Answer:
You have to weigh 0.079 g of Na₂S₂0₃ and dissolve it in water at a final volume of 250 ml.
Explanation:
A 0.00200 M Na₂S₂O₃ solution has 0.002 mol of Na₂S₂O₃ per liter of solution. As we know that 1 L = 1000 ml, and that the molecular weight of Na₂S₂O₃ IS 158 g/mol, we can calculate the mass of Na₂S₂O₃ to weigh as follows:
mass= [tex]\frac{0.002 mol Na2S2O3}{1000 ml solution}[/tex] x [tex]\frac{158 g}{1 mol Na2S2O3}[/tex] x 250 ml solution
mass= 0.079 g
To prepare the solution, we have to weigh in a beaker with 0.079 g of Na₂S₂O₃ by employing an analytical balance. Then, we have to dissolve the mass in a volume of aproximately 230 ml water. For this, we measure the 230 ml of water in a graduated cylinder, we add the volume to the beaker with the mass and we agitate until total disolution. Finally, we tranfer the total amount of the solution in the glass to a volumetric flask with a capacity of 250 ml. We add water until we reach the capacity, and then we homogenize the solution.
The quantity of energy released or absorbed as heat during a chemical reaction is called the 28. a. temperature. c. entropy d. free energy enthalpy of reaction.
Answer: enthalpy of reaction.
Explanation:
Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like [tex]^0C[/tex] and [tex]K[/tex]
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
Free energy is the amount of energy that can be converted into useful work.
Enthalpy of the reaction is the difference between the energy of products and the energy of reactants. it is either the heat released or absorbed during the reaction. It is either positive or negative.
According to the following thermochemical equation, if 951.1 g of NO 2 is produced, how much heat is released at constant pressure? 2NO(g) + O2(g) 2NO 2(g); DH -114.4 kJ
Answer:
The amount of released is 1,182 kJ.
Explanation:
When heat is released at constant pressure, this change in energy is known as enthalpy (ΔH°) of the reaction. Enthalpy is an extensive property, so it depends on the amount of reacting material. Let's take a look at the provided equation:
2 NO(g) + O₂ ⇄ 2 NO₂(g) ΔH° = -114.4 kJ
Since this equation is balanced with 2 moles of NO₂(g), we can say that 114.4 kJ are released every 2 moles of NO₂(g) produced. By convention, when enthalpies are negative, it means that energy is released and the reaction is exothermic. Conversely, positive enthalpies mean energy is absorbed and the reaction is endothermic.
We can calculate the amount of energy released taking into account the previous relationship (-114.4 kJ/2 moles of NO₂(g)), the mass of NO₂(g) produced (951.1g) and its molar mass (46.00g/mol). The calculations would be:
[tex]951.1g.\frac{1molNO_{2} }{46.00g} .\frac{-114.4kJ}{2molesNO_{2} } =-1,182kJ[/tex]
The amount of heat released at constant pressure when 951.1 g of [tex]NO_2[/tex] is produced is approximately -544.5 kJ.
The molar mass of [tex]NO_2[/tex] can be calculated from its molecular weight:
Nitrogen (N) has an atomic mass of approximately 14.01 g/mol,
Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
Thus, the molar mass of [tex]NO_2[/tex] is:
[tex]\[ \text{Molar mass of NO}_2 = 14.01 \text{ g/mol} + 2 \times 16.00 \text{ g/mol} = 46.01 \text{ g/mol} \][/tex]
Now, we can calculate the number of moles of [tex]NO_2[/tex] produced:
[tex]\[ \text{moles of NO}_2 = \frac{\text{mass of NO}_2}{\text{molar mass of NO}_2} = \frac{951.1 \text{ g}}{46.01 \text{ g/mol}} \approx 20.67 \text{ mol} \][/tex]
The thermochemical equation given is:
[tex]\[ 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g); \quad \Delta H = -114.4 \text{ kJ} \][/tex]
This equation tells us that 2 moles of NO2 are produced for every -114.4 kJ of heat released at constant pressure. Therefore, the heat released for the production of 20.67 moles of NO2 is:
[tex]\[ \text{Heat released} = \text{moles of NO}_2 \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \][/tex]
[tex]\[ \text{Heat released} = 20.67 \text{ mol} \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \approx -544.5 \text{ kJ} \][/tex]
Which of the following is a definition of an acid? O a. releases heat when dissolved in water. O b. turns cloudy when dissolved in water. O c. donates OH-ions when dissolved in water O d. donates H ions when dissolved in water.
