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.
What happens to the outlet temperatures of the hot and cold fluid when the both the hot fluid is at a maximum and the cold fluid is at a minimum flowrate?
a- Remains same as that of inlet temperature
b- Hot water temperature decreases and cold water temperature increase
c- Hot water remains almost same but cold water temperature increases drastically.
Answer:
c- Hot water remains almost same but cold water temperature increases drastically.
Explanation:
In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.
Where:
m= mass flowrate
Cp=specific-heat-capacity
dT= difference between the inlet and outlet temperatures
We can state the equality of this equation for both fluids:
(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) ) for hot medium = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) ) for cold medium
Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.
Be sure to answer all parts. Barium carbonate decomposes upon heating to barium oxide and carbon dioxide. Enter and balance the equation (including the physical states) Calculate the number of g of carbon dioxide produced by heating 71.0 g of barium carbonate. g CO2
Final answer:
The balanced equation for the decomposition of barium carbonate is BaCO3(s) → BaO(s) + CO2(g). Heating 71.0 g of barium carbonate will produce 16.06 g of carbon dioxide.
Explanation:
The balanced equation for the decomposition of barium carbonate (BaCO3) is:
BaCO3(s) → BaO(s) + CO2(g)
From the equation, it can be observed that one mole of barium carbonate produces one mole of carbon dioxide.
To calculate the number of grams of carbon dioxide produced by heating 71.0 g of barium carbonate, we need to use the molar mass of barium carbonate (197.34 g/mol). Since one mole of barium carbonate produces one mole of carbon dioxide, we can use the molar mass of carbon dioxide (44.01 g/mol) to calculate the mass of carbon dioxide produced.
71.0 g BaCO3 x (1 mol CO2/197.34 g BaCO3) x (44.01 g CO2/1 mol CO2) = 16.06 g CO2
Therefore, heating 71.0 g of barium carbonate will produce 16.06 g of carbon dioxide.
Calculate the osmotic pressure at 41 degrees C of a 0.078 M solution of glucose. Enter your answer using 2 decimal places!!!!
Answer:
Osmotic pressure is: 2,01 atm
Explanation:
Osmotic pressure is the minimum pressure that you needs to be apply to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. The formula of osmotic pressure is:
π = M×R×T
Where:
M is molar concentration of dissolved species (units of mol/L). 0,078M
R is ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹, ).
T is the temperature on the Kelvin scale. 41°C +273,15 = 314,15 K
Replacing you have:
π = 0,078M×0.08206 L atm mol⁻¹ K⁻¹×314,15 K
π = 2,01 atm
I hope it helps!
Hydrogen sulfide, H2S, has the characteristic odor of rotten eggs. If a sample of H2S gas at 755.0 torr and 25.0°C in a 2.00-L container is allowed to expand into a 13.0-L container at 25.0°C, what is the pressure in the 13.0-L container? Enter your answer in the provided box. atm
Answer:
0.153 atm or 116.1 torr
Explanation:
Hello, in the attached photo you will find the numerical procedure to obtain the pressure in the second state.
- Take into account that in the first part, the moles are needed based on the ideal gas equation, because they are subsequently used in the computation of the second state's pressure.
Best regards.
Using Boyle's Law, the new pressure of the gas when it's allowed to expand to a 13.0-L container is calculated to be 116.15 torr or 0.15 atm.
Explanation:The subject of this question involves the concept of gas laws, specifically Boyle's Law which states that at a constant temperature, the pressure and volume of a gas are inversely related. The initial pressure of the Hydrogen Sulfide gas is 755.0 torr and its initial volume is 2.00L. When the volume increases to 13.0L, the pressure will reduce accordingly.
To calculate the new pressure, the formula of Boyle's Law which is P1V1 = P2V2 can be used, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. Thus, P2 becomes (P1V1/V2) = (755.0 torr * 2.00L) / 13.0L = 116.15 torr.
However, as the question specifies that the answer is needed in atm, the result should be converted from torr to atm. As 1 atm is approximately equal to 760 torr, the final converted pressure is 116.15 torr / 760 torr/atm = 0.15 atm.
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Describe in detail how to prepare 500.0 mL of a 2.5 M calcium chloride solution. Calcium chloride is purchased as a solid. TO pipare weigh the deied wtight a 500.0 ml of a 2.3 M Calciom chlonde sdotion, frat
Answer:
In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the solute gets completely soluble add more water up till the mark of 500 ml.
Explanation:
Concentration of calcium chloride = 2.5 M
Volume of the solution = 500.0 ml = 0.5000 L
Moles of calcium chloride = n
[tex]c=\frac{n}{V(L)}[/tex]
n = moles of solute
c = concentration of solution
V = volume of the solution in L
[tex]2.5 M=\frac{n}{0.5000 L}[/tex]
[tex]n=2.5 M\times 0.5000 L = 1.2500 mol[/tex]
[tex]n=\frac{\text{Mass of calcium chloride}}{\text{Molar mass of calcium chloride}}[/tex]
[tex]1.2500 mol=\frac{\text{Mass of calcium chloride}}{111 g/mol}[/tex]
Mass of calcium chloride = 111 g/mol × 1.2500 mol = 138.75 g
In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the the solute gets completely soluble add more water up till the mark of 500 ml.
