Answer:
Weight of solution produced = 5135 kg
Amount of water removed = 4865 kg
Explanation:
For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg
mi = 0.38x10000 = 3800 kg
The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)
mo = 0.74xmt
0.74xmt = 3800
mt = 3800/0.74
mt = 5135 kg
This is the mass of solution produced.
The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :
wi = 10000 - 3800 = 6200 kg
wo = 5135 - 3800 = 1335 kg
wr = wi - wo
wr = 6200 - 1335
wr = 4865 kg
From the data provided and the concept of balance of mass,, the mass of solution produced is 5135 kg and the water removed is 4865 kg
What is evaporation?
Evaporation is the process by which molecules of a liquid turn to gas.
To calculate the weight of solution produced and amount of water removed:
Using the concept of balance of mass, the incoming mass of sugar must be equal to the outgoing mass.
Incoming mass (Mi) = 38% of 10000 kg
Mi = 3800 kg
Therefore, the outgoing mass (Mo)= 3800 kg
Mo= 74% of the total mass, Mt
Mo = 0.74 x Mt
Mt = 3800/0.74
Mt = 5135 kg
Thus, the mass of solution produced is 5135 kg
The amount of water removed (Wr) = water incoming (Wi) - water outgoing (wo).
Wi = 10000 - 3800 = 6200 kg
Wo = 5135 - 3800 = 1335 kg
Wr = Wi - Wo
Wr = 6200 - 1335
Wr = 4865 kg
Therefore, the water removed is 4865 kg
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Carbon disulfide, a poisonous, flammable liquid, is an excellent solvent for phosphorus, sulfur, and some other nonmetals. A kinetic study of its gaseous decomposition reveals these data: Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s) 1 0.100 2.7 × 10−7 2 0.080 2.2 × 10−7 3 0.055 1.5 × 10−7 4 0.044 1.2 × 10−7 (a) Choose the rate law for the decomposition of CS2
Answer:
Rate law: = [tex]k[CS_2]^1[/tex]
Explanation:
Given:
t [tex][CS_2][/tex]
0.100 [tex]2.7 \times 10^{-7}[/tex]
0.080 [tex]2.2 \times 10^{-7}[/tex]
0.055 [tex]1.5\times 10^{-7}[/tex]
0.044 [tex]1.2\times 10^{-7}[/tex]
Rate law for the given reaction: [tex]k[CS_2]^n[/tex]
Where, n is the order of the reaction.
Divide rate 1 with rate 3
[tex]\frac{0.100}{0.055} =\frac{k[CS_2 (1)]^n}{k[CS_2 (3)]^n} \\\frac{0.100}{0.055} =\frac{k[2.7 \times 10^{-7}]^n}{k[1.5\times 10^{-7}]^n}\\1.81=[1.8]^n\\ n=1[/tex]
So, rate law = [tex]k[CS_2]^1[/tex]
Answer:
[tex]r=-3.73x10^{5}s^{-1} [CS_2][/tex]
Explanation:
Hello,
In this case, a linealization helps to choose the rate law for the decomposition of CS₂ as it is generalized via:
[tex]r=-k[CS_2]^{n}[/tex]
Whereas [tex]n[/tex] accounts for the order of reaction, which could be computed by linealizing the given data using the following procedure:
[tex]-ln(r)=ln(k[CS_2]^{n})\\-ln(r)=ln(k)+ln([CS_2]^{n})\\-ln(r)=ln(k)+n*ln([CS_2])[/tex]
Therefore, on the attached picture you will find the graph and the lineal equation wherein the slope is the order of the reaction and the y-axis intercept the natural logarithm of the rate constant. In such a way, the order of reaction is 1 and the rate constant is:
[tex]ln(k)=12.83\\k=exp(12.83)\\k=3.73x10^{5}s^{-1}[/tex]
Best regards.
Of the following bonds, which one is most polar? Select one: а. С-N b.C-C С. С-F d. C-H
Answer:
c. C-F.
Explanation:
Hello, to know the type of bond, you must subtract the electronegativities for each element forming the bond:
a. C-N=3.04-2.55=0.49
b. C-C=2.55-2.55=0
c. C-F=3.98-2.55=1.43
d. C-H=2.55-2.2=0.35
Thus, since for an electronegativity difference between 0.7 and 1.7, the bond is polar, the highest number (most polar), based on calculations is 1.43, that is for c. C-F.
Best regards.
