above formula, we get :
[tex]s_{N_2}=0.375atm\times 6.70\times 10^{-4}mole/L.atm[/tex]
[tex]s_{N_2}=2.5125\times 10^{-4}mole/L[/tex]
Therefore, the solubility of nitrogen in water at an atmospheric pressure will be, [tex]2.5125\times 10^{-4}mole/L[/tex]
To calculate the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm, we can use Henry's Law. Multiply the mole fraction of nitrogen by the atmospheric pressure to calculate the partial pressure. Then, multiply the Henry's Law constant for nitrogen by the partial pressure to find the solubility of nitrogen in water.
Explanation:To calculate the solubility of nitrogen in water at a given atmospheric pressure, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The equation for Henry's Law is:
C = kH × P
where C is the concentration of the gas in the liquid, kH is the Henry's Law constant, and P is the partial pressure of the gas above the liquid.
In this case, we need to use the mole fraction of nitrogen and the Henry's Law constant for nitrogen to calculate the solubility. The mole fraction of nitrogen is 7.81 x 10^-1 and the Henry's Law constant for nitrogen is 6.70 x 10^-4 mol/(L×atm).
Convert the mole fraction of nitrogen to partial pressure using the given atmospheric pressure of 0.480 atm. Multiply the mole fraction by the atmospheric pressure: 7.81 x 10^-1 ×0.480 atm = 3.7572 x 10^-1 atm.Use the Henry's Law constant for nitrogen and the partial pressure of nitrogen calculated in step 1 to find the solubility of nitrogen in water. Multiply the Henry's Law constant by the partial pressure: 6.70 x 10^-4 mol/(L×atm) ×3.7572 x 10^-1 atm = 2.5108 x 10^-4 mol/L.Therefore, the solubility of nitrogen in water at an atmospheric pressure of 0.480 atm is 2.5108 x 10^-4 mol/L.
A brine is used to measure the absolute permeability of a core plug. The rock sample is 4 cm long and 3 cm 2 in cross section. The brine has a viscosity of 1.0 cp and is flowing a constant rate of 0.5 cm3/sec ( cube centimeter per second) under a 2.0 atm pressure differential.
Calculate the absolute permeability.
Answer : The absolute permeability is, 0.111 darcys.
Explanation : Given,
Volumetric constant rate = [tex]0.5cm^3/s[/tex]
Cross section area = [tex]3cm^2[/tex]
Pressure = 2 atm
Viscosity = 1.0 cp
Length of rock = 4 cm
From the Darcy equation we conclude that,
[tex]\text{Volumetric constant rate}=\frac{(\text{Absolute permeability}\times A\times P)}{\eta \times l}[/tex]
where,
A = cross section area
P = pressure
[tex]\eta[/tex] = viscosity
l = length of rock
Now put all the given values in the above formula, we get:
[tex]0.5cm^3/s=\frac{(\text{Absolute permeability}\times 3cm^2\times 2\text{ atm})}{1.0cp \times 4cm}[/tex]
[tex]\text{Absolute permeability}=0.111\text{ darcys}[/tex]
Therefore, the absolute permeability is, 0.111 darcys.
Notice that "SO4" appears in two different places in this chemical equation. SO42− is a polyatomic ion called "sulfate." What number should be placed in front of CaSO4 to give the same total number of sulfate ions on each side of the equation? ?CaSO4+AlCl3→CaCl2+Al2(SO4)3
Answer : The number placed in front of [tex]CuSO_4[/tex] should be, three (3).
Explanation :
Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.
The given unbalanced chemical reaction is,
[tex]CaSO_4+AlCl_3\rightarrow CaCl_2+Al_2(SO_4)_3[/tex]
This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of chloride and sulfate ion are not balanced.
In order to balanced the chemical reaction, the coefficient 3 is put before the [tex]CuSO_4[/tex], the coefficient 2 is put before the [tex]AlCl_3[/tex] and the coefficient 3 is put before the [tex]CaCl_2[/tex].
Thus, the balanced chemical reaction will be,
[tex]3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3[/tex]
Therefore, the number placed in front of [tex]CuSO_4[/tex] should be, three (3).
The number which should be placed in front of CaSO4 to give the same total number of sulfate ions on each side of the equation is; 3.
According to the question;
We are required to determine what number should be placed in front of CaSO4 to give the same number of each side of the equation.The equation given is;
?CaSO4 + AlCl3 → CaCl2 + Al2(SO4)3
There are 3 moles of SO4 on the right hand side of the equation and as such, there should be the same number of SO4 on the left too.
In essence, the number that should be added in front of CaSO4 is; 3.