Answer: Option (d) is the correct answer.
Explanation:
According to Bronsted-Lowry an acid is defined as the specie which is able to donate hydrogen ions when dissolved in water.
For example, [tex]HCl \rightarrow H^{+} + Cl^{-}[/tex]
On the other hand, bases are the species which are able to donate hydroxide ions when dissolved in water.
For example, [tex]NaOH \rightarrow Na^{+} + OH^{-}[/tex]
Thus, we can conclude that a definition of an acid is that it donates H ions when dissolved in water.
Mass transfer rate in convection is ..... mass transfer in conduction
a) more than
b) less than
c) equal to
d) no relation to
Answer:
(a) more than
Explanation:
Conduction:
Transfer of heat due to direct contact between two mediums at different temperatures, without having any of the bodies traveling. Therefore, conduction heat transfer occurs by the transfer of momentum (molecular) from always the same group of molecules in one medium to another group of molecules in another medium.
Example: Heat transfer INSIDE a solid.
Convention:
Transfer of heat or mass due to at least one traveling medium, where the transfer of momentum is not bounded anymore to the same groups of molecules. Molecules moving to transfer their momentum and keep flowing to the next group, also allowing other molecules behind to do the same. Example: heat transfer by the wind.
Hence, the Mass transfer rate in convection is more than mass transfer in conduction
Final answer:
The mass transfer rate in convection is generally more than (option a) that in conduction due to the macroscopic movement of mass that facilitates faster heat transfer.
Explanation:
The question asks to compare the mass transfer rate in convection to the mass transfer rate in conduction. The correct comparison is that the mass transfer rate in convection is generally more than the mass transfer rate in conduction. This is because conduction is the process by which heat is directly transmitted through a substance when there is a difference of temperature, without movement of the material. In contrast, convection involves the movement of mass (typically a fluid such as air or water) which contributes to a faster transfer of heat. An empirical equation for the rate of heat transfer by forced convection can be represented by Q/t = hA(T₂ – T₁), showing that this rate is dependent on the temperature difference (T₂ – T₁), contact surface area (A), and the convective heat transfer coefficient (h).
Steve holds the world record for the 100-meter dash. At his fastest, Bolt ran 100 meters in 9.58 seconds. Convert his speed into miles per hour. Report your answer using 3 significant figures.
Conversion factors: 1 mile = 1609 meters 60 seconds = 1 minute 60 minutes = 1 hour
Use appropriate Sig Figs in answer!
Answer:
23.4 mph
Explanation:
The conversion factors are multiplied so that the units cancel out:
(100 meters / 9.58 sec) x (1 mile / 1609 meters) x (60 sec / min) x (60 min /1 hr) = 23.4 mph
Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acetate concentration of 250 mM and a pH of 5.0. What concentrations of acetic acid and sodium acetate should you use? Assuming you wish to make 2 liters of this buffer, how many moles of acetic acid and sodium acetate will you need? How many grams of each will you need (molecular weights: acetic acid 60. 05 g mol 1, sodium acetate, 82. 03 g mol 1)?
Answer:
0.182 moles of acetic acid are needed, this means 10.93 g.
0.318 moles of sodium acetate are needed, this means 26.08 g.
Explanation:
The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:
pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.
From the problem we know that
[A⁻] + [HA] = 250 mM = 0.250 M eq. 1
We use the H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:
5.0 = 4.76 + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]
[tex]0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}][/tex] eq.2
Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:
1.74 [HA] + [HA] = 0.250 M
[HA] = 0.091 M
Then we calculate [A⁻]:
[A⁻] + 0.091 M = 0.250 M
[A⁻] = 0.159 M
Using the volume, we can calculate the moles of each substance:
moles of acetic acid = 0.091 M * 2 L = 0.182 molesmoles of sodium acetate = 0.159 M * 2 L = 0.318 molesUsing the molecular weight, we can calculate the grams of each substance:
grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 ggrams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 gThe diffusion flux is defined as: a) Mass, per unit surface area, per unit time b) Surface area, per unit time c) Time, per unit surface area. d) The product of time and surface area, per unit mass
Answer:
The correct answer is option a.
Explanation:
Diffusion flux is defined as movement of mas of atoms diffusing through the unit area in per unit time.It measured in ([tex]kg/m^2 s[/tex]).
[tex]J=\frac{1}{A}\frac{dM}{dt}[/tex]
J = diffusion flux
A = Unit area A through which atoms moves.
M = mass of atoms passes in t interval of time.