The concentration of H30 in your swimming pool is 4 x 10^-5 M. What is the pH of your pool? O a.-4.4 Ob.4.4 Oc. 0.00004 O d.-0.6 O e.0.6
Answer:
b. 4.4
Explanation:
pH is related to the concentration of H₃O⁺ through the following equation:
pH = -log([H₃O⁺]) = -log(4 x 10⁻⁵)
pH = 4.4
why is it that when you use an expression for work you don't use the -VdP but when you use an expression for dH you use PdV and VdP in it?
Answer:
Work is given by -PdV.
It clearly means that when volume of the system changes, work is done either on the system or by the system depending upon volume change.
So, pressure change term -VdP is not included because it is indicating a constant volume.
But in case of enthalpy,
H = U + PV
So, for enthalpy change we can write
dH = dU + PdV + VdP
Because enthalpy change is the sum of all energy changes in the system that includes internal energy, P-V work and effect of pressure change.
Final answer:
The expression for work in thermodynamics is typically shown as -pdV to denote work done by the system, and is path-dependent, while the enthalpy change includes PdV and VdP terms to represent heat and work interactions at constant pressure, since enthalpy is a state function.
Explanation:
The question relates to why the expression for work in thermodynamics is given as work done by the system (-pdV) rather than the work done on the system, and why in the expression for enthalpy change (dH) both PdV and VdP terms appear. In thermodynamics, work done by a system during expansion is often expressed as W = PextΔV, where Pext is the external pressure and ΔV is the change in volume. The negative sign in -pdV signifies that the system does work on its surroundings, thus losing energy. This convention is such that energy leaving the system is negative, aligning with the conventional definition of heat flow being negative when it leaves the system.
Enthalpy (H) is a state function, which unlike work or heat, is not path-dependent. The differential change in enthalpy, dH, for a small, reversible process is given by dH = dU + PdV + VdP, involving both pressure-volume work and the work done on the system due to pressure changes at constant volume. This dH formula accurately describes the total heat and work interactions of a system at constant pressure. The inclusion of VdP in the expression accounts for the work involved when there is a change in pressure at constant volume.
It is also important to note that the exact expression for work can vary depending on the specific thermodynamic process, such as isothermal (PV=nRT) or adiabatic processes and can become complex when accounting for non-ideal behaviors of gases, hence the significance of defining a when and b to zero for an ideal gas equation.
The current global population is 7.7 billion people, and the WHO estimates the minimum water need per person per day is 15 L. An estimate of accessible freshwater available as groundwater, lakes, and rivers is 6.34 x 100 km². How long will this water sustain the current population? Does the annual terrestrial precipitation meet the drinking water needs? If not, where might additional drinking water come from. A liter of water is 1 x 10-km².
Answer:
a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population. ( t = 5.489 E-13 day )
b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population
Explanation:
Let P = 7.7 billion people = 7.7 E12 person
∴ water needed for the total population for one day:
⇒ water amount = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day
⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year
∴ freshwater available:
freshwater = 6.34 E2 Km² H2O
how long will this water sustain the current population?
⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day
this amount of freshwater does not even meet the need of the current global population.
∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature
⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )
⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year
⇒ Py > water amount
the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.
A chemical process requires 7.300 g-mol of calcium sulfate (CaSO.) for a reaction. How many grams are in 7.300 g-mol of calcium sulfate? Caso
Answer: The mass of copper sulfate is 1165 grams.
Explanation:
We are given:
Number of gram-moles of copper sulfate = 7.300 g-mol
We know that:
1 g-mol = 1 mol
So, number of moles of copper sulfate = 7.300 mol
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of copper sulfate = 159.6 g/mol
Moles of copper sulfate = 7.300 moles
Putting values in above equation, we get:
[tex]7.300mol=\frac{\text{Mass of copper sulfate}}{159.6g/mol}\\\\\text{Mass of copper sulfate}=1165g[/tex]
Hence, the mass of copper sulfate is 1165 grams.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
The work done and heat absorbed are both -8,1 kJ
Explanation:
The work done in an isobaric process is defined as:
W = -P (Vf - Vi)
Where P is pressure ( 10 atm)
Vf = 10 L
Vi = 2 L
Thus, W = -80 atm×L ≡ -8,1 kJ
This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.
I hope it helps!
Draw a four (4) carbon hydrocarbon (contains only Carbon and Hydrogen) with one carbon-carbon bond having a bond angle of 120°.
Explanation:
The hybridization of the carbon in which it forms 120° with the other atoms is [tex]sp^2[/tex].
In [tex]sp^2[/tex] hybridized carbon, it forms a geometry of trigonal planar which has angle of 120°. This corresponds to doubly bonded carbon.
To draw:
One carbon-carbon double bond in 4 membered carbon chain.
The structure is shown in image below.
