Answer:
a). P = 688 atm
b). P = 1083.04 atm
c).Δ G = 16.188 J/mol
Explanation:
a). Fugacity 'f' can be calculated from the following equations :
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$[/tex]
where, P = pressure , Z = compressibility
Now, the virial equation is :
[tex]$PV=R(1+6.4\times 10^{-4} P)$[/tex] ........(1)
Also, PV=ZRT for real gases .......(2)
∴ [tex]$ZRT=RT(1+6.4\times 10^{-4} P)$[/tex]
[tex]$Z=1+6.4\times 10^{-4} P$[/tex]
So from the fugacity equation ,
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$[/tex]
[tex]$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$[/tex]
[tex]$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$[/tex]
[tex]$f=Pe^{6.4 \times 10^{-4} P}$[/tex]
Putting the value of P = 500 atm in the above equation, we get,
f = 688 atm
b). Given f = 2P
[tex]$2P=PE^{6.4 \times 10^{-4}P}$[/tex]
[tex]$2=E^{6.4 \times 10^{-4}P}$[/tex]
[tex]$\ln 2 = 6.4 \times 10^{-4}P$[/tex]
∴ P = 1083.04 atm
c). dG = Vdp -S dt at constant temperature, dT = 0
Therefore, dG = V dp
[tex]$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $[/tex]
[tex]$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$[/tex]
[tex]$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$[/tex]
[tex]$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$[/tex]
[tex]$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$[/tex]
Δ G = 16.188 J/mol
At point a the object has 30 joules of energy. how much energy does the object have at point b?
Answer:
It’s 10
Explanation:
I looked it up
To solve this we must be knowing each and every concept related to energy and its calculations. Therefore, at point b, 10joule of energy is present.
What is energy?Energy is defined in physics as the ability to accomplish work or heat items. It's a scalar measurement with magnitude but no direction. Energy is maintained, which implies it can change forms but is not generated or destroyed. potential energy transforms into kinetic energy.
One type of energy can be transferred into another without breaking a thermodynamic rule. Not all of these energy sources are equally beneficial in practical applications. At point a the object has 30 joules of energy. At point b, 10joule of energy is present.
Therefore, at point b, 10joule of energy is present.
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Which of the following pairs of elements would most likely form a molecular covalent bond?
Answer:
Oxygen and chlorine
Explanation:
oxygen and chlorine are most likely to form covalent compounds
What are the products of the chemical reaction of baking soda and vinegar and what is the identity of the gas
Answer:
In the reaction between vinegar and baking soda, the reactants are vinegar (acetic acid) and baking soda (sodium bicarbonate). The products are sodium acetate, water, and carbon dioxide gas.
John pushes a 10 pound box 5 feet. How much work is done?
500 ft-lb
1,000 ft-lb
5 ft-lbs
50 ft-lbs
Answer:
50 I think not 100% sure
Explanation:
Because you just do 10*5 which equals 50.
You have cyclohexene, water, cyclohexanol, and a little sulfuric acid catalyst in a test tube at 60 oC and everything is at equilibrium. Indicate what happens to that equilibrium and why with the following changes. (Note that adding water to an alkene is exothermic) 1) You heat it up to 80 oC. 2) You cool it down to 40 oC. 3) You remove some of the water. 4) You add a little more water. Use your answers to explain how to most successfully make cyclohexanol from cyclohexene, and how to most successfully make cyclohexene from cyclohexanol.
Answer:
See explanation
Explanation:
We can convert cyclohexanol to cyclohexene in the presence of a strong acid such as sulfuric acid catalyst in a test tube at 60 oC by heating up the mixture to about 80 oC. This is a dehydration reaction so water is removed to yield the alkene. A drying agent is used to remove any trace amount of water left in the system. This overall reaction is endothermic.
Also, the reverse is the case when we want to carry out the hydration of cyclohexene to yield cyclohexanol. The overall reaction is exothermic and involves the addition of more water to the alkene and then cooling down the system to about 40 oC.
HELPPP PLZz nowww!!!
A. C
B. B
C. A
D. D
Answer:
b
Explanation:
Answer:
answer b I think so I'm.not clear about it ,I think its CRT ans plz follow me
The volume of 350. mL of gas at 25°C is decreased to 250. mL at
constant pressure. What is the final temperature of the gas?
V1/T1 = V2/T2
V1 = 350. mL
T1 = 25 + 273 = 298 K
V2 = 250. mL
T2 = ?
Solve the equation for T2: T2 = V2T1/V1 = (250. mL)(298 K)/(350. mL) = 213 K = -60°C
two compounds of iron a and b contains 1.278 and 1.904 g of chlorine respectively for each gram of iron show that these amount are in the ratio of 2:3
Answer:
Ratio = 2 : 3 (Approx.)
Explanation:
Given:
Compound A = 1.278 g
Compound B = 1.904
Find:
Ratio
Computation:
Ratio = Compound A / Compound B
Ratio = 1.278 / 1.904
Ratio = 2 : 3 (Approx.)
How many particles does 2 moles of H2O have?