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Predict the sign for AS for each of the following systems: Water freezing Water evaporating Crystalline urea dissolving Assembly of the plasma membrane from individual lipids. Assembly of a protein from individual amino acids.
Answer: 1. Water freezing : [tex]\Delta S[/tex] is -ve.
2. Water evaporating : [tex]\Delta S[/tex] is +ve.
3. Crystalline urea dissolving : [tex]\Delta S[/tex] is +ve.
4. Assembly of the plasma membrane from individual lipids: [tex]\Delta S[/tex] is -ve.
5. Assembly of a protein from individual amino acids: [tex]\Delta S[/tex] is -ve.
Explanation:
Entropy is defined as the measurement of degree of randomness in a system.
It is represented by symbol S and we can only measure a change in entropy which is given by [tex]\Delta S[/tex].
If there is decrease in randomness , the sign for [tex]\Delta S[/tex] is -ve and If there is increase in randomness , the sign for [tex]\Delta S[/tex] is +ve.
1. Water freezing: Entropy decreases as we move from liquid state to to solid state and thus [tex]\Delta S[/tex] is -ve.
2. Water evaporating : Entropy increases as we move from liquid state to gaseous state and thus [tex]\Delta S[/tex] is +ve.
3. Crystalline urea dissolving : The molecules convert from solid and ordered state to aqueous phase and random state. Thus the entropy increases and thus [tex]\Delta S[/tex] is +ve.
4. Assembly of the plasma membrane from individual lipids: random lipids are associating to form a single large polymer and thus entropy decreases and thus [tex]\Delta S[/tex] is -ve.
5. Assembly of a protein from individual amino acids: random amino acids are associating to form a single large polymer and thus entropy decreases and thus [tex]\Delta S[/tex] is -ve.
A hydrogen molecule (diameter 1.38 × 10-8 cm), traveling at the rms speed, escapes from a 4200 K furnace into a chamber containing cold argon atoms (diameter 3.43 × 10-8 cm) at a density of 4.43 × 1019 atoms/cm3(a) What is the speed of the hydrogen molecule
Answer: The speed of hydrogen molecule is [tex]7.199\times 10^3m/s[/tex]
Explanation:
The equation used to calculate the root mean square speed of a molecule is:
[tex]V_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
where,
[tex]V_{rms}[/tex] = root mean square speed of the molecule = ?
R = Gas constant = 8.314 J/mol.K
T = temperature = 4200 K
M = molar mass of hydrogen molecule = [tex]2.02g=2.02\times 10^{-3}kg[/tex] (Conversion factor: 1 kg = 1000 g)
Putting values in above equation, we get:
[tex]V_{rms}=\sqrt{\frac{3\times 8.314\times 4200}{2.02\times 10^{-3}}}\\\\V_{rms}=7.199\times 10^3m/s[/tex]
Hence, the speed of hydrogen molecule is [tex]7.199\times 10^3m/s[/tex]
Select all the true statements. When an atom gains an electron, it becomes a cation. Anions carry a positive charge. The Cl− and Br− ions have the same number of electrons. The K+ ion is formed when a potassium atom loses one electron. The Fe2+ and Fe3+ ions have the same number of protons. The Cu+ and Cu2+ ions have the same number of electrons.
Answer:
The statements 4 and 5 are true.
Explanation:
1. When an atom gains an electron it becomes negatively charged. This negatively charged species is called anion.
A + e⁻ → A⁻ (anion)
Therefore, the statement 1 is false.
2. An anion is formed when an atom gains an electron and becomes negatively charged. Therefore, an anion is a negatively charged species.
A + e⁻ → A⁻ (anion)
Therefore, the statement 2 is false.
3. The atomic number of chlorine atom Cl is 17 and atomic number of bromine atom Br is 35.
Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.
Therefore, the number of electrons in Cl atom is 17 and the number of electrons in Br atom is 35.
When the Cl atom gains one electron it forms Cl⁻ ion and when the Br atom gains one electron it forms Br⁻ ion.
Therefore, the number of electrons in Cl⁻ ion is 17 + 1 = 18 electrons
and the number of electrons in Br⁻ ion is 35 + 1 = 36 electrons
Therefore, Cl⁻ and Br⁻ ions do not have the same number of electrons.
Therefore, the statement 3 is false.
4. When potassium atom (K) loses one electron it forms a positively charged species called potassium cation (K⁺).
K → K⁺ + e⁻
Therefore, the statement 4 is true.
5. The atomic number of Fe atom is 26.
Since, the atomic number of an atom is equal to the number of protons present in that atom.