Liquid n-pentane and liquid n-octane are mixed to form a stream flowing at a rate of 1,496.5 lbm/hr. An in-line density measurement determines the mixture density is 0.654 g/mL. Calculate the volumetric flow rate of the liquid n-pentane feed in cm3/sec. The specific gravities of n-pentane and n-octane are 0.631 and 0.708, respectively. Assume ideal mixing.
Answer:
202.20 cm^3/s
Explanation:
Hello,
I'm sending a photo showing the solution for this exercise.
The solution of the 2x2 system could be achieved by using a calculator or any available algebraic method.
Best regards.
Using the steam tables, determine the enthalpy change for 1lb of water:
Condensing a saturated vapor at 212 Fahrenheit and 1 atm abs.
Answer:
change in enthalpy for 1 lb water is - 970.10872 Btu
Explanation:
Given data:
weight of water 1 lb
pressure 1 atm
temperature 212 Fahrenheit
From steam table
initial enthalpy [tex]h_1 = 1150.288 Btu/lb[/tex]
final enthalpy[tex] h_2 = 180.18 Btu/lb[/tex]
change in enthalpy is given as
[tex]\Delta h = m (h_2 -h_1)[/tex]
[tex]\Delta h = 1\times ( 180.18020 - 1150.288)[/tex]
[tex]\Delta h = - 970.10872 Btu[/tex]
change in enthalpy for 1 lb water is - 970.10872 Btu
How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of the solution is reduced to 2.5 i) Less AgCH3CO0(s) will dissolve if AgNO; (s) is added to the aqueous solution iii) More AgCH3CoO(s) will dissolve if NaOH (aq) is added to the aqueous solution a) 0 b) c) 2 d) 3
Answer:
All three statements are true
Explanation:
Solubility equilibrium of silver acetate:
[tex]AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}[/tex]
If pH is increased then concentration of [tex]H^{+}[/tex] increases in solution resulting removal of [tex]CH_{3}COO^{-}[/tex] by forming [tex]CH_{3}COOH[/tex]. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more [tex]AgCH_{3}COO[/tex] will dissolveIf [tex]AgNO_{3}[/tex] is added then concentration of [tex]Ag^{+}[/tex] increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less [tex]AgCH_{3}COO[/tex] will dissolveInsoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of [tex]Ag^{+}[/tex]. So, more [tex]AgCH_{3}COO[/tex] will dissolveHence all three statements are true
In the titration of 50.0 mL of HCl of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.5 mL of 0.130 M Ca(OH)2 is added. Show the calculation of the molarity of the HCI.
Answer:
The molarity of HCl is 0.138 M
Explanation:
The titration reaction is as follows:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:
At the endpoint:
moles Ca(OH)₂ = moles HCl / 2
Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:
mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂
mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂
The number of moles of HCl will be:
mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl
This number of moles was present in 50.0 ml, then, in 1000 ml:
mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol
Then:
Molarity HCl = 0.138 M
what are the characteristics of waste water from paper and pulp industry?
Explanation:
The pulp and the paper mill industry is intensive consumer of the water and the natural resources like wood and on return, discharging an ample variety of the gaseous, liquid and solid wastes to environment.
The wastewater which is being eliminated from pulp and paper industry usually contains high and tremendous levels of concentrations of the biochemical oxygen demand and the chemical oxygen demand and thus, shows high level of toxicity and a strong and deep black-brown color.
Explain the difference between a physical and a chemical property. Give at least 2 examples of each type of property. Chemistry seeks to explain both the physical and chemical properties of matter.
Answer:
Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed. [tex]H_2+\frac{1}{2}O_2\rightarrow H_2O[/tex]
Example: 1. reactivity of substances
2. Toxicity of substances
Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc. [tex]H_2O(s)\rightarrow H_2O(l)[/tex]
Example: 1. Melting Point
2. Density
Final answer:
Physical properties describe the characteristics of matter that can be observed or measured without changing its identity, while chemical properties describe how matter changes its chemical structure or composition.
Explanation:
Physical properties are characteristics that describe matter, such as size, shape, color, and mass, that can be observed or measured without changing the identity of the matter. Examples of physical properties include the melting point of ice and the boiling point of water. Chemical properties, on the other hand, describe how matter changes its chemical structure or composition. Examples of chemical properties include flammability, reactivity, and the ability to undergo oxidation. For instance, the ability of gasoline to burn is a chemical property.
