Answer:
[tex]m_{CO2}=260.7 g CO2[/tex]
Explanation:
First of all we need to calculate the energy required:
[tex]KE= 0.5*m*v^2[/tex]
where:
[tex]m=2700kg[/tex]
[tex]v=67 mph=29.95 m/s[/tex]
[tex]KE= 0.5*2700kg*(29.95)^2[/tex]
[tex]KE= 1210953 J=1210.953 kJ[/tex]
Octane's combustion enthalpy: [tex]\Delta H_{comb}=- 5450 kJ/mol[/tex]
The reaction:
[tex]C_8H_{18} + 25/2 O_2 longrightarrow 8 CO_2 +9 H_O[/tex]
Mass of CO2:
[tex]m_{CO2}=\frac{1210.953 kJ}{5450mol}*\frac{1}{0.3}*\frac{8 mol CO2}{1 mol}*\frac{44 g CO2}{mol CO2}[/tex]
[tex]m_{CO2}=260.7 g CO2[/tex]
The values between 1 and 14. The formula for pH is y = − log 10 x, where y is the pH value and x is the concentration of hydrogen ions. A Golden Delicious apple has an average concentration of hydrogen atoms of 2.51 × 10 − 4. What is the pH value of a Golden Delicious apple?
Answer:
3.6
Explanation:
The pH of a solution can be defined as the negative logarithm to base 10 of the concentration of hydrogen ions in the solution. In the equation form, it can be written as :
pH = -Log [H+]
In the question, we have been given the concentration of the hydrogen ions but we do not know the value of the pH. We then insert the value of the concentration into the equation.
This means [H+] = 2.5 * 10^-4
pH = -Log [ 2.5 * 10^-4] = 3.6
The cost, C C, of producing x x Totally Cool Coolers is modeled by the equation C = 0.005 x 2 − 0.25 x + 12 C=0.005x2-0.25x+12 How many coolers need to be produced and sold in order to minimize the cost? (Round to the nearest whole number.)
Answer:
25 coolers are need to be produce and sell in order to minimize the cost.
Explanation:
[tex]C = 0.005x^2-0.25x+12[/tex] ..[1]
Differentiating the given expression with respect to dx.
[tex]\frac{dC}{dx}=\frac{d(0.005x^2-0.25x+12)}{dx}[/tex]
[tex]\frac{dC}{dx}=0.01x-0.25+0[/tex]
Putting ,[tex]\frac{dC}{dx}=0[/tex]
[tex]0=0.01x-0.25+0[/tex]
[tex]0.01x=0.25[/tex]
x = 25
Taking second derivative of expression [1]
[tex]\frac{d^2C}{dx^2}=\frac{d(0.01x-0.25)}{dx}=0.01[/tex]
[tex]\frac{d^2C}{dx^2}>0[/tex] (minima)
25 coolers are need to be produce and sell in order to minimize the cost.
What is the hybridization of the central atom of molecules?
Explanation:
The number of electron density regions that encircle an atom is calculated by the steric number, which also defines the hybridisation of the central atom. In that scenario, there are six hybridised orbitals in the central atom, leading to a [tex]\displaystyle sp3d^2[/tex]hybridization.
I am joyous to assist you anytime.
If you heat the air inside a rigid, sealed container until its Kelvin temperature doubles, the air pressure in the container will also double. Is the same thing true if you double the Celsius temperature of the air in the container? Explain.
Answer:
No
Explanation:
If you heat the air inside a rigid, sealed container, then you are keeping it's mass and volume constant.
Then, Pressure of the container is directly proportional to the temperature.
Gay-Lussac's Law: This law states that the pressure of a given amount of gas when at constant volume is directly proportional to gas's absolute temperature in Kelvin.
So, if Kelvin temperature doubles, the air pressure in the container will also double.
The relationship between Celsius temperature and Kelvin temperature is not linear.
[tex]T_{K} = T_{C} + 273.15[/tex]
If [tex]T_{K}[/tex] becomes [tex]2*T_{K}[/tex], then [tex]T_{C}[/tex] will become [tex](2 * T_{K})- 273.15[/tex] .
i.e, it is increasing for sure, but doesn't double.
Given the coefficients of the balanced equation, the enthalpy of the reaction can be expressed in terms of the amount of ammonia as follows: ΔHrxn=−906 kJ4 mol NH3 Next, find the number of moles of NH3 that react to determine the heat associated with 345 g of NH3.
Answer:
q = -4588.89 kJ
Explanation:
First, we need to write the equation taht is taking place. In this case, is the combustion of NH3 so the reaction is as follow:
4NH3 + 5O2 --------> 4NO + 6H2O ΔHrxn = -906 kJ
Now, in order to get the number of moles that react here and determine the heat, we first calculate the experimental moles of NH3 within the 345 g of ammonia:
moles = m/MM
The molar mass of NH3 is 17 g/mol so:
moles = 345 / 17 = 20.29 moles
Now, we will stablish a relation between the theorical moles and the experimental moles:
q = 20.29 moles * (-906) kJ / 4 moles
q = -4,588.89 kJ
You need to make 3-bromo-5-isopropyltoluene starting with m-isopropyltoluene. Supply the reagent needed in the order it is needed and draw the resulting product.