Diffusion flux is the mass of gas passing through a unit surface area per unit time, influenced by the concentration gradient, surface area, and particle travel distance.
Explanation:The diffusion flux can be defined as the amount of gas passing through a given area over a unit of time. Therefore, the correct answer is a) Mass, per unit surface area, per unit time. The rate of diffusion is dictated by several factors such as the concentration gradient, the surface area available for diffusion, and the distance the gas particles must travel. In addition, it's important to note that the time required for diffusion is inversely proportional to the diffusion rate.
Learn more about Diffusion Flux here:https://brainly.com/question/33792345
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Oil with density of 0.8505 and velocity of 4m/s in a pipe that has a downward slope of 1:50 (sine). At a certain point in the pipe, a pressure gauge shows a pressure of 80kPa.
Determine the pressure at a point 200m downstream of the gauge if:
(a) The flow losses are ignored (ideal case);
(b) There is a flow loss equal to 10% of the total initial head.
Answer:
(a) Pressure at point 200 m downstream: 113.4 kPa
(b) Pressure at point 200 m downstream: 105.4 kPa
Explanation:
The point of reference, that will be named Point A, is the pressure gage.
The point 200 m downstream will be named Point B.
With a slope of 1:50, the height difference between A and B is (1/50)*200=4m.
(a) In the case there is no friction losses, we can write
[tex]p_a+\frac{\rho V^2}{2} +\rho*g*h_a=p_b+\frac{\rho V^2}{2} +\rho*g*h_b[/tex]
Because ther is no change in the diameter of the pipe, V=constant.
Rearranging
[tex]p_a+\rho*g*h_a=p_b+\rho*g*h_b\\\\\Delta p = \rho * g * \Delta h\\\\\Delta p = 0.8505 * 1000\frac{kg}{m^3} *9.81\frac{m}{s^2}*4 m\\\\ \Delta p = 33.373.62 kg/(m*s^2)=33.373.62 Pa=33.4 kPa[/tex]
Then we have an increase in of 33.4 kPa, and pressure in Point B (downstream) is:
[tex]p_b=p_a+\Delta p = 80 kPa + 33.4 kPa = 113.4 kPa[/tex]
(b) If we consider 10% of the total initial head as friction loss, we have
[tex]p_a+\rho*g*h_a=p_b+hf +\rho*g*h_b\\\\hf=0.1*p_a\\\\\Delta P = \rho*g*\Delta h-0.1*p_a\\\\\Delta P = 0.8505*1000 kg/m^3*9.81m/s2*4m-0.1*80kPa\\\\\Delta P = 33.4kPa-8kPa=25.4kPa[/tex]
In this case the rise in pressure is 25.4 kPa, due to the friction losses.
The pressure at point B is
[tex]p_b=p_a+\Delta p = 80 kPa + 25.4 kPa = 105.4 kPa[/tex]
A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base (B) solution. Show all your work.
pKb for the weak base = 4.82.
Molar mass of the weak base = 82.0343g/mole.
Note: pKa = -logKa
pKb = -logKb
pH + pOH = 14
[H+ ] [OH- ] = 10^-14
Answer:
11.39
Explanation:
Given that:
[tex]pK_{b}=4.82[/tex]
[tex]K_{b}=10^{-4.82}=1.5136\times 10^{-5}[/tex]
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.805\ g}{82.0343\ g/mol}[/tex]
[tex]Moles= 0.022\ moles[/tex]
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.022}{0.055}[/tex]
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
[tex]K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}[/tex]
[tex]1.5136\times 10^{-5}=\frac {x^2}{0.4-x}[/tex]
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
pH = 14 - pOH = 14 - 2.61 = 11.39
How many cubic centimeters are there in 200 milliliters?
Answer:
200 mL = 200 cm³
Explanation:
The relationship between cm³ and mL is 1:1.
1 cm³ = 1 mL
Thus, 200 mL is converted to cm³ as follows:
(200 mL)(1 cm³/1 mL) = 200 cm³
Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
Answer:
The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.