According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams of chlorine gas? 3 C12(g) + 3 H2O(1) ► 5 HCl(aq) + HCO3(aq) grams HC103
Answer:
[tex]m_{HClO_3}=12.7gHClO_3[/tex]
Explanation:
Hello,
Considering the reaction:
[tex]3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3[/tex]
The molar masses of chlorine and chloric acid are:
[tex]M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol[/tex]
Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:
[tex]m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3[/tex]
Best regards.
The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish-brown. 2NOBr(g) → 2NO(g) + Br2(g) Use the data below to make the following determinations: (a) the average rate of decomposition of NOBr over the entire experiment. (b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds. Time (s) [NOBr] (mol/L) 0.00 0.0100 2.00 0.0071 4.00 0.0055 6.00 0.0045 8.00 0.0038 10.00 0.0033 The rates of decomposition of NOBr are
The average rate of decomposition of NOBr over the entire experiment is -6.7 x 10^(-4) mol/(L*s), and the average rate of decomposition between 2.00 and 4.00 seconds is -8.0 x 10^(-4) mol/(L*s).
Explanation:The average rate of decomposition of NOBr over the entire experiment can be determined by calculating the change in concentration of NOBr over the change in time. From the given data, the initial concentration of NOBr is 0.0100 mol/L and the final concentration is 0.0033 mol/L after 10 seconds. Therefore, the average rate of decomposition is (0.0033 mol/L - 0.0100 mol/L) / (10 s - 0 s) = -6.7 x 10^(-4) mol/(L×s).
The average rate of decomposition of NOBr between 2.00 and 4.00 seconds can be calculated using the same method. The initial concentration at t=2.00 s is 0.0071 mol/L and the concentration at t=4.00 s is 0.0055 mol/L. Therefore, the average rate of decomposition between 2.00 and 4.00 seconds is (0.0055 mol/L - 0.0071 mol/L) / (4.00 s - 2.00 s) = -8.0 x 10^(-4) mol/(L×s).
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The average rate of decomposition of NOBr over the entire experiment is 0.00067 M/s, while the average rate between 2.00 and 4.00 seconds is 0.0008 M/s.
(a) the average rate of decomposition of NOBr over the entire experiment
We need to calculate the average rate of decomposition of NOBr over the entire given time period. The formula to calculate the average rate of decomposition is:
Rate = - Δ[NOBr] / Δt
Where Δ[NOBr] is the change in concentration and Δt is the change in time.
Initial concentration, [NOBr]0 = 0.0100 M
Final concentration, [NOBr] = 0.0033 M
Initial time = 0.00 s
Final time = 10.00 s
Δ[NOBr] = [NOBr] - [NOBr]0 = 0.0033 M - 0.0100 M = -0.0067 M
Δt = 10.00 s - 0.00 s = 10.00 s
Average Rate = - (-0.0067 M) / (10.00 s) = 0.00067 M/s
(b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds
Initial concentration, [NOBr]2.00s = 0.0071 M
Final concentration, [NOBr]4.00s = 0.0055 M
Initial time = 2.00 s
Final time = 4.00 s
Δ[NOBr] = [NOBr]4.00s - [NOBr]2.00s = 0.0055 M - 0.0071 M = -0.0016 M
Δt = 4.00 s - 2.00 s = 2.00 s
Average Rate = - (-0.0016 M) / (2.00 s) = 0.0008 M/s
The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.190 Ms−1 at 150 ∘C. The rate of reaction is ________ Ms−1.
(A) 0.0361
(B) 0.0950
(C) 0.0860
(D) 2.63 0.380
Answer:
Rate of reaction is [tex]0.0950M.s^{-1}[/tex]
Explanation:
Applying law of mass action for this reaction: [tex]rate=-\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [H_{2}]}{\Delta t}=\frac{\Delta [Br_{2}]}{\Delta t}[/tex][tex]-\frac{\Delta [HBr]}{\Delta t}[/tex] represents rate of disappearance of HBr, [tex]\frac{\Delta [H_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]H_{2}[/tex] and [tex]\frac{\Delta [Br_{2}]}{\Delta t}[/tex] represents rate of appearance of [tex]Br_{2}[/tex]Here, [tex]-\frac{\Delta [HBr]}{\Delta t}=0.190M.s^{-1}[/tex]So, rate of reaction = [tex]\frac{1}{2}\times (0.190M.s^{-1})=0.0950M.s^{-1}[/tex]A chemist must prepare 400 mL of 1.00M aqueous aluminum sulfate working solution. He'll do this by pouring out 1.82 mol/L aqueous aluminum sulfate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the aluminum sulfate stock solution should the chemist pour out?
To prepare the 1.00M aqueous aluminum sulfate working solution, the chemist should pour out approximately 219.78 mL of the aluminum sulfate stock solution.