To synthesize 3-bromo-5-isopropyltoluene from m-isopropyltoluene, brominate using Br2/FeBr3, then rearrange with NaOH. Product: 3-bromo-5-isopropyltoluene with bromine at meta position and isopropyl group at para position.
To synthesize 3-bromo-5-isopropyltoluene starting with m-isopropyltoluene, we need to perform two main reactions: bromination and rearrangement. Here are the steps and reagents needed:
1. **Bromination**: We need to brominate the m-isopropyltoluene at the meta position. We can achieve this using bromine (Br2) in the presence of a Lewis acid catalyst like iron (III) bromide (FeBr3).
[tex]\[ \text{m-isopropyltoluene} + \text{Br}_2 + \text{FeBr}_3 \rightarrow \text{3-bromomethyl-isopropylbenzene} \][/tex]
2. **Rearrangement**: Next, we need to rearrange the 3-bromomethyl-isopropylbenzene to form the desired product, 3-bromo-5-isopropyltoluene. This rearrangement can be achieved using a strong base like sodium hydroxide (NaOH).
[tex]\[ \text{3-bromomethyl-isopropylbenzene} + \text{NaOH} \rightarrow \text{3-bromo-5-isopropyltoluene} \][/tex]
Now, let's draw the resulting product, 3-bromo-5-isopropyltoluene:
In this structure, the bromine atom (Br) is attached to the third carbon atom of the benzene ring, and the isopropyl group (-CH(CH3)2) is attached to the fifth carbon atom of the benzene ring.
if a test yields consistent results every time it is used, it has a high degree of...
Answer:Reliability
Explanation:
Reliability of a test refers to how consistently a test measures a characteristic under the same conditions.
Reliability can be defined as the degree of consistency of which a chemical test gives a similar result. measure. A test is said to be highly reliable when it gives the same repeated result under the same conditions of measure.
But when a test gives different results under the same condition of measure it has a low reliability.
Hence, If a test yields consistent results every time it is used, it has a high degree of reliability.
Isotopes are atoms of the same element with different numbers of
Answer:
neutron
Explanation:
which affects mass number
Duralin is an alloy of aluminium containing 4.5% manganese, 3.5% copper, and 0.45% magnesium. How much magnesium is present in a 88 g sample of this alloy?
Answer: The mass of magnesium in given amount of alloy is 0.396 g
Explanation:
We are given:
Percent mass of manganese = 4.5 %
Percent mass of copper = 3.5 %
Percent mass of magnesium = 0.45 %
Percent mass of aluminium = (100 - 4.5 - 3.5 - 0.45) = 91.55 %
Mass of alloy = 88 g
We need to calculate the mass of magnesium in 88 g of alloy, we get:
[tex]\Rightarrow 88\times \frac{0.45}{100}=0.396g[/tex]
Hence, the mass of magnesium in given amount of alloy is 0.396 g
. The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion 802 kJ/mol CH4) in the bomb. The temperature changed by 10.8C. a. What is the heat capacity of the bomb? b. A 12.6-g sample of acetylene, C2H2, produced a temperature increase of 16.9C in the same calorimeter. What is the energy of combustion of acetylene (in kJ/mol)?
The heat capacity of the bomb calorimeter is approximately 31.391 kJ/°C, and the energy of combustion of acetylene is approximately 1096.5 kJ/mol.
Explanation:The heat capacity of a bomb calorimeter can be determined using the formula: q = CΔT, where q represents the energy absorbed or released, C is the heat capacity, and ΔT is the change in temperature.
First, we will determine the amount of energy absorbed by the bomb calorimeter from the combustion of methane. Given that the molar mass of methane is 16.04 g/mol, 6.79 g of methane corresponds to 0.423 mol. Therefore, the amount of energy absorbed is 0.423 mol x 802 kJ/mol = 339.046 kJ. Subsequently, we can calculate the heat capacity of the bomb calorimeter by rearranging the initial formula to: C = q/ΔT. So, C = 339.046 kJ/10.8 °C, which gives a heat capacity of approximately 31.391 kJ/°C.
Secondly, the energy of combustion of acetylene is given by the formula: q = CΔT. We already know the heat capacity of the bomb calorimeter (31.391 kJ/°C) and the change in temperature (16.9 °C). Therefore, the energy of combustion of acetylene is 31.391 kJ/°C x 16.9 °C = 530.7 kJ. As the molar mass of acetylene (C2H2) is 26.04 g/mol, 12.6 g corresponds to 0.484 mol. Thus, the energy of combustion of acetylene per mole is given by: 530.7 kJ/0.484 mol, which is approximately 1096.5 kJ/mol.