On a weekend trip to Winnipeg (Canada) you see that gasoline sells for 103.6 cents Cnd per liter. What is the cost of gasoline in US dollars per gallon? Should you fill up in Winnipeg or wait until you get back to Pembina? (a border city in the US - gas price is $2.64 USS/gal) Note: 1 dollar Cnd 0.754 dollars US b) If the legal speed limit in Canada is 100 km/hr, what is the limit in miles/hr and m/sec?
Answer:
a) gas price = 2.96 Dollar US/gal; we could wait until we return to Pembina.
b ) v = 100 Km/h = 62.1371 miles/h = 27.777 m/s
Explanation:
a) gas price = 103.6 cents Cnd / L * ( L / 0.264172 gal ) = 392.17 cents/gal
⇒ gas price = 392.17 cents Cnd/gal * ( dollar Cnd/100cents ) = 3.9217 dollar Cnd/gal
⇒ gas price = 3.9217 dollar Cnd/gal * ( 0.754 dollar US/dollar Cnd)
⇒ gas price = 2.96 dollars US/gal
the difference between the price of gas in Winnipeg and Pembina is 0.32 dollars US; we could wait until we return to Pembina to fill the tank.
b) velocity = 100 Km/h
⇒ 100 Km/h * ( 0.621371 miles/Km ) = 62.1371 miles/h
⇒ 100 Km/h * ( 1000 m/Km ) * ( h/3600s) = 27.777 m/s
The cost of gasoline in US dollars per gallon in Winnipeg is $5.20 US per gallon, which is higher than the gas price in Pembina ($2.64 US per gallon). Therefore, it is advisable to wait until you get back to Pembina to fill up your gas tank. The speed limit in Winnipeg is 62.14 miles per hour or 27.78 meters per second.
Explanation:To convert the cost of gasoline from Canadian dollars per liter to US dollars per gallon, we need to consider the exchange rate and volume conversion.
First, we need to convert Canadian cents to Canadian dollars. Since there are 100 cents in a dollar, the cost of gasoline in Canadian dollars is 103.6 cents / 100 = 1.036 Canadian dollars per liter.
To convert Canadian dollars per liter to US dollars per gallon, we need to consider the exchange rate. Since 1 Canadian dollar is equivalent to 0.754 US dollars, the cost of gasoline in US dollars per liter is 1.036 Canadian dollars / 0.754 US dollars = 1.373 US dollars per liter.
Since there are roughly 3.79 liters in 1 gallon, the cost of gasoline in US dollars per gallon is 1.373 US dollars per liter * 3.79 liters = 5.20 US dollars per gallon.
Comparing the cost of gasoline in Winnipeg (103.6 cents Cnd per liter) to Pembina ($2.64 US per gallon), it is cheaper to fill up in Winnipeg as the cost in US dollars per gallon is 5.20 US dollars per gallon, which is less than $2.64 US per gallon in Pembina.
To convert the speed limit from kilometers per hour to miles per hour, we can use the conversion factor 1 kilometer = 0.6214 miles. The speed limit in miles per hour is 100 km/hr * 0.6214 miles/km = 62.14 miles/hr.
To convert the speed limit from kilometers per hour to meters per second, we can use the conversion factor 1 kilometer = 1000 meters and 1 hour = 3600 seconds. The speed limit in meters per second is 100 km/hr * (1000 m/km) / (3600 s/hr) = 27.78 m/s.
Write 2,469,100 in Engineering Notation with 3 significant figures.
Answer: [tex]2.47\times 10^{6}[/tex]
Explanation:
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
Now converting the given value of 2,469,100 into engineering notation, we get [tex]2.47\times 10^{6}[/tex]
Hence, the correct answer is, [tex]2.47\times 10^{6}[/tex]
An FM radio transmitter broadcasts at 98.4 MHz with a power of
45 kW. How many photons does it generate per second?
To determine the number of photons emitted per second by a 45 kW FM radio transmitter at 98.4 MHz, the energy of one photon is calculated using Planck's constant and the frequency, and then the power output is divided by this energy. The result is approximately 6.90 × 1028 photons per second.
To calculate the number of photons emitted per second by an FM radio transmitter, we need to know the energy of one photon and the total energy emitted per second (power output of the transmitter).
The energy of a photon, E, can be calculated using the Planck's equation: E = h * f, where h is Planck's constant (6.626 × 10-34 J·s), and f is the frequency of the photon in hertz. In this case, the frequency (f) is 98.4 MHz, or 98.4 × 106 Hz.
The total power output in joules per second is given by the transmitter's power, which is 45 kW, or 45,000 J/s. By dividing the total energy emitted per second by the energy of one photon, we obtain the number of photons emitted per second.