Explanation:To prepare a 1.00M aqueous aluminum sulfate working solution, the chemist needs to dilute the 1.82 mol/L aluminum sulfate stock solution. The final volume required is 400 mL.
Using the formula for dilution, (stock concentration) × (stock volume) = (final concentration) × (final volume),
we can rearrange the formula to solve for the stock volume:
Stock volume = (final concentration) × (final volume) / (stock concentration)
Substituting the given values,
Stock volume = (1.00 M) × (400 mL) / (1.82 mol/L) = 219.78 mL
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The chemist should pour out 220 mL of the aluminum sulfate stock solution.
To prepare the desired 400 mL of 1.00M aluminum sulfate solution, the chemist needs to use the concept of dilution, which is described by the equation:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
-[tex]\( C_1 \)[/tex] is the concentration of the stock solution (1.82 mol/L),
- [tex]\( V_1 \)[/tex] is the volume of the stock solution needed (unknown),
- [tex]\( C_2 \)[/tex] is the concentration of the working solution (1.00M),
-[tex]\( V_2 \)[/tex] is the volume of the working solution (400 mL).
First, we need to ensure that all volumes are in the same unit, so we'll convert 400 mL to liters:
[tex]\[ 400 \text{ mL} = 0.400 \text{ L} \][/tex]
Now we can plug the known values into the dilution equation:
[tex]\[ (1.82 \text{ mol/L})V_1 = (1.00 \text{ mol/L})(0.400 \text{ L}) \][/tex]
Solving for [tex]\( V_1 \):[/tex]
[tex]\[ V_1 = \frac{(1.00 \text{ mol/L})(0.400 \text{ L})}{1.82 \text{ mol/L}} \][/tex]
[tex]\[ V_1 = \frac{0.400}{1.82} \text{ L} \][/tex]
[tex]\[ V_1 \approx 0.220 \text{ L} \][/tex]
Converting liters back to milliliters:
[tex]\[ 0.220 \text{ L} = 220 \text{ mL} \][/tex]
Therefore, the chemist should pour out 220 mL of the 1.82 mol/L aluminum sulfate stock solution to make 400 mL of a 1.00M solution.
Convert the following temperatures to Kelvin:
A. 100°C
B. 600°R
C. 98°F
Convert the following temperatures to °C:
D. 77.4°F
E. 77.4 K
F. 77.4°R
Explanation:
A. 100°C to Kelvins
[tex]T(K)=T(^oC)+273.15[/tex]
[tex]T(K)=100(^oC)+273.15=373.15 K[/tex]
B 600°R to Kelvins
[tex](T)^oK=((T)^oR)\times 1.8[/tex]
[tex](T)^oK=600\times 1.8 K = 1080 K[/tex]
C. 98°F to Kelvins
[tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]
[tex](T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K[/tex]
D. 77.4°F to degree Celsius
[tex]((T)^oC)=((T)^oF-32)\times \frac{5}{9}[/tex]
[tex](T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC[/tex]
E. 77.4 K to degree Celsius
[tex]T(^oC)=T(^K)-273.15[/tex]
[tex]T(^oC)=77.4(K)-273.15=-195.75^oC[/tex]
F. 77.4°R to degree Celsius
[tex](T)^oC=((T)^oR-491.67)\times \frac{5}{9}[/tex]
[tex](T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC[/tex]
What is the percent s character in an sp? hybridized orbital? 25% 33% 50% 67% 75%
Answer:
50%
Explanation:
The percent s - character , in a particular hybridization can be calculated from the following formula , i.e. ,
% s character = 1 / ( 1 + x ) * 100
Where , x = number of p orbitals .
Hence , from the question ,
The % s character in sp hybrid orbital is calculated as -
% s character = 1 / ( 1 + x ) * 100
x = 1 ( since , one p orbital is present in the sp hybridization )
% s character = 1 / ( 1 + 1 ) * 100
% s character = 1/ 2 * 100
% s character = 50 %.
Hence , the % s character in sp hybrid orbital = 50 % .
In sp hybridization, one s and one p atomic orbital combine to form two new hybrid orbitals. Each of these sp hybrid orbitals has 50% s character and 50% p character, so the percent s character in a sp hybridized orbital is 50%. This kind of hybridization is fundamental to understanding molecular geometry in chemistry.
Explanation:The term sp hybridization refers to the mathematical combination of one s and one p atomic orbital to form two new hybrid orbitals. Each hybrid orbital in a sp configuration contains 50% s character and 50% p character, therefore, the percent s character in a sp hybridized orbital is 50%. This orbital hybridization concept is a fundamental aspect of understanding molecular geometry in chemistry.