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The heat capacity of the bomb calorimeter is about 50229 kJ/°C. The energy of combustion of acetylene is found to be approximately 84710 kJ, which can then be converted back to kJ/mol from g.
Explanation:The subject question refers to a bomb calorimetry experiment in which different substances, namely methane and acetylene, are being combusted. The heat capacity of the bomb calorimeter can be calculated using the formula q = CΔT, where q represents heat, C is heat capacity and ΔT is the change in temperature. The energy of combustion for the second substance can be calculated using the same formula rearranged as q/ΔT = C.
By substituting the values from the methane combustion (q = 6.79g * 802 kJ/mol, ΔT = 10.8°C), we find that the heat capacity is about 50229 kJ/°C. For the acetylene combustion, we can use the same formula rearranged to find q (q = CΔT), which gives approximately 84710 kJ. This value can then be converted back to kJ/mol from g to get the energy of combustion of acetylene.Learn more about Bomb Calorimetry here:https://brainly.com/question/33792718
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Which set of these comparisons is INCORRECT? (More stable means it has a more negative energy.) a) The 1s orbital in H is more stable than the 1s orbital in He+ b) The 2s orbital in He atom is less stable than the 2s orbital in He+ c) The 2s subshell in Li is more stable than the 2p subshell in Li
Answer:
a.
Explanation:
By the Linus Pauling's diagram, we know that the order of the energy of the subshell at the same atom, from the less to the higher, is:
1s < 2s < 2p < 3s < 3p < 4s < 4d ...
As larger is the atom, the attraction energy between the electron and the nuclei is weak. The size of the atom increase from right to left in the periods, and from the top to the bottom in the families. The cations are smaller than the neutral atoms (the attraction is more effective with fewer electrons), and the anions are larger than the neutral atoms. So, in order of size, and also the energy of the similar subshells:
He⁺ < He < H
How higher the energy, less negative it is, so less stable is the orbital.
a. As shown above, the energy of the orbital at H is higher than the energy at the orbital at He⁺, so the 1s orbital is more stable at He⁺, and the sentence is incorrect.
b. As shown above, the energy of the orbital at He is higher than the energy of the orbital He⁺, so the 2s orbital at He is less stable than the 2s orbital at He⁺, and the sentence is correct.
c. From the Linus Pauling's diagram, the energy of 2p is higher than the energy of the 2s, so the 2s subshell is more stable, and the sentence is correct.
Name the following structure:
Answer: Ethyl ethanoate
Explanation:
An ester composed of the alkyl section (Ethyl) derived from ethanol and the alkanoate section (ethanoate) derived from ethanoic acid.
Answer:
The correct answer is d) ethyl ethanoate.
Explanation:
This organic compound corresponds to an ester, also called ethyl acetate. It is synthesized by the esterification of Fischer using acetic acid and ethanol in the presence of a catalyst.
An 8.6 grams piece of aluminum, heated to 100°C, is placed in a coffee cup (styrofoam) calorimeter that contains 402.4 grams of water at 25°C. The specific heat of water is 4.18 J/g°C and the specific heat of aluminum is 0.900 J/g°C.
What will be the final equilibrium temperature for both the aluminum and the water? (Assume no heat is lost to the surroundings and the coffee cup absorbs a negligible amount of heat).
Final answer:
The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.
Explanation:
To find the final equilibrium temperature for both the aluminum and the water, we can use the principle of thermal equilibrium and the conservation of energy.
First, let's calculate the heat gained or lost by the aluminum and the water:
The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the aluminum:
Qalu = malu * calu * ΔTalu = (8.6 g) * (0.900 J/g°C) * (T - 100°C)
For the water:
Qwater = mwater * cwater * ΔTwater = (402.4 g) * (4.18 J/g°C) * (T - 25°C)
Since the total heat gained by the aluminum is equal to the total heat lost by the water:
Qalu = -Qwater
Substituting the values and solving for T:
(8.6 g) * (0.900 J/g°C) * (T - 100°C) = -(402.4 g) * (4.18 J/g°C) * (T - 25°C)
Simplifying the equation:
(7.74 T - 774) = -(1681.832 T - 42045)
1765.832 T = 41271
T ≈ 23.39°C
The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.
The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm, what is the concentration (molarity) of dissolved nitrogen?
Answer:
Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L
Explanation:
More the pressure of the gas, more will be its solubility.
So, for two different pressure, the relation between them is shown below as:-
[tex]\frac {P_1}{C_1}=\frac {P_2}{C_2}[/tex]
Given ,
P₁ = 1 atm
P₂ = 0.76 atm
C₁ = 6.8 × 10⁻⁴ mol/L
C₂ = ?
Using above equation as:
[tex]\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}[/tex]
[tex]C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L[/tex]
Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L
Which is not a characteristic of metals?
a. Most metal oxides are ionic solids that are basic.
b. They have a shiny luster and various colors, although most are silvery.
c. Solids are malleable and ductile.
d. They are poor conductors of heat and electricity.