Explanation:
Suppose there are 100 moles in solution:
Moles of sulfuric acid = 6% of 100 moles = 6 moles
Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g
Moles of water = 100%- 6% = 94%= 94 moles
Mass of water = 94 mol × 18 g/mol = 1692 g
Specific gravity of the solution ,S.G= 1.07
Density of solution = D
[tex]S.G=\frac{D}{d_w}[/tex]
[tex]d_w[/tex] = density of water = 1 g/mL
[tex]D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL[/tex]
Mass of the solution = 588 g + 1692 g = 2280 g
Volume of the solution = V
Volume = [tex]\frac{Mass}{Density}[/tex]
[tex]=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L[/tex]
1 mL = 0.001 L
[tex]Molarity = \frac{n}{V(L)}[/tex]
n = number of moles of compound
V = volume of the solution in L
here we have ,n = 6 moles of sulfuric acid
V = 2.13084 L
So, the molarity of the solution is :
[tex]Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L[/tex]
Why is it important to have an energy balance on a chemical process facility
Final answer:
The importance of an energy balance in a chemical process facility lies in ensuring efficient energy use, maintaining safety, optimizing chemical processes for increased yield, and reducing environmental impact.
Explanation:
An energy balance is crucial in a chemical process facility because it ensures the efficient use of energy, minimizes waste, and helps maintain the safety of the operation. Since energy is neither created nor destroyed (law of conservation of energy), it's important to know where energy is being consumed and produced within chemical processes. This understanding allows engineers to optimize the process, increase the yield of the desired product, and reduce environmental impact by minimizing wasted energy and reducing unwanted by-products.
An energy balance aids in maintaining the operation within the desired range of conditions, preventing unsafe levels of energy which could lead to accidents. Moreover, certain reactions require specific amounts of energy to produce raw materials or to synthesize products, and tracking energy use is essential for economic and environmental reasons.
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.501 L flask at 1043 K. At equilibrium, the flask contains 0.169 mol of CO gas, 0.257 mol of H2 gas, and 0.255 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1043 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the K expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into K expression, solve for the unknown. (MTS 5/16/2018)
Answer:
[ H2O ]eq = 0.298 mol/L
Explanation:
CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)∴ V = 0.501 L
∴ T = 1043 K
at equilibrium:
∴ n CH4 = 0.255 mol
⇒ [ CH4 ]eq = 0.255 mol / 0.501 L = 0.509 mol/L
∴ n CO = 0.169 mol
⇒ [ CO ]eq = 0.169 / 0.501 = 0.337 mol/L
∴ n H2 = 0.257
⇒ [ H2 ] eq = 0.257 / 0.501 = 0.513 mol/L
∴ Kc = [ H2 ]³ * [ CO ] / [ CH4 ] * [ H2O ] = 0.30
⇒ [ H2O ] = [ H2 ]³ * [ CO ] / [ CH4 ] * 0.30
replacing the value of the concentration in Kc:
⇒ [ H2O ] = ( 0.513 )³ * ( 0.337 ) / ( 0.509 ) * 0.30
⇒ [ H2O ] = 0.298 mol/L
The equilibrium concentration of water vapor in the conversion of methane to other fuels at 1043 K is determined to be 0.6196 M by using the equilibrium constant expression and the known concentrations of the other substances.
Explanation:The student's question involves determining the water concentration at equilibrium for the conversion of methane to other fuels involving a reaction of methane (CH4) and water vapor (H2O). First, we need to write a balanced chemical equation for the reaction:
CH4(g) + 2H2O(g) ⇌ CO2(g) + 4H2(g)
To find the water concentration, we use the equilibrium constant expression (Kc):
Kc = [tex][CO]^1[H2]^4 / [CH4]^1[H2O]^2[/tex]
Given that Kc = 0.30 at 1043 K, and the equilibrium amounts of the substances are CH4 = 0.255 mol, CO = 0.169 mol, and H2 = 0.257 mol in a 0.501 L flask, we can calculate their concentrations and solve for [H2O]:
[CH4] = 0.255 mol / 0.501 L = 0.509 M
[CO] = 0.169 mol / 0.501 L = 0.337 M
[H2] = 0.257 mol / 0.501 L = 0.513 M
Plugging these into the Kc expression and solving for [H2O], we get:
0.30 = (0.337)[([tex]0.513)^4] / [(0.509)([H2O]^2)][/tex]
After calculations, the water concentration at equilibrium, [H2O], is found to be approximately 0.6196 M.
out of 2.00 50 mL of water is added to 30 mL of a 0.50 M salt solution. What is the new concentration? (To write your answer using scientific notation use 1.0E-1 instead of 1.0 x 10-1)
Answer:
The concentration of the new solution will be 0.31 M
Explanation:
The number of moles per liter of the solution is 0.50 mol. Then, in 50 ml there will be:
50 ml · (0.50 mol / 1000 ml) = 0.025 mol.
If we add 30 ml and assuming that the solution is an ideal solution, the final volume will be 80 ml.