As an example, consider carbon. In its ground state, carbon has two unpaired electrons in separate 2p orbitals. If one of its 2s electrons is excited to a 2p orbital, the atom can then hybridize these three orbitals (one 2s orbital and two 2p orbitals) to form an sp hybridized state. The result of this is two equivalent sp orbitals arranged linearly at 180° to each other. This sp hybridization in carbon is seen in molecules like carbon dioxide (CO2).
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If a sodium chloride (NaCl) solution in water at 40°C has reached a concentration 50% calculate the quantity of NaCl crystals that will form once crystallization has been started in per 100 g water. The solubility of NaCl at 40°C is 36.6 g/100 g water.
Answer:
13.4g of NaCl will crystallize per 100g of water at 40°C.
Explanation:
Solubility stands for the maximum amount of solute that may stay dissolved per 100 g of water at a specific temperature. A solution with this concentration is known as a saturated solution. At 40°C, every 100 g of water, 36.6 g of NaCl can remain dissolved. If more solute is added to the solution the excess will crystallize and precipitate. Since prior to criystallization there are 50 g of NaCl dissolved, what will precipitate will be:
50g - 36.6g = 13.4g
Once crystallization is initiated in a 50% sodium chloride solution at 40°C, 13.4 g of NaCl crystals will form per 100 g of water. This is calculated by subtracting the solubility limit at 40°C (36.6 g) from the amount initially present in the solution (50 g).
If a sodium chloride (NaCl) solution in water at 40°C has a concentration of 50%, for every 100 g of water, there is 50 g of NaCl dissolved. Since the solubility of NaCl at 40°C is 36.6 g per 100 g of water, the solution is supersaturated because it contains more solute than can remain in solution at that temperature. When crystallization starts, the excess NaCl above its solubility limit will precipitate out.
To calculate the quantity of NaCl crystals that will form, we subtract the maximum solubility at 40°C from the total amount of NaCl currently dissolved:
Amount of NaCl present initially = 50 g
Maximum solubility at 40°C = 36.6 g
NaCl crystals that will form = Amount of NaCl present initially - Maximum solubility at 40°C
NaCl crystals that will form = 50 g - 36.6 g = 13.4 g
Therefore, 13.4 g of NaCl crystals will form per 100 g of water once crystallization has been started.
A turbine operates on superheated steam that enters at 80 bar absolute and 500˚C, and leaves the turbine at 7.5 bar absolute and 250˚C. The flow rate of the steam is 400 kg/min and the turbine operates adiabatically. The steam exiting the turbine goes to a heater, which is supplied with 2000
2kW of energy. The steam leaves the heater at 5 bar absolute. All of the pipes through which the
steam flows are 0.5 m (id) in diameter.
a. How much power is produced by the turbine? (4 marks)
b. What is the temperature of the steam leaving the heater? (4 marks)
c. What is the velocity (m/s) of the steam entering the turbine? (3 marks)
d. To what temperature can the steam at 80 bar be cooled before a liquid will appear?
The reaction is as follows: CH4 + 202 + CO2 + 2H2O The enthalpy of reaction between methane ( 2 ) and oxygen ( 2 ) is exothermic (-882 kJ/mol) with respect to the dissipation of methane. It is of interest to generate 126799 kJ of energy from the combustion. How many kg of methane will need to be combusted? Molecular weight: C-12 kg/kmol H-1 kg/kmol 0 - 16 kg/kmol Select one: O a. 2.300 O b. 2300.209 O c. 1.725 O d. 0.009
Answer:
2300j
Explanation:
How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is Ba2+ + SO42- → BaSO4(s)? x mL
Answer:
1.7927 mL
Explanation:
The mass of solid taken = 4.75 g
This solid contains 21.6 wt% [tex]Ba(NO_3)_2[/tex], thus,
Mass of [tex]Ba(NO_3)_2[/tex] = [tex]\frac {21.6}{100}\times 4.75\ g[/tex] = 1.026 g
Molar mass of [tex]Ba(NO_3)_2[/tex] = 261.337 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.026\ g}{261.337\ g/mol}[/tex]
[tex]Moles= 0.003926\ mol[/tex]
Considering the reaction as:
[tex]Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3[/tex]
1 moles of [tex]Ba(NO_3)_2[/tex] react with 1 mole of [tex]H_2SO_4[/tex]
Thus,
0.003926 mole of [tex]Ba(NO_3)_2[/tex] react with 0.003926 mole of [tex]H_2SO_4[/tex]
Moles of [tex]H_2SO_4[/tex] = 0.003926 mole
Also, considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Molarity = 2.19 M
So,
[tex]2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}[/tex]
Volume = 0.0017927 L
Also, 1 L = 1000 mL
So, volume = 1.7927 mL
As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.05. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)
Answer:
We will need 147.772 mL of KH2PO4 to make this solution
Explanation:
For this case we can give the following equation:
H2PO4 - ⇄ H+ + HPO42-
With following pH- equation:
pH = pKa + log [HPO42-]/[H2PO4-]
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
-0.16 = log [HPO42-]/[H2PO4-]
10^-0.16 = [HPO42-]/[H2PO4-]
0.6918 = [HPO42-]/[H2PO4-]
Let's say the volume of HPO42-= x then the volume of H2PO4- will be 250 mL - x
Since both have a concentration of 1M = 1 mol /L
If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]
0.6918 = x / (250 - x)
0.6918*250 - 0.6918x = x
172.95 = 1.6918x
x = 102.228 mL
The volume of HPO42- = 102.228 mL
Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL
To control this we can plug this in the pH equation
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
7.05= 7.21 + log (102.228 / 147.772) = 7.05
We will need 147.772 mL of KH2PO4 to make this solution
To produce the buffer solution, use the Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid, ensuring that the concentration of KH2PO4 is higher than K2HPO4, and calculate the necessary volumes of 1.00 M stock solutions.