Answer: d: They are poor conductors of heat and electricity .
Explanation:
Although is common knowledge that metals are characterized by high electrical conductivity and heat conductivity, these properties can be justified by looking at the metallic bonding.
In a metal, the bonding electrons are delocalized over the entire crystal. In fact, metal atoms in a crystal can be imagined as an array of positive ions immersed in a sea of delocalized valence electrons.The mobility of the delocalized electrons makes metals good conductors of heat and electricity.
A 2.15g sample of benzene (C_6H_6) is burned in a bomb calorimeter, and the temperature rises from 22.46 degree C to 34.34 degree C. Calculate the heat capacity of the bomb calorimeter. Note the following thermochemical equation: C_6H_6(I) + 15/2 O_2 (g) rightarrow 6CO_2 (g) + 3H_2O (g) Delta H degree = -3267.5 kJ
Answer:
The heat capacity of the bomb calorimeter is 7.58 J/°C.
Explanation:
[tex]C_6H_6(I) + \frac{15}{2} O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (g) ,\Delta H^o = -3267.5 kJ[/tex]
First, we will calculate energy released on combustion:
[tex]\Delta H[/tex] = enthalpy change = -3267.5 kJ/mol
q = heat energy released
n = number of moles benzene= [tex]\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{2.15 g g}{78 g /mol}=0.02756 mol[/tex]
[tex]\Delta H=-\frac{q}{n}[/tex]
[tex]q=\Delta H\times n =-3267.5 kJ/mol\times 0.02756 mol=-90.0657 kJ[/tex]
q = -90.0657 kJ = -90,065.7 J
Now we calculate the heat gained by the calorimeter let it be Q.
Q = -q= -(-90,065.7 J) = 90,065.7 J (conservation of energy)
[tex]Q=c\times (T_{final}-T_{initial})[/tex]
where,
Q = heat gained by calorimeter
c = specific heat capacity of calorimeter =?
[tex]T_{final}[/tex] = final temperature = [tex]34.34^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]22.46^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]90,065.7 J=c\times (34.34-22.46)^oC[/tex]
[tex]c=\frac{90,065.7 J}{(34.34-22.46)^oC}=7.58 J/^oC[/tex]
The heat capacity of the bomb calorimeter is 7.58 J/°C.
A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m . Assume that the container is completely filled to the top with liquid nitrogen, that the temperature is 23.5?C, and that the atmospheric pressure is 1.2atm .Calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated into the closet. The closet is ventilated such that the temperature and pressure remain constant through this process. (Liquid nitrogen has a density of 0.807 g/mL.)
Answer:
26.99 % of air that will be displaced
Explanation:
Step 1: Data given
A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m
Temperature = 23.5 °C
Atmospheric pressure = 1.2 atm
Liquid nitrogen has a density of 0.807 g/mL
Molar mass of N2 = 28 g/mol
Step 2: Calculate mass of nitrogen
Mass of nitrogen = density * volume
Mass of nitrogen = 0.807 g/mL * 1200 mL
Mass of nitrogen = 968.4 grams
Step 3: Calculate moles of N2
Moles N2 = mass N2 / molar mass N2
Moles N2 = 968.4 grams /28 g/mol
Moles N2 = 34.586 moles
Step 4: Calculate volume
p*V = n*R*T
⇒ p = the the pressure = 1.2 atm
⇒ V = the volume of N2 = TO BE DETERMINED
⇒ n = the number of moles = 34.586 moles
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 23.5 °C = 296.65
V = (n*R*T)/p
V = (34.586 * 0.08206 * 296.65)/1.2
V = 701.61 L
Step 5: Calculate the total volume of the chamber
1.0 m * 1.3 m * 2 m = 2.6 m³ = 2600 L
Step 6: Calculate the percent volume displaced
(701.61 L / 2600 L) * 100% = 26.99%
26.99 % of air that will be displaced
The percent (by volume) of air that would be displaced is 26.99 %.
Given information:Temperature = 23.5 °C
Atmospheric pressure = 1.2 atm
Liquid nitrogen has a density of 0.807 g/mL
Molar mass of N2 = 28 g/mol
The calculation of the percent of air:Mass of nitrogen = density × volume
[tex]= 0.807 g/mL \times 1200 mL[/tex]
= 968.4 grams
Now
Moles N2 = mass N2 ÷ molar mass N2
[tex]= 968.4 grams \div 28 g/mol[/tex]
= 34.586 moles
Now volume
[tex]p\times V = n \times R \times T[/tex]
Here p = the the pressure = 1.2 atm
n = the number of moles = 34.586 moles
R = the gas constant = [tex]0.08206 L\times atm/K\times mol[/tex]
T = the temperature = 23.5 °C = 296.65
[tex]V = (n\times R\times T)\div p\\\\= (34.586 \times 0.08206 \times 296.65)\div 1.2[/tex]
= 701.61 L
Now the total volume of the chamber is
[tex]= 1.0 m \times 1.3 m \times 2 m \\\\= 2.6 m^3[/tex]
= 2600 L
Now finally the percent volume displaced
[tex]= (701.61 L \div 2600 L) \times 100\%[/tex]
= 26.99%
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A bomb calorimetric experiment was run to determine the enthalpy of combustion of methanol. The reaction is CH3OH(l)+3/2O2(g)→CO2(g)+2H2O(l) The bomb calorimeter has a heat capacity of 250.0 J/K. Burning 0.028 g of methanol resulted in a rise in temperature from 21.50 ∘C to 23.41 ∘C. Calculate the change in internal energy for the combustion of methanol in kJ/mol.