When the Fe atom loses two electrons to form Fe²⁺ and when the Fe atom loses three electrons to form Fe³⁺ ion, the number of protons remains the same.
Therefore, the ions Fe²⁺ and Fe³⁺ have the same number of protons.
Therefore, the statement 5 is true.
6. The atomic number of copper atom Cu is 29.
Since, for neutral atom, the atomic number of an atom is equal to the number of electrons present in that atom.
Therefore, the number of electrons in Cu atom is 29
When the Cu atom loses one electron it forms Cu⁺ ion and when the Cu atom loses two electrons it forms Cu²⁺ ion.
Cu → Cu⁺ + e⁻ and Cu → Cu²⁺ + 2e⁻
Therefore, the number of electrons in Cu⁺ ion is 29 - 1 = 28 electrons
and the number of electrons in Cu²⁺ ion is 29 - 2 = 27 electrons
Therefore, Cu⁺ ion and Cu²⁺ ion do not have the same number of electrons.
Therefore, the statement 6 is false.
Atom is the smallest constituent of any chemical species and contains protons and electrons. The true statements are, [tex]\rm K^{+}[/tex] ion is formed when a potassium atom loses one electron. The [tex]\rm Fe^{2+}[/tex] and [tex]\rm Fe^{3+}[/tex] ions have the same number of protons.
What are cations and anions?When an atom acquires electrons from some other atom then they become negatively charged and are called anions.
Cations and anions are formed when the atom loses and gains an electron from another species. When an atom acquires an electron they are called an anion and when loose electrons are called a cation.
When potassium atom (K) relinquishes an electron then a positively charged species formed is called a cation. It can be shown as,
[tex]\rm K \rightarrow K ^{+} + e^{-}[/tex]
The atomic number of an iron atom is 26 and is equal to the number of protons. The number of protons remains the same when the iron atom relinquishes two electrons yields ferrous ions and when loses three electrons yields ferric ion. Hence, ferrous and ferric have the same number of protons.
Therefore, option 4. K+ ion is formed when a potassium atom loses one electron and option 5. ferrous and ferric ions have the same number of protons are correct.
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The lowest pressure attainable using the best available vacuum techniques is about 10−12 N/m2. Part A At such a pressure, how many molecules are there per cm3 at 2 ∘C?
Hey there!:
We are given P = 10⁻²² N/m2 ,
Volume, V = 1 cm3= 10⁻⁶ m³ and
T= 13ºC= 13+273= 286 K
Using gas law, P*V = n R T, where n is number of moles and R=8.314JK⁻¹.
n = PV/(RT) => n = 10-12*10⁻⁶/ (8.314*286) = 4.205*10₋²² moles
Hence number of molecules = number of moles * Avogadro's number
= 4.205*10-22 moles * 6.023*1023 molecules/mole
= 253
Number of molecules = 250 (upto 2 significant figures)
Number of molecules: N = 263.31
Further explanationSome of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated),:
Boyle's law at constant T,
[tex] \displaystyle P = \dfrac {1} {V} [/tex]
Charles's law, at constant P,[tex] \displaystyle V = T [/tex]
Avogadro's law, at constant P and T,[tex] \displaystyle V = n [/tex]
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
[tex] \large {\boxed {\bold {PV = nRT}}} [/tex]
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / m
m = mass
M = relative molecular mass
Known
P = 10−12 N / m2
V = 1 cm3 = 10-6 m3
T = 2 ºC = 2 + 273 = 275 K
R = 8,314 J / mol. K
[tex]\rm n=\dfrac{PV}{RT}\\\\n=\dfrac{10^{-12}\times 10^{-6}}{8.314\times 275}\\\\n=\boxed{\bold{4.374\times 10^{-22}}}[/tex]
then the number of molecules (N):
N = n x No
N = 4,374.10⁻²² x 6.02.10²³
N = 263.31
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Which law relates to the ideal gas law
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Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess thionyl chloride, the residue was yellow, and had a mass of 2.3921 g.
(a) Was the nickel chloride sample a hydrate? ___________________
(b) If the data indicates a hydrate, calculate x, the number of waters of hydration.
Show your work.
Answer:
a) yes, it was an hydrate
b) the number of waters of hydration, x = 6
Explanation:
a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.
b) NiCl2. xH2O
mass if dehydrated NiCl2 = 2.3921 grams
mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.
NiCl2.xH2O
mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole
mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole
Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each
for NiCl2 = 0.01846/0.01846 = 1
for H2O = 0.11072/0.01846 = 5.9976 = 6
thus the hydrated sample was NiCl2. 6H2O