Then, 0.025 mol will be present in 80 ml solution. In 1 l there will be:
1000 ml · (0.025 mol / 80 ml) = 0.31 mol.
The concentration of the new solution will be 0.31 M.
Which form of bond between two carbons is capable of rotation without breaking the bond? * single bond double bond triple bond
Answer:
single bond
Explanation:
Single bond is the chemical bond in which 2 valence electrons are involved. These are the sigma bond which are formed formed by the head-on overlapping between the orbitals of the two atoms involving in the bond. The bond lies on the inter nuclear axis and thus , the bond can be rotated without the breaking of bond.
On the other hand, double and triple bonds contains one sigma bond and also bonds which are formed by the sideways overlapping of the orbitals and thus, these do not lie on inter nuclear axis and breaks on rotation.
If the molality of a NaBr(aq) solution is 2.50 m, what is the weight percent of NaBr? The molar mass of NaBr is 1029 g/mol 20.5% 25.0% 25.7% 34.6% 65.4% ent Navigator J K L
Answer:
The weight percent of NaBr is 25,7%
Explanation:
Molality is a way to express the concentration of a chemical in terms of moles of substances per kg of solution. Weight percent is other way to express concentration in terms of mass of substance per mass of solution per 100
Thus, to obtain weight percent you must convert moles of NaBr to grams with molar mass and these grams to kilograms, thus:
2,50 mol / kg solution × (102,9 g / 1mol) × (1 kg / 1000 g) =
[tex]\frac{0,257 kg NaBr}{kg solution }[/tex] × 100 = 25,7 % (w/w)
I hope it helps!
Write dissociation equations for the following soluble salts dissolving in water. Then draw a particle - level representation for each (omit water molecules). (a) NaBr (b) ZnCl2
Answer:
The dissociation equations for NaBr gives Na+ and Br-
The dissociation equations for ZnCl2 gives Zn2+ and 2 Cl-
Explanation:
The following pictures shows that the dissociation of one particle of NaBr produces one particle of Na+ (sodium cation) and one particle of Br- (bromine anion).
The dissociation of one particle of ZnCl2 produces one particle of Zn+2 (Zinc cation) and two particles of Cl- (chlorine anion).
An unstable isotope that naturally decays CAN emit which of the following? Choose all correct responses, there are at least two. {Hint: probably are more than two}
Question options:
A) Alpha particle
B) Beta particle
c) Radiation
D) Positron
E) Gamma rays
F) Visible light
Answer: Isotopes are the species which have same number of protons but differs in the number of neutrons. Or in other words isotopes have same atomic number but different mass number. Isotopes are generally unstable and decays to an stable isotope. When the isotope is unstable it releases energy in the form of radiations. It emits alpha particle, gamma rays, and beta particles.
Therefore, the correct option is (A),(B) and (C)
The airspeed velocity of an unladen swallow is 10.00 meters per second. What is it in miles per hour? You must use appropriate significant figures to receive full credit (3 pts) 1 in = 2.54 cm, 100 cm = 1m, 12 in =1ft, 5280 ft = 1 mile, 60 s = 1 min, 60 min = 1 hour
Answer : The speed in miles per hour is 22 mile/hr.
Explanation :
The conversion used from meters to miles is:
[tex]1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}[/tex]
The conversion used from second to hour is:
[tex]1s=\frac{1}{60}min\times \frac{1hr}{60min}[/tex]
The conversion used from meter per second to mile per hour is:
[tex]1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}[/tex]
[tex]1m/s=2.2mile/hr[/tex]
As we are given the speed 10.00 meter per second. Now we have to determine the speed in miles per hour.
As, [tex]1m/s=2.2mile/hr[/tex]
So, [tex]10.00m/s=\frac{10.00m/s}{1m/s}\times 2.2mile/hr=22mile/hr[/tex]
Therefore, the speed in miles per hour is 22 mile/hr.
Liquid water boiling in an open pan on a stove is our system. Is the system open or closed? What boundary transfers are happening? Is the system at steady state?
Answer:
It's an open system, tranfering heat through a rigid, diathermal wall and matter through an imaginary and permeable wall, and it is not at steady state.
