Answer:
The compound is [tex]Mn_{3}O_{4}[/tex]
Explanation:
The mass percentage of Mn is 72.1% and the mass percentage of O is 27.9%.
The mass percentage of a compound is given by:
[tex]percentage_{A}=\frac{n*MM_{A}}{MM_{C}} *100[/tex]
where:
n is its coefitient in the compund formula
MMa=Molar mass of the element A
MMc=Molar mass of the compound
So, we can figure out which compound is by dividing the percentage by its molar mass
Mn=72.1÷54.938045=1.31239
O=27.9÷15.9994=1.74382
Then, we divide each result by the smaller one (Mn)
Mn=1.31239÷1.31239=1
O=1.74382÷1.31239=1.3287
Each the realation of Mn:O is 1:1.3287
Then we multiply each result by 3:
Mn=1×3=3
O=1.3287×3=3.986≈4
Finally we figure out that the compound has 3 atoms of Mn and 4 atoms O. Result= [tex]Mn_{3}O_{4}[/tex]
Mn3O4 is sometimes used as a starting material in the production of soft ferrites e.g. manganese zinc ferrite, and lithium manganese oxide, used in lithium batteries.
The empirical formula of a compound with 72.1% Mn and 27.9% O is MnO.
Explanation:The empirical formula of a compound can be determined based on its mass percentage composition of different elements. In this case, the compound is composed of 72.1% Mn (manganese) and 27.9% O (oxygen).
To find the empirical formula, we need to consider the relative number of atoms in the compound. Assuming we have 100g of the compound, we have 72.1g of Mn and 27.9g of O.
Using the molar masses of Mn (54.938 g/mol) and O (15.999 g/mol), we can calculate the number of moles of each element:
Moles of Mn = 72.1g / 54.938 g/mol = 1.314 molMoles of O = 27.9g / 15.999 g/mol = 1.744 molNow, we divide the number of moles by the smallest value to get the mole ratio:
Mn:O = 1.314 mol / 1.314 mol = 1:1O:Mn = 1.744 mol / 1.314 mol = 1.326:1Since we want whole-number ratios, we round the ratios to the nearest whole number:
Mn:O = 1:1Therefore, the empirical formula of the compound with 72.1% Mn and 27.9% O is MnO.
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
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.
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Give the electron configurations for the following ions: P, P. Ar, and Te plain what interaction is in the hydrogen fluoride (HF) and hydrogen chloride (HCI. Which one do you espect to have a higher bolling emperature? Why?
Explanation:
Atomic number of P = 15
Common oxidation state of P = -3, +3 and +5
Electronic configuration of P^3+: [tex]1s^2 2s^22p^63s^2[/tex]
Electronic configuration of P^3-: [tex]1s^2 2s^22p^63s^23p^6[/tex]
Electronic configuration of P^5+: [tex]1s^2 2s^22p^6[/tex]
Atomic number of Ar = 18
Argon has stable octet, so it generally does not exist as ions.
Electronic configuration of Ar: [tex]1s^2 2s^22p^63s^23p^6[/tex]
Atomic number of Te = 52
Common oxidation states of Te = -2, +2, +4, +6
Electronic configuration of Te^2-: [tex][Kr]4d^10 5s^2 5p^6[/tex]
Electronic configuration of Te^2+: [tex][Kr]4d^10 5s^2 5p^2[/tex]
Electronic configuration of Te^4+: [tex][Kr]4d^10 5s^2[/tex]
Electronic configuration of Te^6+: [tex][Kr]4d^10[/tex]
Boiling point of HF is more as compared to HCl
Reason:
Molecular weight of HF is low, but is boiling point is high because of presence of hydrogen bonding. Whereas in case of HCl, its molecular weight is high but has only weak van der Waals intercations. As hydrogen bonding is stronger than van der Waals interactions, therefore, boiling point of HF is more.
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
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.
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
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.
Hydrocyanic acid has a Ka of 4.0
x10-10. What is the percent ofionization of
a 1.0 molar solution?
Answer:
0.002 %
Explanation:
Given that:
[tex]K_{a}=4.0\times 10^{-10}[/tex]
Concentration = 1.0 M
Consider the ICE take for the dissociation of Hydrocyanic acid as:
HCN ⇄ H⁺ + CN⁻
At t=0 1.0 - -
At t =equilibrium (1.0-x) x x
The expression for dissociation constant of Hydrocyanic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]
[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]
x is very small, so (1.0 - x) ≅ 1.0
Solving for x, we get:
x = 2×10⁻⁵ M
Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]
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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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.