Answer:
The change in internal energy during the combustion reaction is- 545.71 kJ/mol.
Explanation:
First we have to calculate the heat gained by the calorimeter.
[tex]q=c\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat = [tex]250.0 J/^oC[/tex]
[tex]T_{i}[/tex] = Initial temperature = [tex]21.50^oC=294.65 K[/tex]
[tex]T_{f}[/tex] = Final temperature = [tex]23.41^oC=296.56 K[/tex]
Now put all the given values in the above formula, we get:
[tex]q=250.0 J/K\times (296.56 -294.65 )K[/tex]
[tex]q=477.5 J [/tex]
Now we have to calculate the enthalpy change during the reaction.
[tex]\Delta H=-\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat gained = -477.5 J
n = number of moles methanol = [tex]\frac{\text{Mass of methanol}}{\text{Molar mass of methanol}}=\frac{0.028 g}{32 g/mol}=0.000875 mol[/tex]
[tex]\Delta H=-\frac{477.5 }{0.000875 mol}=-545,714.28 J/mol=-545.71 kJ/mol[/tex]
Therefore, the change in internal energy during the combustion reaction is- 545.71 kJ/mol.
The development of which polymer enabled the British to have effective radar equipment and thus contributed to the British success in the Battle of Britain in World War II? A) polystyrene B) polyvinyl chloride C) polyacetylene D) polyethylene
Answer: Polyethylene
Explanation:
The waxy solid had been formed during high-pressure experiments conducted in 1933 by ICI scientists Reginald Gibson and Eric Fawcett. They had heated a mixture of ethylene and benzaldehyde to 170°C (338°F), using apparatus that could submit materials to a pressure of 1,900 atmospheres (1,925 bars). But the reactions were explosive and safety concerns prompted the now defunct ICI, which merged into Dutch-based Akzo Nobel, to halt the research.
A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and the assembly expands from an initial volume of 0.85 L to a final volume of 5.88 L against an external pressure of 1.0 atm.a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.b) Calculate w. Express your answer using two significant figures.c) Calculate q. Express your answer using two significant figures.
We can calculate the change in internal energy of the system by multiplying the power of the light bulb by the time it's on, resulting in 7200 Joules. The work done can be calculated using the formula for work done against pressure, resulting in -508.64 Joules. The heat energy is determined using the first law of thermodynamics, resulting in 6691.36 Joules.
Explanation:The student is asked to calculate the change in internal energy (ΔU), work done (w), and the heat transferred (q) in a system involving a moving piston and a light bulb. Let's address each part individually:
(a) The power (P) of the light bulb is given as 100 W. Power is energy per unit time, so we can find the energy by multiplying the power by the time it's used. Time is given as 2.0×10−2 hour, which needs to be converted to seconds for consistency: 2.0×10−2 hour × 3600 s/hour = 72 seconds. Thus, the change in internal energy (ΔU) is P × t = 100 W × 72 s = 7200 Joules.
(b) Work done by the system (w) equals -P*ΔV where P is the external pressure and ΔV is change in volume. The system expands, so work is done against the external pressure. Here, ΔV = final volume - initial volume = 5.88 L - 0.85 L = 5.03 L. We must convert volume to m³ (cubic meters), 1 L = 1.0e-3 m³, so, ΔV = 5.03e-3 m³. Also convert pressure to Pa from atm, 1 atm = 1.013e5 Pa, so P = 1.013e5 Pa. Inserting these values into the equation gives w = -(1.013e5 Pa)(5.03e-3 m³) = -508.64 Joules.
(c) Finally, to determine the heat energy (q) we utilize the first law of thermodynamics that states ΔU = q - w. Solving for q, we get, q = ΔU + w = 7200 Joules + (-508.64 Joules) = 6691.36 Joules.
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Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single σ bond or around a double σ plus π bond, or would they be the same?
Answer:
it is easier to rotate and single bond rather than a double bond made of a sigma bond and pi bond
Explanation:
Rotation around a single bond happens easily but it is very limited around a double bond because of the overlapping electron cloud above and below the imaginary axis between the two atoms.
It is easier to rotate around a single sigma (σ) bond than around a double bond that includes both a sigma and a pi bond. This is because single bonds are not dependent on the orientation of the atom's orbitals, unlike double bonds. In the latter, any rotation can break the bond.