Explanation:To prepare a potassium dihydrogen phosphate buffer solution with a pH of 7.05, when the pKa of H2PO4− is 7.21, the Henderson-Hasselbalch equation can be used, which is pH = pKa + log([A−]/[HA]), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. Since the desired pH is slightly less than the pKa, you'll need a bit more acid (KH2PO4) than the base (K2HPO4). To find the exact amounts, set up the equation like this: 7.05 = 7.21 + log([K2HPO4]/[KH2PO4]). Solving for the ratio [K2HPO4]/[KH2PO4], we get a value less than 1, which indicates that the concentration of KH2PO4 must be higher than that of K2HPO4 in our buffer solution. Then, you can calculate the volume of each stock solution needed to reach the final total volume of 250 mL, remembering the molarities are both 1.00 M and that volumes are additive.
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How does a molecular formula differ from a structural formula?
Explanation:
Molecular formulas show correct and accurate number of each type of the atoms which are present in molecule.
On the other hand, structural formulas show arrangement of atoms and covalent bonds between them.
For example,
The molecular formula for carbon dioxide is [tex]CO_2[/tex]
The structural formula is O = C = O
A molecular formula gives the simplest whole number ratio of atoms in a compound, while a structural formula shows the actual arrangement of atoms and bonding in a molecule.
Explanation:A molecular formula is a way to represent a compound using the symbols of the elements present and the number of atoms of each element in the compound. It gives the simplest whole number ratio of atoms in the compound. For example, the molecular formula of glucose is C6H12O6, indicating that it contains 6 carbon, 12 hydrogen, and 6 oxygen atoms.
A structural formula shows how the atoms in a molecule are connected to each other. It represents the actual arrangement of the atoms in the compound, including the bonding between them. For example, the structural formula of glucose shows the specific arrangement of carbon, hydrogen, and oxygen atoms in the molecule.
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How many grams of disulfur decafluoride are present in 2.15 moles of this compound? grams
Answer
Take account the molar mass of this compound (first of all u should know the formula, S2F10) so when u have the molar mass, you can get how many grams of it, are present in 2.15 moles. The right mass is 546,3 g.
Explanation:
Molar mass in the S2F10 should be, molar mass in S (32.06 x2) + molar mass of F (18.998 x 10) = 254,1 g/m
A voltaic electrochemical cell consists of a copper electrode in a Cu2SO4(aq) solution, and a palladium electrode in a PdSO4(aq) solution at 25°C. The salt bridge consists of a solution of KCl(aq).
What is the concentration of the Cu+if the concentration of the PdSO4 is 0.498 M and the measured cell potential is 0.447 V?