Number of Photons per Second = Power Output (J/s) / Energy per Photon (J)
Now, let's perform the calculations:
Energy per Photon, E = (6.626 × 10-34 J·s) × (98.4 × 106 Hz) = 6.522 × 10-26 J
Photons per Second = 45,000 J/s / 6.522 × 10-26 J = 6.90 × 1028 photons/s
What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K
Answer: The de-Broglie's wavelength of a hydrogen molecule is [tex]3.26\AA[/tex]
Explanation:
Kinetic energy is the measure of temperature of the system.
The equation used to calculate kinetic energy of a particle follows:
[tex]E=\frac{3}{2}kT[/tex]
where,
E = kinetic energy of the particles = ?
k = Boltzmann constant = [tex]1.38\times 10^{-23}J/K[/tex]
T = temperature of the particle = 30 K
Putting values in above equation, we get:
[tex]E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J[/tex]
Calculating the mass of 1 molecule of hydrogen gas:Conversion factor used: 1 kg = 1000 g
1 mole of hydrogen gas has a mass of 2 grams or [tex]2\times 10^{-3}kg[/tex]
According to mole concept:
[tex]6.022\times 10^{23}[/tex] number of molecules occupy 1 mole of a gas.
As, [tex]6.022\times 10^{23}[/tex] number of hydrogen molecules has a mass of [tex]2\times 10^{-3}kg[/tex]
So, 1 molecule of hydrogen will have a mass of = [tex]\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg[/tex]
To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:[tex]\lambda=\frac{h}{\sqrt{2mE_k}}[/tex]
where,
[tex]\lambda[/tex] = De-Broglie's wavelength = ?
h = Planck's constant = [tex]6.624\times 10^{-34}Js[/tex]
m = mass of 1 hydrogen molecule = [tex]3.32\times 10^{-27}kg[/tex]
[tex]E_k[/tex] = kinetic energy of the particle = [tex]6.21\times 10^{-22}J[/tex]
Putting values in above equation, we get:
[tex]\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}[/tex]
[tex]\lambda=3.26\times 10^{-10}m=3.26\AA[/tex] (Conversion factor: [tex]1\AA=10^{-10}m[/tex] )
Hence, the de-Broglie's wavelength of a hydrogen molecule is [tex]3.26\AA[/tex]
Explain why a strong acid must have a weak conjugate base
Explanation:
The specie produced when a strong acid is dissociated in water along with the proton is the conjugate base of that acid.
Consider a strong acid such as HCl which donates the proton so readily as it is a strong base and there is no tendency for conjugate base, which is [tex]Cl^-[/tex] to re accept that proton.
A strong base is the one which accepts proton and holds it firmly. Thus, the strong acid has weak conjugate base.
A strong acid must have a weak conjugate base because once the strong acid donates its proton, the remaining base does not readily recapture it. This inverse relationship ensures that the acid can fully dissociate. For instance, HCl, a strong acid, has a weak conjugate base [tex]Cl^-[/tex].
A strong acid is characterized by its ability to donate protons easily. This means that once the acid has donated a proton, the resulting conjugate base is left behind. If this conjugate base were strong, it would readily attract and recapture the proton, preventing the acid from fully dissociating.
For instance, when HCl (a strong acid with a small [tex]pK_a[/tex] of -7) dissociates, it forms [tex]Cl^-[/tex] ions. The [tex]Cl^-[/tex] ions do not easily regain a proton, making them a weak base. Therefore, we can say that the conjugate base of a strong acid is invariably weak.
Understanding the Relationship with [tex]pK_a[/tex]
Acids with small or negative [tex]pK_a[/tex] values are strong acids. Their corresponding conjugate bases are weak bases because they do not have a strong tendency to grab protons back. This is why HCl, with a very low [tex]pK_a[/tex], has a conjugate base ([tex]Cl^-[/tex] ) that is very weak.
Acid-Base Equilibrium
An important concept in acid-base chemistry is that an acid-base reaction favors the formation of the weaker acid and base. This means in an equilibrium reaction, the side with the stronger acid and stronger base converts into the weaker acid and weaker base.
To summarize:
A strong acid has a weak conjugate base.The strength of an acid is inversely related to the strength of its conjugate base.Acid-base equilibria favor the formation of weaker acids and bases.A citrus dealer in Florida sells boxes of 100 oranges at a roadside stand. The boxes routinely are packed with one to three extra oranges to help ensure that customers are happy with their purchases. The average weight of an orange is 7.2 ounces, and the average weight of the boxes in which the oranges are packed is 3.2 pounds. Determine the total weight of five of these 100-orange boxes.