Explanation:
An open system is that that interacts with its surroundings exchanging energy and matter. In an open pan with boiling water you have an open system because steam (matter) is leaving the system, as well as heat (energy) through the pan/stove. A boundary is what separates the system from its surroundings, there are many types of boundaries, based on how they transfer energy they can be diathermal (conducting heat) or adiabatic (insulating), on their rigidity they can be rigid, flexible, imaginary or movable and based on their permeability. For the system described we have an imaginary boundary on top that is also permeable allowing matter to go out or in the system, and another wall (the stove/pan itself that is rigid and impermeable avoiding the loss of matter and diathermal, allowing the conduction of heat. It is said that a system is at a steady state when the variables that define that system remain constant over time. In an open pan, you can't fully control those variables, you'll have matter and energy scaping from it with no way to regulate it.I hope you find interesting and useful this information! good luck!
what is the osmotic pressure of a .25 M solution of NaCl at 25 degrees Celsius?
Answer:
6.113 atm
Explanation:
Data provided in the question:
Molarity of the solution = 0.25 M
Temperature, T = 25° C = 25 + 273 = 298 K
Now,
Osmotic pressure (π) is given as:
π = MRT
where,
M is the molarity of the solution
R is the ideal gas constant = 0.082057 L atm mol⁻¹K⁻¹
on substituting the respective values, we get
π = 0.25 × 0.082057 × 298
or
π = 6.113 atm
a For each of the salts on the left, match the salts on the right that can be compared directly, using Ksp values, to estimate solubilities. (If more than one salt on the right can be directly compared, include all the relevant salts by writing your answer as a string of characters without punctuation, e.g, ABC.)
1. silver iodide ___ [need answer here]
2. manganese(II) hydroxide ___ [need answer here]
A. Fe(OH)2
B. CaSO3
C. NiCO3
D. Ba3(PO4)2
Write the expression for K in terms of the solubility, s, for each salt, when dissolved in water.
silver iodide Ksp = ___ [need answer here]
manganese(II) hydroxide Ksp = ___ [need answer here]
Note: Multiply out any number and put it first in the Ksp expression. Combine all exponents for s.
Answer:
a.
silver iodide BC
manganese(II) hydroxide A
b.
silver iodide Ksp = S²
manganese(II) hydroxide Ksp = 4S³
Explanation:
Ksp (Solubility products) are the equilibrium constants for poorly soluble compounds. As every equilibrium constant, it is formed by the product of the products raised to their stoichiometric coefficients divided by the product of reactants raised to their stoichiometric coefficients. We only include in the constant gases and aqueous species. So, to solve this task, we need to write each reaction and its Ksp.
1. Silver Iodide
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)
When AgI is put in water, the concentration of Ag⁺ and I⁻ that actually dissolve is known as solubility (S). So the concentration of both Ag⁺ and I⁻ would be S.
AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)
S S
We can replace this in the Ksp expression:
Ksp = [Ag⁺].[I⁻] = S.S = S²
We can follow the same steps to find out the relationship between Ksp and S for each compound.
2. Mn(OH)₂(s) ⇄ Mn²⁺(aq) + 2OH⁻(aq)
S 2S
In this case, the concentration of OH⁻ is 2S because 2 moles are produced along with 1 mole of Mn²⁺.
Ksp = [Mn²⁺].[OH⁻]² = S.(2S)² = 4S³
A. Fe(OH)₂(s) ⇄ Fe²⁺(aq) + 2OH⁻(aq)
S 2S
Ksp = [Fe²⁺].[OH⁻]² = S.(2S)² = 4S³
B. CaSO₃(s) ⇄ Ca²⁺(aq) + SO₃²⁻(aq)
S S
Ksp = [Ca²⁺].[SO₃²⁻] = S.S = S²
C. NiCO₃(s) ⇄ Ni²⁺(aq) + CO₃²⁻(aq)
S S
Ksp = [Ni²⁺].[CO₃²⁻] = S.S = S²
D. Ba₃(PO₄)₂(s) ⇄ 3 Ba²⁺(aq) + 2 PO₄³⁻(aq)
3S 2S
Ksp = [Ba²⁺]³.[PO₄³⁻]²= (3S)³.(2S)²= 108S⁵
a. The salts that can be compared using the same Ksp expressions are:
silver iodide BC
manganese(II) hydroxide A
b.
silver iodide Ksp = S²
manganese(II) hydroxide Ksp = 4S³
None of the listed salts can be compared to silver iodide using Ksp values. Manganese(II) hydroxide can be compared to Fe(OH)2. The Ksp expressions in terms of solubility, s, are s^2 for silver iodide and 4s^3 for manganese(II) hydroxide.
Explanation:We can directly compare salts by considering their solubility product constants, or Ksp values
. 1. Silver iodide: No salts on the right share the same ions with silver iodide, so none can be directly compared.