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
Calculate the volume of 42.0 kg of a substance whose density is 8.96 g/mL. Express your answer in milliliters using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
4687.5 mL
Explanation:
Given that the density of the substance = 8.96 g/mL
Mass = 42.0 kg
The conversion of kg into g is shown below as:
1 kg = 1000 g
So, mass = 42.0 × 1000 g = 42000 g
Volume = ?
So, volume:
[tex]Volume=\frac {{Mass}}{Density}[/tex]
[tex]Volume=\frac {42000\ g}{8.96\ g/mL}[/tex]
The volume of the 42.0 kg of the substance = 4687.5 mL
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]
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.
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.
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.
A culture of E.coli bacteria doubles every 40 minutes. If there are 50 bacteria initially, how many will there be after 12 hours? Round your answer to the nearest whole number
Answer:
There will be 13107200 number of bacteria after 12 hours
Explanation:
1 hour = 60 minutes
So, 12 hours = ([tex]12\times 60[/tex]) minutes = 720 minutes
Initially there are 50 bacteria.
As number of bacteria doubles in every 40 minutes therefore rate of increase in number of bacteria will be similar to a first order reaction.
Hence, number of bacteria after 720 minutes = [tex]50\times (2)^{\frac{720}{40}}[/tex] = 13107200
1 kg of Oxygen gas is present in a 500 L tank. Find the specific and molar volumes of the gas.
Answer:
specific volume O2 = 0.5 Kg/m³
molar volume O2 = 8 E-3 m³/mol
Explanation:
specific volume (Sv):
∴ Sv = 1 / ρ
∴ ρ = mass / volume
∴ V = 500 L * ( m³/1000 L) = 0.5 m³
∴ ρ = 1 Kg / 0.5m³ = 2 Kg/m³
⇒ Sv = 1 / 2 Kg/m³ = 0.5 m³/Kg
molar volume ( Vm ):
∴ Vm = volume/mol
∴ Mw O2 = 1000g O2 * ( mol/16g O2) = 62.5 mol O2
⇒ Vm = 0.5 m³ / 62.5 mol
⇒ Vm = 8 E-3 m³/mol
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]
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]
Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume
ik its not B
Answer:
option D= temperature and volume
Explanation:
Definition:
Charle's law stated that ,
" At constant pressure, the volume of given amount of gas is directly proportional to its absolute temperature"
V∝ T
V= kT
OR
V/T = k (k is proportionality constant)
if the temperature is changed from T1 to T2 then the volume of gas is also changed from V1 to V2. Then expression will be:
V1/T1= k and V2/T2=k
V1/T1=V2/T2
suppose a cylinder is filled with a gas having volume V1 at temperature T1. When the gas is heated its temperature raises from T1 to T2 and its volume also increased with increase of temperature from V1 TO V2.
In Charles' law, the two changing properties are temperature and volume, with pressure maintained constant. The law indicates a direct proportionality between temperature and volume for a given amount of gas.
Explanation:The two changing properties in Charles' law are temperature and volume. Charles' law states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. This means if the temperature of the gas increases, its volume increases as well, provided the pressure and the amount of gas remain constant. In other words, doubling the absolute temperature at constant pressure will double the volume. It is important to note that in gas laws, temperatures must always be expressed in kelvins.
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%
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 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]
Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.
Answer: Option (c) is the correct answer.
Explanation:
Entropy is defined as the degree of randomness that is present within the particles of a substance.
As [tex]NH_{4}NO_{3}[/tex] is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ([tex]NH^}{+}_{4}[/tex]) and nitrate ions ([tex]NO^{-}_{3}[/tex]).
Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.
Thus, we can conclude that ammonium nitrate ([tex]NH_{4}NO_{3}[/tex]) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.
One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer in L using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
28.52 L
Explanation:
First, let's calculate the density of the ocean, which is the mass divided by the volume:
d = m/V
d = 35.06/1
d = 35.06 g/L
So, for a mass of 1.00 kg = 1000.00 g
d = m/V
35.06 = 1000.00/V
V = 1000.00/35.06
V = 28.52 L
How all the data are expressed with two significant figures, the volume must also be expressed with two.
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]
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
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.