Explanation:The matter at hand relates to the ease of rotation around different types of bonds - a single sigma (σ) bond versus a double bond made of one sigma (σ) and one pi (π) bond. Simplifying the science, it is easier to rotate around a single sigma (σ) bond than it is around a double bond that includes both a sigma and a pi bond. This is because the orbital overlap, which forms these bonds among atoms, doesn't rely on the relative orientation of the orbitals on each atom for a single bond. So, twisting or rotating doesn't affect the bonding. Whereas, with a double bond, the σ and π bonds form from different types of orbital overlaps. With this type of bond, the stable configuration is planar (flat) as seen in ethene molecules. Any rotation would misalign their overlapping, unstable, and effectively break the pi bond.
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Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin () into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.
Answer:
4C3H5N3O9 (l) ---------> 12CO2 (g) + H20 (g) + 6N2 (g) + 6O2 (g)
Explanation:
Nitroglcerin is a drug basically used to treat chest pain. It is a dense, colourless and explo9sive liquid. Its molecular formula is C3H5N3O9.
It decomposes to gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon di oxide
The equation for its decomposition is shown below;
4C3H5N3O9 (l) ---------> 12CO2 (g) + H20 (g) + 6N2 (g) + 6O2 (g)
A Sample of 61.8g of hyroboric acid(H3BO3), a weak acid, is dissolved in 1,000g of water to make a 1.0 molal solution. Which would be the best procedure to determine the molarity if the solution?
Answer:
Molarity = 1.0 M
Explanation:
Step 1: Data given
Mass of H3BO3 = 61.8 grams
Mass of water = 1,000 grams = 1kg
Molality = 1.0 molal
Step 2: Calculate number of moles
1.0 molal = 1 mol of solute/ 1kg of solvent
Number of moles = molality * mass of the solvent = 1.0 mol/kg * 1kg = 1 mol
This means 61.8 grams of H3BO3 should be 1 mol
We can control this:
Number of moles = mass / molar mass
Number of moles H3BO3 = 61.8/ 61.83 g/mol = 1 mol
Step 3: Calculate the volume of water
Let's suppose the density of water is 1g/mL
Volume of water = mass / density
Volume = 1,000 grams / 1g/mL
Volume = 1,000 mL = 1.00 L
Step 4: Calculate molarity
Molarity = number of moles / volume
Molarity = 1 mol / 1.00 L
Molarity = 1.0 M
For dilute AQUEOUS solutions molality ≅ molarity
A 1.00-L gas sample at 100.°C and 500. torr contains 52.0% helium and 48.0% xenon by mass. What are the partial pressures of the individual gases?
Answer:
Partial pressure of He=486 torr, partial pressure of Xe= 14 torr
Explanation:
Using the equation, PV=nRT----------------------------------------(1)
Making n the subject of the formula;
n= PV/RT---------------------(2)
Where n= number of moles, v= volume, T= temperature, P= volume.
n= (500 torr/760 torr × 1 atm)× 1L ÷ 373K ×0.082 L atmK^-1. Mol^-1
n= 0.6579 atm.L/ 30.6233
n= 0.0215 mol.
Let the total mass of the gas= 2b(in g).
Mass of helium gas = b (in g) = mass of Xenon gas
Mole of helium gas= b(in g) / 4 gmol^-1
=b/4 mol
Mole of Xenon= b g/131.3 gmol^-1
= b/131.3 mol.
Solving for b, we have;
b/4+b/131.3 = 0.0215 mole
(131.3+4)b/525.2= 0.0215
Multiply both sides by 1/135.3.
b= 11.2918/135.3
b= 0.0835 g
Mole of He gas= 0.0835/4= 0.0209
Mole of Xe gas= 0.0215- 0.0209
= 0.0006 mol
Mole fraction of He = 0.0209/0.0215
= 0.972
Mole fraction of Xe= 0.0006/0.0215
= 0.028
Partial pressure of He gas= 0.972× 500 torr= 486 torr
Partial pressure of Xe gas= 0.028 ×500 torr = 14 torr
Final answer:
Dalton's Law is used to find the partial pressures of He (388.7 torr) and Xe (111.3 torr) in a 1.00 L, 500 torr mixture by calculating their mole fractions based on mass percentages and multiplying by the total pressure.
Explanation:
To determine the partial pressures of helium and xenon in a 1.00-L gas sample at 100.°C and 500. torr, we can use Dalton's Law of Partial Pressures. First, we calculate the mass of each gas based on the given percentage by mass. Then, we convert the mass of each gas to moles using their respective molar masses. Finally, using Dalton's Law, the partial pressure of each gas is calculated as the mole fraction of that gas multiplied by the total pressure.