Given: Cu+(aq) + e- ↔ Cu(s) E°=+0.521 V
and Pd2+(aq) + 2e- ↔ Pd(s) E°=+0.987 V
Answer:
[tex]\large \boxed{\textbf{1.48 mol/L}}[/tex]
Explanation:
We must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF}lnQ[/tex]
1. Calculate E°
Anode: Pd²⁺ (0.498 mol·L⁻¹) + 2e⁻ ⇌ Pd; E° = +0.987 V
Cathode: Cu ⇌ Cu⁺ (x mol·L⁻¹) + e⁻; E°= - 0.521 V
Overall: Pd²⁺(0.498 mol·L⁻¹) + 2Cu ⟶ Pd + 2Cu⁺ (x mol·L⁻¹); E° = 0.466 V
2. Calculate Q
[tex]\begin{array}{rcl}0.447 & = & 0.466 - \dfrac{8.314\times 298}{2 \times 96 485} \ln Q\\\\-0.019& = & -0.01284 \ln Q\\\ln Q & = & 1.480\\Q & = & e^{1.480}\\ & = & 4.392\\\end{array}[/tex]
3. Calculate [Cu⁺]
[tex]\begin{array}{rcl}Q & = & \dfrac{\text{[Cu$^{+}$]}^{2}}{\text{[Pd]}}\\\\4.392 & = & \dfrac{{x}^{2}}{0.498}\\\\x^{2}& = & 2.187\\x & = & 1.48\\\end{array}\\\text{The concentration of Cu$^{+}$ is $\large \boxed{\textbf{1.48 mol/L}}$}[/tex]
Aluminum chloride, AICh is an inexpensive reagent used in many industrial processes. It is made by treating scrap aluminum with chlorine according to the following equation. 2 Aln + 3 Cla) → 2 AlCl3() a) If 13,49 g of Al and 35.45 g of Cl2 are allowed to react, how much AlCl is produced? b) How many grams of the excess reactant is left?!
Answer:
a) 44.442 grams of aluminium chloride is produced.
b) 8.996 grams of the excess reactant that id aluminum is left unreacted.
Explanation:
[tex]2Al(s) + 3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
Moles of aluminum = [tex]\frac{13.49 g}{27 g/mol}=0.4996 mol[/tex]
Moles of chlorine gas = [tex]\frac{35.45}{35.5 g/mol}=0.4993 mol[/tex]
According to reaction, 1 mol aluminum reacts with 3 moles of chlorine gas.
Then 0.4996 moles of aluminum will react with:
[tex]\frac{3}{1}\times 0.4996 mol=1.4988 mol[/tex] of chlorine gas
Then 0.4993 moles of chlorine gas will react with:
[tex]\frac{1}{3}\times 0.4993 mol=0.1664 mol[/tex] of aluminum
As we can see that moles of an aluminium are in excess. Hence, aluminium is an excess reagent.
So moles of aluminum chloride will depend upon moles of chlorine gas.
According to reaction 3 moles of chlorine gas give 2 moles of aluminium chloride.
Then 0.4993 mole of chlorine gas will give:
[tex]\frac{2}{3}\times 0.4993 mol=0.3329 mol[/tex] of aluminium chloride .
Mass of 0.3329 moles of aluminium chloride :
=0.3329 mol × 133.5 g/mol= 44.442 g
44.442 grams of aluminium chloride is produced.
Moles of aluminium left unreacted = 0.4996 mol - 0.1664 mol = 0.3332 mol
Mass of 0.3332 moles of aluminum :
0.3332 mol × 27 g/mol = 8.996 g[/tex]
8.996 grams of the excess reactant that id aluminum is left unreacted.
Using stoichiometry, we calculate how much aluminum chloride is formed from given masses of aluminum and chlorine and determine the excess reactant remaining. This requires finding the limiting reactant, using the balanced chemical equation, and converting between moles and grams.
Explanation:To answer the question about how much aluminum chloride is produced when 13.49 g of Al and 35.45 g of Cl2 react, we need to use stoichiometry, which is the method of quantitatively relating the products and reactants in a chemical reaction. Here is the step-by-step explanation:
Determine the molar masses of the reactants and product (Al = 26.98 g/mol, Cl2 = 70.90 g/mol, AlCl3 = 133.34 g/mol).Calculate the number of moles of each reactant (n = mass/molar mass)Identify the limiting reactant by comparing the stoichiometric ratios from the balanced equation with the number of moles of each reactantUse the molar ratio from the balanced equation to find out how many moles of product can be formed by the limiting reactant.Convert the moles of AlCl3 back into grams using the molar mass of AlCl3 (AlCl3's mass = moles x molar mass).To find the excess reactant remaining:
Subtract the amount of limiting reactant that reacted, from the initial amount, to find the amount of excess reactant.
The value of the equilibrium constant for the following chemical equation is Kr 40 at 25°C. Calculate the solubility of Al(OH)s(s) in an aqueous solution buffered at pH 11.00 at 25°C. Al(OH)(s)OH (aq)Al(OH)l'(aq) + a) 4.0 x 1010 M b) 6.3 M d) 0.37 M c) 0.040 M
Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]
The concentration of OH⁻ is:
pOH = 14 - pH = 3
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]
0,04M = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!