Answer:
[tex]\large \boxed{\text{250 lb}}[/tex]
Explanation:
1. Weight of oranges
An average box contains 102 oranges, so
Five boxes contain 510 oranges
[tex]\text{Weight of oranges} = \text{510 oranges} \times \dfrac{\text{7.2 oz}}{\text{1 orange}} = \text{3672 oz}\\\\\text{Weight} = \text{3672 oz} \times \dfrac{\text{1 lb}}{\text{16 oz}} = \text{230 lb}[/tex]
2. Weight of boxes
[tex]\text{Weight} = \text{5 boxes} \times \dfrac{\text{3.2 lb}}{\text{1 box}} = \text{16 lb}[/tex]
3. Total weight
[tex]\begin{array}{rcr}\text{Oranges} & = & \text{230 lb}\\\text{Boxes} & = & \text{16 lb}\\\text{TOTAL} & = & \textbf{250 lb}\\\end{array}\\\text{On average, the total weight is $\large \boxed{\textbf{250 lb}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the average weight of an orange.
What is the change in enthalpy when 4.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas? 250 2(g) + O2(g) ® 250 31 g): D Hº = 198 kJ A. -396 kJ B. 198 kJ C.-198 kJ D. 396 kJ E. 792 kJ
Answer: A. -396 kJ
Explanation:
The standard enthalpy of formation or standard heat 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.
[tex]2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)[/tex] [tex]\Delta H^0=198kJ[/tex]
Reversing the reaction, changes the sign of [tex]\Delta H[/tex]
[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)[/tex] [tex]\Delta H^0=-198kJ[/tex]
On multiplying the reaction by 2, enthalpy gets multiplied by 2:
[tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex] [tex]\Delta H=2\times -198kJ=-396kJ[/tex]
Thus the enthalpy change for the reaction [tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex] is -396 kJ.
The change in enthalpy when 4.00 mol of sulfur trioxide decomposes into sulfur dioxide and oxygen gas is 792 kJ.
Explanation:The question is asking for the change in enthalpy when 4.00 mol of sulfur trioxide decomposes into sulfur dioxide and oxygen gas. Looking at the provided data, the change in enthalpy (ΔHº) for the decomposition of one mole of sulfur trioxide is 198 kJ. Enthalpy is an extensive property, meaning it depends on the amount of substance involved. Therefore, if the reaction involves 4.00 mol of sulfur trioxide, the total change in enthalpy will be 4 times 198 kJ, which is 792 kJ. Hence, the correct answer is E. 792 kJ.
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If an object has a mass of 30.0 lbm and weighs 28.9 lbt, what is the local gravity where it is located? W = m*g
Explanation:
As it is given that W = [tex]m \times g[/tex] = 28.9 lbf
Mass is given as 30.0 lbm
Now, calculate [tex]g_{c}[/tex] as follows.
[tex]g_{c} = \frac{weight given}{mass}[/tex]
= [tex]\frac{28.9 lbf}{30.0 lbm}[/tex]
= 0.9633 lbf/lbm
Since, it is known that 1 lbf = lbm \times [tex]32.2 ft/s^{2}[/tex]. Therefore, local gravity (g') will be calculated as follows.
[tex]0.9633 \times 32.2 ft/s^{2}[/tex]
= [tex]31.02 ft/s^{2}[/tex]
Thus, we can conclude that the value of local gravity is [tex]31.02 ft/s^{2}[/tex].
What symbol is used to express the factor, 10-6? A) B) M C) n D) m
Answer:
The correct answer is that letter. (μ)
Explanation:
Hello!
According to the international system of units, we use prefixes to indicate the number of places we ran the point.
The letter μ indicates [tex]10^{-6}[/tex]
The correct answer is that letter. (μ)
It is not in the options but it is correct.
A. Define the terms 'oxidation' and 'reduction.
B. The oxidation number of S
inS2O32- is:
C. The oxidation number of S
inS4O62- is:
How were the assignations made?
Answer:
A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.
B. The oxidation number of S in S₂O₃²⁻ is +2.
C. The oxidation number of S in S₄O₆²⁻ is +5/2.
Explanation:
A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.
The overall charge on a molecule is equal to the sum of the oxidation states of all the atoms in a molecule.
B. S₂O₃²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.
Therefore, the oxidation state of sulfur (x) can be calculated by
2x + 3 (-2) = -2
2x + (-6) = -2
2x = -2 + 6 = 4
x = 4 ÷ 2= +2
Therefore, the oxidation number of S is +2.
C. S₄O₆²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.
Therefore, the oxidation state of sulfur (x) can be calculated by
4x + 6 (-2) = -2
4x + (-12) = -2
4x = -2 + 12 = 10
x = 10 ÷ 4 = +5/2
Therefore, the oxidation number of S is +5/2.
Two hundred kg of liquid contains 30% butane, 40% pentane, and the rest hexane (mass %) Determine: The mole fraction composition of the liquid The mass fraction composition on hexane free basis 1. 2.
Answer:
The mole fraction composition of the liquid is :
Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.