First, let's convert the total mass percentage to actual mass assuming we have 100 grams of the mixture:
Mass of helium (He) = 52.0% of 100 g = 52.0 g
Mass of xenon (Xe) = 48.0% of 100 g = 48.0 g
Next, we'll convert mass to moles using the molar mass of helium (4.00 g/mol) and xenon (131.29 g/mol):
Moles of He = 52.0 g / 4.00 g/mol = 13.0 moles
Moles of Xe = 48.0 g / 131.29 g/mol = 0.365 moles
The total moles of gas is the sum of the moles of He and Xe:
Total moles of gas = Moles of He + Moles of Xe
Total moles of gas = 13.0 moles + 0.365 moles = 13.365 moles
Next, we calculate the mole fraction for each gas:
Mole fraction of He = Moles of He / Total moles of gas = 13.0 / 13.365
Mole fraction of Xe = Moles of Xe / Total moles of gas = 0.365 / 13.365
Lastly, we determine each gas's partial pressure:
Partial pressure of He = Mole fraction of He × Total pressure
Partial pressure of Xe = Mole fraction of Xe × Total pressure
Substituting the values we get:
Partial pressure of He = (13.0 / 13.365) × 500 torr
Partial pressure of Xe = (0.365 / 13.365) × 500 torr
Partial pressures calculated are for He and Xe respectively.
NaOH(s) dissolves readily in water to form Na+(aq) and OH-(aq). If 35.1 g of NaOH is dissolved in 199 g of water in a coffee cup calorimeter, the temperature of the resulting solution increases from 23.0 oC to 62.6 oC. Calculate the enthalpy change per mole of NaOH dissolved. Assume the specific heat capacity of the solution is 4.18 J/oC*g.
Answer:
ΔH = -44,13 kJ/mol
Explanation:
For the following reaction:
NaOH(s) → Na⁺(aq) + OH⁻(aq)
It is possible to obtain the heat produced using:
Q = 4,18J/g°C×mass×ΔT
Where the mass is 35,1g + 199g = 234,1g
And ΔT is FinalT - InitialT → 62,6°C - 23,0°C = 39,6°C
Replacing:
Q = 38750 kJ = 38,75kJ
The enthalpy change is obtained using:
ΔH = -Q/moles of NaOH
Moles of NaOH are:
35,1g×[tex]\frac{1mol}{40g}[/tex]= 0,878 moles
Thus:
ΔH = -38,75 kJ/0,878mol = -44,13 kJ/mol
I hope it helps!
The value of ΔH for the reaction below is -72 kJ. __________ kJ of heat are released when 1.0 mol of HBr is formed in this reaction. H 2 (g) + Br 2 (g) → 2HBr (g)
The study of chemicals and bonds is called chemistry. There are different types of elements are there and these are metal and nonmetal.
The correct answer is 36 KJ of heat is released when 1.0 mole of HBr is formed.
What is Hess's law?The heat of any reaction ΔH for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:[tex]H_2 (g) + Br_2 (g) ---> 2HBr (g) \ \ \ \ H = -72 KJ[/tex]
This is the energy released when 2 moles of HBr is formed from one mole each of H2 and Br2. Therefore, Heat released for the formation of 1 mol HBr would behalf on this.
ΔHreq = -36 kJ
36 KJ of heat is released when 1.0 mole of HBr is formed.
Hence, the correct answer is 36KJ.
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The energy change ΔH for the reaction of H2 with Br2 to form 2 moles of HBr is -72 kJ. Therefore, when 1 mole of HBr is formed, half of this energy, which is -36 kJ, is released.
Explanation:In your question, the chemical equation shows that 2 moles of HBr are formed from the reaction of H2 with Br2, with the release of -72 kJ of heat. This energy change, ΔH, represents the amount of heat released during the reaction, indicating it's an exothermic reaction. Given that energy change corresponds to the formation of 2 moles of HBr, the heat released when 1 mole of HBr is formed would be -72 kJ divided by 2.
This results in -36 kJ released per mole of HBr formed. Thus, when 1.0 mol of HBr is formed in this reaction, -36 kJ of heat is released.
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2. Which of the following is true about the total number of reactants and the total number of
products in the reaction shown below?
CsH(?)+802(8) 750,(8) +6H2008)
(1 point)
9 moles of reactants chemically change into 11 moles of product.
9 grams of reactants chemically change into 11 grams of product.
9 liters of reactants chemically change into 11 liters of product.
9 atoms of reactants chemically change into 11 atoms of product.
9 moles of reactants chemically change into 11 moles of product.
Explanation:
In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .
For example 8O₂ depicts 8 moles of two oxygen atoms. The number of moles is a unit for quantifying particles. You can liken it to a dozen, gross or a score. Since a mole of a substance contains avogadro number of particles. We can relate the number of moles to other parameters.Learn more:
Moles https://brainly.com/question/2272966
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Answer:
A) 9 moles of reactants chemically change into 11 moles of product.
Explanation:
C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)
The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.
This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.
Number of moles of Carbon atoms
Reactant = 5
Product = 5
Number of moles of Hydrogen atoms
Reactant = 12
Product = 12
Number of moles of Oxygen atoms
Reactant = 16
Product = 10 + 6 = 16
Since the equation is balanced, we can now compare the moles.