2. Manganese(II) hydroxide: It contains Mn2+ and OH- ions. Thus, it can be compared to the Fe(OH)2 salt which contains Fe2+ and OH- ions, making the matching answer A.
For the solubility expressions for each salt, when Multiplied we put the number first and then combine all exponents:
For silver iodide (AgI), the equilibrium is AgI ↔ Ag+ + I-, so the Ksp expression is Ksp = s x s = s^2.
For manganese(II) hydroxide (Mn(OH)2), the equilibrium is Mn(OH)2 ↔ Mn2+ + 2OH-, so the Ksp expression is Ksp = 4s^3.
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What is the molar concentration of chloride ions in a
solutionprepared by mixing 100mL of 2.0 M KCL with 50 L of a 1.5
MCaCl2 solution?
Answer:
The concentration of chloride ions in the final solution is 3 M.
Explanation:
The number of moles present in a solution can be calculated as follows:
number of moles = concentration in molarity * volume
In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻
For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:
number of moles of Cl⁻ = 2 * number of moles of CaCl₂
number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻
The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.
Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:
Concentration = number of moles of Cl⁻ / volume
Concentration = 150.2 mol / 50.1 l = 3.0 M
To find the molar concentration of chloride ions in the solution, calculate the moles of chloride ions in each solution and add them together, then divide by the total volume of the final solution.
Explanation:To determine the molar concentration of chloride ions in the solution, we need to calculate the number of moles of chloride ions and divide it by the total volume of the solution.
First, calculate the moles of chloride ions in KCl solution:
Moles of chloride ions = molarity of KCl solution * volume of KCl solution
Next, calculate the moles of chloride ions in CaCl2 solution:
Moles of chloride ions = (molarity of CaCl2 solution * volume of CaCl2 solution) * 2 (since CaCl2 has 2 chloride ions per formula unit)
After finding the moles of chloride ions in each solution, add them together and divide by the total volume of the final solution to get the molar concentration of chloride ions.
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Compute 4.62 x 4.48697. Round the answer appropriately. Express your answer numerically using the proper number of sian
Answer:
20.7
Explanation:
20.7298014 rounded off to 3 sig fig =20.7
Calculate the pressure, in atmospheres, required to compress a sample of helium gas from 27.3 L (at 1.00 atm) to 3.36 L at constant temperature. Enter your answer in the box provided. atm
Answer:
8.125 atm
Explanation:
Hello,
Boyle's law states that at constant temperature:
[tex]P_1V_1=P_2V_2[/tex]
In this case:
[tex]P_1=1 atm, V_1=27.3L, V_2=3.36L[/tex]
Solving for [tex]P_2[/tex]:
[tex]P_2=\frac{P_1V_1}{V_2}=\frac{1atm*27.3L}{3.36L}\\P_2=8.125atm[/tex]
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Solve for t In(A/B) -kt where B-1.65 x 102 M A 1.00 x 10 M. k-4.80 x 104s* and t=
Answer:
[tex]t=5.84\times 10^{-5}s[/tex]
Explanation:
First rearrange the whole equation by keeping t in one side and rest parameters in another side of equationThen plug-in all the given values for each parameters and get the solution[tex]ln(\frac{A}{B})=-kt[/tex]
or, [tex]t=\frac{1}{-k}\times ln(\frac{A}{B})[/tex]
Now plug-in all given values:
[tex]t=\frac{1}{-4.80\times 10^{4}s^{-1}}\times ln(\frac{1.00\times 10M}{1.65\times 10^{2}M})[/tex]
So, [tex]t=5.84\times 10^{-5}s[/tex]
Glucose is a carbohydrate that contains carbon, hydrogen, and oxygen. The empirical formula of glucose is CH2O and its molar mass is 180.12 g/mol. Find the molecular formula of glucose.
Answer: The molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]
Explanation:
We are given:
Empirical formula of the compound = [tex]CH_2O[/tex]
Empirical mass of the compound = [tex][(1\times 12)+(2\times 1)+(1\times 16)]=30g/mol[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
We are given:
Mass of molecular formula = 180.12 g/mol
Mass of empirical formula = 30 g/mol
Putting values in above equation, we get:
[tex]n=\frac{180.12g/mol}{30g/mol}=6[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 6)}H_{(2\times 6)}O_{(1\times 6)}=C_6H_{12}O_6[/tex]
Hence, the molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]
For the following reaction, 101 grams of magnesium nitride are allowed to react with 144 grams of water. Mg3N2 (5) + 6 H20 (1) — 3 Mg(OH)2 (aq) + 2 NH2 (aq) What is the FORMULA for the limiting reagent? What is the maximum amount of magnesium hydroxide that can be formed? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining
Answer : The formula of limiting reagent is, [tex]Mg_3N_2[/tex].