Explanation:
Mass of the liquid mixture = 200 g
Percentage of butane = 30%
Mass of butane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]
Moles of butane = [tex]n_1=\frac{60 g}{58 g/mol}=1.0345 mol[/tex]
Percentage of pentane= 40%
Mass of pentane= [tex]\frac{40}{100}\times 200 g=80 g[/tex]
Moles of pentane= [tex]n_2=\frac{80 g}{58 g/mol}=1.1111 mol[/tex]
Percentage of hexane = 100% - 30% - 40% = 30%
Mass of hexane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]
Moles of hexane = [tex]n_2=\frac{60 g}{86 g/mol}=0.6977 mol[/tex]
Mole fraction of butane, pentane and hexane : [tex]\chi_1, \chi_2 \& \chi_3[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908[/tex]
[tex]\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454[/tex]
The distance from Earth to the Moon is approximately 240.000 mi Review Constants Periodic Table Part D Earth s ound the Sun at an average speed of 29.783 km/s Convert this speed to miles per hour Express your answer using live significant figures VOAEG V v * un x-100
Answer:
66622.653 mi/hr
Explanation:
The average speed at which the Earth rotates around the Sun = 29.783 km/s
Also, The conversion of km to miles is shown below:
1 km = 0.621371 miles
The conversion of s to hr is shown below:
1 s = 1 / 3600 hr
So,
[tex]29.783\ km/s=\frac {29.783\times 0.621371\ miles}{\frac {1}{3600}\ hr}[/tex]
Thus,
The average speed at which the Earth rotates around the Sun in miles per hour = 66622.653 mi/hr
briefly explain the difference between an isotope and two species that are isoelectronic provide an example for an isotope and isoelectronic species to argon-40
Explanation:
Isotope are the species which have same atomic number (number of the protons) but number of neutrons different. which corresponds to different mass of isotopes.
For example,
[tex]^6_{12}C, ^6_{13}C, ^6_{14}C[/tex] are the three isotopes of carbon.
Isoelectronic are the species which have the same number of electrons.
For example,
Ne is isoelectronic with [tex]Na^+[/tex]
Isotopes of [tex]^{18}_{40}Ar[/tex] are:
[tex]^{18}_{36}Ar, ^{18}_{38}Ar[/tex]
The number of the electrons in [tex]^{18}_{40}Ar[/tex] = 18
Number of the electrons in [tex]K^+[/tex] = 18
Thus, [tex]K^+[/tex] is isoelectronic to Argon-40
Understand the Relationship Between the Structure of Alcohol and Their physical Properties Question Which of the following alcohols most easily forms hydrogen bonds? Select the correct answer below: O ethanol O 3-pentanol O2-butanol O2-propanol FEEDBACK MORE INSTRUCTION SUBMIT Content attribution
Answer:
Ethanol most easily forms hydrogen bonds.
Explanation:
The difference among the alcohols in this question is the size of carbonic chain and the position of the -OH group.
Ethanol has 2 carbons and the -OH group is terminal. The other alcohols have more carbons and the -OH group is not terminal. This means that the approximation of molecules will be facilitated for ethanol, and the interaction through hydrogen bons will be easier. However, for the other molecules, there will be steric hindrance, which will make it more difficult for the molecules to make hydrogen bonds.
The figure attached shows the alcohol structures.
What is the fluid pressure (in units of ܲܽPascals) a distance 3.2 meters below the surface of Lake Superior?
Answer:
fluid pressure = 132590 pascal
Explanation:
Fluid pressure can be calculated by using following relation:
[tex]P_o - P_g = g\time p\times (Z_b - Z_a )[/tex]
Where
Pb = fluid pressure at a distance of 3.2 m below from surface
Pa = fluid pressure at surface = atmospheric pressure = 101325 Pascal
g = gravitational constant = 9.8 m/sec2
[tex]\rho = density of fluid = 997 kg/m³[/tex]
Z = location depth from surface of lake
Therefore , [tex]Z_a = 0 meter[/tex]
[tex]Z_b = 3.2 meter[/tex]
[tex]P_b - P_a =g\times\rhop\times (Z_a - Z_b)[/tex]
[tex]= 9.8 \times 997\times (3.2-0)[/tex]
[tex]= 31265 kg/m-sec^2 [/tex]
1 Pascal = 1 Kg/(m.Sec)
[tex]P_b = P_a +31265[/tex]
= 101325 + 132590 Pascal
= 132590 pascal
Calculate the area of a square with side length 1 = 2.2 cm. Use standard decimal notation for answer input.
Answer: The area of square is [tex]4.84cm^2[/tex]
Explanation:
To calculate the area of square, we use the formula:
[tex]A=l^2[/tex]
where,
A = area of square = ?
l = edge length of square = 2.2 cm
Putting values in above equation, we get:
[tex]\text{Area of square}=(2.2)^2=4.84cm^2[/tex]
Hence, the area of square is [tex]4.84cm^2[/tex]
The vapor pressure of water is 28.3 mm Hg at 28 °C. What mass of water vapor, in mg, would be present in a vapor volume of 600. mL at 28°C? Selected Answer: B. 16.3 Correct Answer B. 16.3
Explanation:
The given data is as follows.
P = 28.3 mm = [tex]\frac{28.3}{760}[/tex], V = 600 mL = [tex]600 ml \frac{0.001 L}{1 ml}[/tex] = 0.6 L
R = 0.082 L atm/mol K, T = (28 + 273) K = 301 K
Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.