From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.
This alone leads us to take option A as our answer.
Let us see how other options are wrong however;
Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.
Option C) This is wrong again due to same reasons mentioned for option B.
Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.
When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?
a.zero
b.negative
c.positive
The work done by a gas on its surrounds is positive when the volume of the gas increases because the gas is doing work to push against the external pressure.
Explanation:The work done by a gas on its surroundings is defined as positive when the gas expands, meaning when the volume of the gas increases. This is because the gas is doing work to push against the external pressure and expand its volume. So, the correct answer to your question 'When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?' would be (c) positive.
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When the volume of a gas increases, the work done by the gas on its surroundings is positive. This occurs because the gas is doing work on its surroundings, resulting in energy loss for the gas. The key equation is w = -Pext ΔV.
So the answer is C)Positive
To determine this, we need to understand the concept of work in thermodynamics. Specifically, the work done by a gas during expansion is given by the equation:
w = -[tex]P_{ext[/tex] ΔV,
where:
w is the work done by the gas.[tex]P_{ext[/tex] is the external pressure opposing the gas expansion.ΔV (Delta V) is the change in volume.When the volume of the gas increases, ΔV is positive because the final volume ([tex]V_{final[/tex]) is greater than the initial volume ([tex]V_{initial[/tex]). According to the equation, since ΔV is positive and there is a negative sign in front, the work (w) done by the gas is negative.
This negative sign indicates that the system (the gas) is doing work on its surroundings, thus losing energy. Therefore, when the volume of a gas increases, the work done by the gas on its surroundings is positive.
The density of an aqueous solution containing 10.0 percent of ethanol (C2H5OH) by mass is 0.984 g/mL. (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.125 mole of ethanol
The solution contains 2.41 m (molality) and 2.14 M (molarity) of ethanol. A volume of 58.5 mL of this solution would contain 0.125 mol of ethanol.
Explanation:(a) Let's calculate the molality of this solution. Mass of ethanol in 1000g of solution is 100g (10% of 1000g). The molar mass of ethanol (C2H5OH) is (2*12.01 + 6*1.008 + 1*16.00 + 1*1.008) = 46.07 g/mol. Hence, the number of moles = 100g / 46.07 g/mol = 2.17 mol. Since these moles are in 900g of water (1000g - 100g), molality = 2.17 mol / 0.900 kg = 2.41 m.
(b) To calculate molarity, let's find the volume of 1000g of solution: Volume = mass/density = 1000g / 0.984 g/mL = 1016.26 mL = 1.016 L. Therefore, molarity = 2.17 mol/1.016 L = 2.14 M.
(c) To find the volume of solution that would contain 0.125 mol ethanol, we'll use molarity: Volume = moles/Molarity = 0.125 mol / 2.14 M = 0.0585 L or 58.5 mL.
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The molality of the solution is 2.41 mol/kg, the molarity is 2.135 M, and to contain 0.125 mole of ethanol, 58.57 mL of the solution is required.
(a) Molality is defined as the number of moles of solute per kilogram of solvent. Given the density of the solution, we can determine the mass of 100.0 mL of the solution:
Density = 0.984 g/mL
Mass of 100 mL of solution = 100 mL * 0.984 g/mL = 98.4 g
Mass of ethanol (10% by mass) = 0.10 * 98.4 g = 9.84 g
Mass of water = 98.4 g - 9.84 g = 88.56 g
To convert grams of ethanol to moles:
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
Moles of ethanol = 9.84 g / 46.07 g/mol = 0.2135 mol
To convert grams of water to kilograms:
88.56 g = 0.08856 kg
Therefore, the molality is:
Molality = 0.2135 mol / 0.08856 kg = 2.41 mol/kg
(b) Molarity is defined as the number of moles of solute per liter of solution. We already calculated the moles of ethanol as 0.2135 mol in 100 mL of solution.
Volume of solution = 0.100 L (since 100 mL = 0.1 L)
Therefore, the molarity is:
Molarity = 0.2135 mol / 0.100 L = 2.135 M
(c) We need to find the volume of solution that contains 0.125 mole of ethanol. Using the molarity calculated:
Molarity = 2.135 M = moles of solute / volume of solution (L)
Volume of solution = moles of solute / M
Volume of solution = 0.125 mol / 2.135 M = 0.05857 L = 58.57 mL
Therefore, the volume of the solution that would contain 0.125 mole of ethanol is 58.57 mL.
How many electrons will fluorine gain or lose in forming an ion
Answer:
1 electron
Explanation:
Fluorine is a nonmetal with a high electronegativity and electron affinity so it tends to gain electrons to reach stability. According to the octet rule, fluorine will gain enough electrons so as to complete its valence shell with 8 electrons. Fluorine is in the Group 17 in the Periodic Table, which means it has 7 electrons in its valence shell. Therefore, it will gain 1 electron to complete its octet and form the anion F⁻.