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
The mass of excess reactant is, 36 grams.
Solution : Given,
Mass of [tex]Mg_3N_2[/tex] = 101 g
Mass of [tex]H_2O[/tex] = 144 g
Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
Molar mass of [tex]Mg(OH)_2[/tex] = 58 g/mole
First we have to calculate the moles of [tex]Mg_3N_2[/tex] and [tex]H_2O[/tex].
[tex]\text{ Moles of }Mg_3N_2=\frac{\text{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=\frac{101g}{101g/mole}=1moles[/tex]
[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{144g}{18g/mole}=8moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react with 6 mole of [tex]H_2O[/tex]
So, given 1 moles of [tex]Mg_3N_2[/tex] react with 6 moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg_3N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Mg(OH)_2[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]Mg_3N_2[/tex] react to give 3 mole of [tex]Mg(OH)_2[/tex]
So, given 1 mole of [tex]Mg_3N_2[/tex] react to give 3 moles of [tex]Mg(OH)_2[/tex]
Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=(3moles)\times (58g/mole)=174g[/tex]
The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.
Now we have to calculate the moles of excess reactant [tex](H_2O)[/tex].
Moles of excess reactant = 8 - 6 = 2 moles
Now we have to calculate the mass of excess reactant.
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
[tex]\text{ Mass of }H_2O=(2moles)\times (18g/mole)=36g[/tex]
The mass of excess reactant is, 36 grams.
Calculate the heat of reaction for the combustion of one mole of benzene molecules (CH) at 25 (Given the AHE: (CO()] = -393.5 kJ/mol, (H.0(0)] = -285.8 kJ/mol, (CH(B)] = 82.6 kl/mol)
Answer:
ΔHf = 3301 kJ/mol
Explanation:
The standard enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
In a reaction ΔHf = ∑ n ΔHproducts - ∑ n ΔHreagents
Where n are moles
For the reaction:
1 C₆H₆ + 7.5 O₂ → 6 CO₂ + 3 H₂O
The ΔHf = -285,8 kJ/mol × 3 - 393,5 kJ/mol×6 - [82,6 kJ/mol×1 -0kJ/mol×7,5]
ΔHf = 3301 kJ/mol
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An aluminum atom has a mass of 4.48 * 10-23 g and a small airplane has a mass of 5000 kg. Use this information to answer the questions below. Be sure your answers have the correct number of significant digits.
(a) What is the mass of 1 mole of aluminum atoms?
(b) How many moles of aluminum atoms have a mass equal to the mass of a small airplane?
Explanation:
(a) According to the mole concept, 1 mole of an atom contains [tex]6.022 \times 10^{23}[/tex] atoms.
Hence, number of atoms present in [tex]4.48 \times 10^{-23}[/tex] g will be as follows.
[tex]4.48 \times 10^{-23} \times 6.022 \times 10^{23}[/tex]
= 26.97 g
or, = 27 g
This means that 1 mole of aluminium atoms contain 27 g.
(b) Mass of air plane is 5000 kg of [tex]5000 \times 1000 g[/tex] (as 1 kg = 1000 g).
As mass of 1 mole aluminium is calculated as 27 g and mass of air plane is given as 5000000 g.
Therefore, calculate the number of moles of aluminium atoms as follws.
No. of moles of Al atoms = [tex]\frac{\text{mass of small air plane}}{\text{mass of 1 mol Aluminium}}[/tex]
= [tex]\frac{5000000g}{27 g}[/tex]
= 185185.18
So, the answer in three significant digits will be 185000 moles of aluminum atoms have a mass equal to the mass of a small airplane.
The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.
The correct answer is described below
What are moles?The mole is the base unit of the amount of substance in the International System of Units. It is defined as exactly 6.02214076×10²³ elementary entities, which may be atoms, molecules, ions, or electronsAccording to the question, the answer of the first question is:-
The number of atoms will be:-
[tex]4.48*10^{-23}*6.022*10^{23}\\\\=26.97[/tex]
The correct answer is 27
The answer to the second question is as follows:-
The mass of the air is [tex]5000*1000g[/tex]. hence, As the mass of 1-mole aluminum is calculated as 27 g and mass of airplane is given as 5000000 g.
No moles will be
:- [tex]\frac{500000}{27} \\\\=185185.18[/tex]
Hence, the correct answer is mentioned above.
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