PV = nRT
[tex]\frac{28.3}{760} \times 0.6 L[/tex] = [tex]n \times 0.082 L atm/mol K \times 301 K[/tex]
n = [tex]9.03 \times 10^{-4}[/tex] mol
As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
[tex]9.03 \times 10^{-4}[/tex] mol = [tex]\frac{mass}{18 g/mol}[/tex]
mass = [tex]162.5 \times 10^{-4}g[/tex]
= [tex]163 \times 10^{-4}g[/tex] (approx)
Since, 1 g = 1000 mg. Therefore, [tex]163 \times 10^{-4}g[/tex] will be equal to [tex]163 \times 10^{-4}g \times \frac{10^{3}mg}{1 g}[/tex]
= 16.3 mg
Thus, we can conclude that mass of water vapor present is 16.3 mg.
To find the mass of water vapor in a given vapor volume at a certain temperature, you can use the ideal gas law. Convert the temperature from Celsius to Kelvin, calculate the number of moles using the ideal gas law equation, and then convert moles to grams using the molar mass of water. Finally, convert grams to milligrams. The mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.
Explanation:To find the mass of water vapor, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can rearrange the equation to solve for n, the number of moles. Once we have the moles, we can use the molar mass of water to convert it to mass in grams. Finally, we can convert grams to milligrams by multiplying by 1000.
Given:
Vapor pressure of water = 28.3 mm Hg
Temperature = 28 °C
Vapor volume = 600 mL
Converting the temperature to Kelvin:
T = 28 °C + 273.15 = 301.15 K
Using the ideal gas law equation to find the number of moles (n):
n = PV / RT
Substituting the values:
n = (28.3 mm Hg * 600 mL) / (62.36 L mm/mol K * 301.15 K)
n = 0.2829 mol
Using the molar mass of water (18.015 g/mol) to find the mass:
mass = n * molar mass
mass = 0.2829 mol * 18.015 g/mol
mass = 5.097 g
Converting grams to milligrams:
mass = 5.097 g * 1000 mg/g
mass = 5097 mg
Therefore, the mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.
Nitrogen monoxide, NO, reacts with bromine, Br2, to give nitrosyl bromide, NOBr. A sample of 0.0524 mol NO with 0.0262 mol Br2 gives an equilibrium mixture containing 0.0311 mol NOBr. What is the composition of the equilibrium mixture?
Answer:
The molar composition of the equilibrium mixture is
NO: 0.338 = 33.8%
Br2: 0.169 = 16.9%
NOBr: 0.493 = 49.3%
Explanation:
The reaction of nitrosyl bromide formation can be written as
[tex]NO+0.5\,Br_2\rightarrow NOBr[/tex]
To form 1 mol of NOBr, we need 1 mol of NO and 0.5 mol of Br2.
A sample of 0.0524 mol NO with 0.0262 mol Br2 gives an equilibrium mixture containing 0.0311 mol NOBr.
Then, to form 0.0311 mol NOBr, were needed 0.0311 mol NO and 0.01555 mol of Br2.
The amount of NO that stay in the same form is (0.0524-0.0311)=0.0213 mol NO.
The amount of Br2 that stay in the same form is (0.0262-0.01555)=0.01065 mol Br2.
The total amount of mol is (0.0213 mol NO + 0.01065 mol Br2 + 0.0311 mol NOBr) = 0.06305.
The molar composition is
NO: 0.0213/0.06305 = 0.338 = 33.8%
Br2: 0.01065/0.06305 = 0.169 = 16.9%
NOBr: 0.0311/0.06305 = 0.493 = 49.3%
The maximum pressure that can be developed for a certain fluid power cylinder is 6000 psi. Compute the force it can exert if the piston diameter is 2 inches.
Answer:
18840 lbf
Explanation:
Considering the physical equation:
[tex]F = PA[/tex]
It is possible to determine the force exerted by the piston if the maximum pressure developed ([tex]P_{max}[/tex]) and the area ([tex]A[/tex]) are previously known.
The area of the piston can be known through:
[tex]A = \pi *r^{2}[/tex]
Being [tex]r = \frac{d}{2} = 1~in[/tex]
Then
[tex]A = \pi * r^{2} = 3,14~in^{2}[/tex]
On the other hand, physical units of pressure in PSI are [tex]\frac{lbf}{in^{2}}[/tex], where the force is given in lbf.
Finally, the exerted force is:
[tex]F = 6000~PSI \cdot 3,14~in^{2} = 18840~lbf[/tex]
What is the sum of the numbers of neutrons and electrons in the ion 208Po Enter the value in the box as a whole number.
Answer:
Explanation:
Hello,
Polonium has the atomic number 83, thus, it has 83 protons and electrons; in such a way, the amount of neutrons is given by the subtraction:
[tex]neutrons=atomic.mass - atomic.number\\neutrons=208-83\\neutrons=125[/tex]
Now, the addition between the neutrons and electrons is the same atomic mass considering the previous formula, thus:
[tex]electrons+neutrons=83+125=208[/tex]
So this procedure could be avoided by just looking at the given information "208Po" which contains the result.
Best regards.
To calculate the sum of neutrons and electrons in the 208Po ion, presume it is a neutral atom due to the lack of charge information, which would result in 124 neutrons and 84 electrons. The sum of these particles is then 208.
Explanation:The question asks for the sum of the number of neutrons and electrons in the ion 208Po. To find the number of neutrons, we subtract the atomic number (number of protons) from the mass number. Polonium (Po) has an atomic number of 84, which is also the number of protons in a neutral atom of Polonium. The mass number of this isotope is 208, so the number of neutrons is 208 - 84 = 124. As for the number of electrons, since it is an ion and not neutral, we need to know the charge to determine this. However, as the charge is not provided in the question, we will assume that 208Po represents a neutral atom. Therefore, the number of electrons would also be 84. The sum of neutrons and electrons in the ion 208Po assuming it is uncharged is 124 + 84 = 208.
Based on the information given, classify each of the pure substances as elements or compounds, or indicate that no such classification is possible because of insufficient information. Analysis with an elaborate instrument indicates that substance A contains two elements. substance B and substance C react to give a new substance D. substance E decomposes upon heating to give substance F and substance G. heating substance H to 1000 ∘C causes no change. substance I cannot be broken down into simpler substances by chemical means. substance J cannot be broken down into simpler substances by physical means. heating substance K to 500 ∘C causes it to change from a solid to a liquid.
The classifications of Substance A: Compound, Substance B: Compound, Substance C: Compound, Substance D: Compound, Substance E: Compound, Substance F: Element or Compound (depends on further information), Substance G: Element or Compound (depends on further information), Substance H: Element (most likely), Substance I: Element, Substance J: Element and Substance K: Pure Element or Compound (depends on further information)
This classification is based on the definitions of elements and compounds in chemistry.
To classify the pure substances as elements or compounds, we can follow these steps based on the information provided:
Substance A: Contains two elements. Since it is a pure substance that can be separated into its components, it is classified as a compound.
Substance B and C: These substances react to form a new substance D. Since they undergo a chemical change to give a new substance, both B and C are also compounds.
Substance D: This is the new substance formed from the reaction of B and C and will also be classified as a compound because it is a product of the chemical reaction.
Substance E: Decomposes upon heating to produce substances F and G, indicating it is made of more than one element. Therefore, it is classified as a compound.
Substance F and G: These are the products resulting from the decomposition of substance E. Without additional information on whether they can be broken down into simpler substances, they are initially considered elements or compounds; however, since they are products of a compound's decomposition, if they consist of one type of atom, they could be elements.
Substance H: Heating it to 1000 °C causes no change, indicating it might be a pure element or possibly a stable compound that does not decompose at this temperature. Without more information, it's uncertain, but it is likely to be an element.
Substance I: Cannot be broken down into simpler substances by chemical means, which classifies it as a pure element.
Substance J: Cannot be broken down into simpler substances by physical means, indicating it is a pure element.
Substance K: Changes from a solid to a liquid upon heating to 500 °C. This phase change suggests that it is a pure substance, but without information, it could be either a pure element (e.g., a metal) or a compound (e.g., a salt).
An aerosol can with a temperature of 292 K and a pressure of 1.25 atm was left outside in the sun on a hot summer day. The pressure inside the can rose to 1.33 atm. What was the resulting temperature in the can?
322K
1,600K
311K
274K
ik its not A
Answer:
310.69K
Explanation:
Given parameters:
Initial temperature T₁ = 292K
Initial pressure P₁ = 1.25atm
Final pressure P₂ = 1.33atm
Unknown:
Final temperature T₂ = ?
Solution
To find the unkown, we need to apply the combined gas law. From the combined gas law, it can be deduced that at constant volume, the pressure of a give mass or mole of gas varies directly with the absolute temperature.
Since the same aerosol can is heated, the volume is constant.
[tex]\frac{P_{1} }{T_{1} }[/tex] = [tex]\frac{P_{2} }{T_{2} }[/tex]
Now, we have to make T₂ the subject of the formula:
T₂ = [tex]\frac{P_{2} x T_{1} }{P_{1} }[/tex]
T₂ = [tex]\frac{1.33 x 292 }{1.25 }[/tex] = 310.69K
The resulting temperature in the can is approximately 311 K.
To solve this problem, we can use Gay-Lussac's Law, which states that the pressure of a given mass of gas is directly proportional to its absolute temperature (in Kelvin) when the volume is kept constant. Mathematically, this can be expressed as:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Rearranging the equation to solve for [tex]\( T_2 \),[/tex] we get:
[tex]\[ T_2 = T_1 \times \frac{P_2}{P_1} \][/tex]
Substituting the given values:
[tex]\[ T_2 = 292 \text{ K} \times \frac{1.33 \text{ atm}}{1.25 \text{ atm}} \] \[ T_2 = 292 \text{ K} \times 1.064 \] \[ T_2 \approx 310.608 \text{ K} \][/tex]
Rounding to the nearest whole number, since temperature in Kelvin is typically expressed without decimals:
[tex]\[ T_2 \approx 311 \text{ K} \][/tex]
Blood alcohol concentration (BAC) is a measure of alcohol in your body, expressed as grams of alcohol per 100ml of blood. For a BAC of 0.05, every 100ml of your blood contains 0.05 grams of alcohol. Your BAC can be measured through your breath, blood, and urine. Drivers with an open licence have to keep under a BAC of 0.05.
Scientists estimate the mass of blood in a human body to be approximately 8 percent of body weight. If we assume that 17 % of the alcohol that a person drinks from beer goes into the bloodstream, how many litres of beer can an 71 kg person drink to get a BAC of 0.05? The alcohol content of beer is 6 % volume, and the blood density is 1025kg/m3, the alcohol density is 789 kg/m3.
Answer:
A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer
Explanation:
The mass of blood in a human body is approximately 8%, so if a person weighs 71 kg, the mass of blood would be:71 kg * 8/100 = 5.68 kg of blood.
Using blood density, we can calculate the volume that 5.68 kg of blood occupies:5.68 kg * [tex]\frac{1m^{3}}{1025kg}[/tex] = 0.0055415 m³
We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:
[tex]0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL[/tex]
Now we calculate the amount of alcohol that would be needed in the bloodstream to get a BAC of 0.05:[tex]\frac{5541.5mL}{100mL}*0.05g[/tex] = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05
The amount of alcohol that needs to be ingested is higher than 2.77 g, due to the fact that only 17% of the alcohol goes into the bloodstream, so:2.77 g [tex]*\frac{100}{17} =[/tex] 16.29 g of alcohol need to be ingested
Then we use the alcohol concentration of beer to calculate the volume of beer needed, using the alcohol density. First we convert the alcohol density to g/L, making sure the units that we want to convert cancel each other:[tex]789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L} =789g/L[/tex]
Now we use the density to calculate the litres of alcohol needed, keeping in mind that 16.29 g of alcohol are needed:[tex]16.29g*\frac{1L}{789g}=[/tex] 0.02065 L of alcohol are needed.
Finally we calculate the litres of beer needed, keeping in mind the concentration of alcohol in beer:[tex]0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =[/tex]0.3442 L of beer are needed.
To have a Blood Alcohol Concentration (BAC) of 0.05, a 71 kg person would need to consume approximately 271.57 liters of beer with a 6% alcohol content, assuming that 17% of the alcohol ends up in the bloodstream.
Explanation:First, let's calculate the total mass of blood in the body. The total weight of the body is 71 kg and blood comprises about 8 percent of body weight. Therefore, the mass of blood = 0.08 * 71 = 5.68 kg. Because the blood density is 1025 kg/m³, this implies that the volume of the blood is 5.68/1025 = 0.00554 m³, or 5.54 liters. When the BAC is 0.05, this means that there are 0.05 grams of alcohol in every 100 ml of blood, or 0.05 grams of alcohol per 0.1 liter of blood. Therefore, to have a BAC of 0.05 in 5.54 liters, there needs to be 0.05 * 5.54 * 10 = 2.77 grams of alcohol in the blood. We know that 17 percent of the alcohol that a person drinks goes into their bloodstream. Therefore, the total mass of beer consumed to get this amount of alcohol is = 2.77 / 0.17 = 16.294 grams of pure alcohol. Beer has an alcohol equivalent to 6 % of its volume, so we find that the total volume of beer consumed = 16.294 / 0.06 = 271.57 liters. Because alcohol density is 789 kg/m³, this would be approximately 0.271 m^3 or 271.57 liters of beer.
Learn more about Blood Alcohol Concentration here:https://brainly.com/question/3166778
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how does erucamide make substances scratch resistant?
Answer:
Explained
Explanation:
Erucamide is plastic additive used as anti slip agent and used as scratch resistant. It is used as scratch resistance because
- It reduces coefficient of friction between the polymer layer.
- Slip additives in Erucamide form coalesce along the surface.
-These additives move from the bulk to the part surface to form closed pack layers.
In this experiment, 0.070 g of caffeine is dissolved in 4.0mL of water. The caffeine is extracted from the aqueous solution three times with 2.0-mL portions of methylene chloride. Calculate the total amount of caffeine that can be extracted into the three portions of methylene chloride (see Technique12, Section12.2). Caffeine has a distribution coefcient of 4.6, between methylene chloride and water.
Answer:
0.068 g
Explanation:
The equation that is used to determine the fraction q remaining in water of volume V₁ after n extractions of volume V₂ is:
qⁿ = (V₁/(V₁ + KV₂))ⁿ
Substituting in the values gives the fraction of caffeine left in the water phase:
q³ = (4.0mL/(4.0mL + 4.6(2.0mL))³ = 0.0278
The fraction of caffeine extracted into the methylene chloride phase is:
1 - 0.0278 = 0.972
The amount of caffeine extracted into the methylene chloride is:
(0.070g)(0.972) = 0.068 g