Hello,
QUESTION)Whatever the chemical entity, one mole of an entity contains 6.02 x 10²³ entities, this is the Avogadro constant (Na).
N = 2 x Na
N = 2 x 6.02 x 10²³ N = 1.204 x 10²⁴There are therefore 1.204 x 10²⁴ molecules of water in 2 moles of water.
what is the total pressure in a 10.0L flask which contains 0.200mol of H2 and 0.215mol of N2
Answer:
0.998 atm.
Explanation:
NOTE: The temperature is missing in the above question and the value is 20 °C
From the question given above, the following data were obtained:
Volume (V) = 10.0 L
Number of mole of H₂ = 0.2 mole
Number of mole N₂ = 0.215 mole
Temperature (T) = 20 °C
Total pressure (P) =?
Next, we shall determine the total number of mole in flask. This can be obtained as follow:
Number of mole of H₂ = 0.2 mole
Number of mole of N₂ = 0.215 mole
Total number of mole (n) =?
n = mole of H₂ + mole of N₂
n = 0.2 + 0.215
n = 0.415 mole
Thus, the total number of mole in the flask is 0.415 mole
Next, we shall convert 20 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 20 °C + 273
Temperature (T) = 293 K
Finally, we shall determine the total pressure in the flask. This can be obtained as follow:
Volume (V) = 10.0 L
Total number of mole (n) = 0.415 mole
Temperature (T) = 293 K
Gas constant (R) = 0.0821 atm.L/Kmol
Total pressure (P) =?
PV = nRT
P × 10 = 0.415 × 0.0821 × 293
Divide both side by 10
P = (0.415 × 0.0821 × 293) / 10
P = 0.998 atm.
Therefore, the total pressure in the flask is 0.998 atm.
WHAT IS THE CHEMICAL FORMULA FOR Rb+ and P3-
Answer:
3Rb3P → 9Rb + P3
Hope it helps!!!
The balanced form of the expression is given as:
[tex]3Rb_3P \rightarrow 9Rb + P3[/tex]
Balancing chemical equationsFrom the given question, we are to balance the equation of Rb and P^3+
[tex]Rb_3P \rightarrow Rb + P3[/tex]
Since the number of moles at the reactant must be equal to that of the product.
The balanced form of the expression is given as:
[tex]3Rb_3P \rightarrow 9Rb + P3[/tex]
Learn more on balancing of equation here: https://brainly.com/question/26227625
define saturated and unsaturated fats
Answer:
i don't no the
answer
i am sorry
Answer:
Saturated fats: Saturated fats are a type of fat-containing a high proportion of fatty acid molecules without double bonds, considered to be less healthy in the diet than unsaturated fat.
Unsaturated fats: Unsaturated fats are a type of fat-containing a high proportion of fatty acid molecules with at least one double bond, considered to be healthier in the diet than saturated fat.
Which is the name of the kind of solid substance
formed in this figure?
How many grams does 5.29 x 10^25 molecules of AgNO3 weigh?
14900 g AgNO₃
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Reading a Periodic TableMolesAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
[Given] 5.29 × 10²⁵ molecules AgNO₃
[Solve] grams AgNO₃
Step 2: Identify Conversions
Avogadro's Number
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
Step 3: Convert
[DA] Set up: [tex]\displaystyle 5.29 \cdot 10^{25} \ molecules \ AgNO_3(\frac{1 \ mol \ AgNO_3}{6.022 \cdot 10^{23} \ molecules \ AgNO_3})(\frac{169.88 \ g \ AgNO_3}{1 \ mol \ AgNO_3})[/tex][DA] Multiply/Divide [Cancel out units]: [tex]\displaystyle 14923 \ g \ AgNO_3[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
14923 g AgNO₃ ≈ 14900 g AgNO₃
Aside from direct harm, what secondary issue do invasive species present?
They may be able to eat all kinds of native organisms.
They might introduce other organisms like viruses and bacteria.
They might damage the habitat that other organisms rely on.
They may not have any predators.
Answer:
B
Explanation:
Usually, when the temperature is increased, what will happen to the rate of dissolving?
05))
A)
It will stop.
B)
It will increase.
e
It will decrease
D)
It will remain the same.
Answer:
B) It will increase.
Explanation:
Solubility tends to increase with an increase in temperature
A student is given an antacid tablet that weighs 5.4630 g. The tablet is crushed and 4.3620 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 13.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.7 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) has been neutralized in the 25.00 mL sample that was titrated
Solution :
It is given that :
Weight of the antacid tablet = 5.4630 g
4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.
25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.
27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]
= 12.27 ml of the original stomach acid.
1. When 0.629 grams of Na3PO4.12H2O and 0.527 g BaCl2.2H2O are mixed with water , how many grams of Ba3(PO4)2 are produced?
The number of grams of Ba₃(PO₄)₂ that is produced is : 0.4328 gram
Given data
mass of Na₃PO₄.12H₂O = 0.629 grams
mass of BaCl₂.2H₂O = 0.527 grams
molar mass of Na₃PO₄.12H₂O = 380.12 g/mol
molar mass of BaCl₂.2H₂O = 244.26 g/mol
Determine the number of grams of Ba₃(PO₄)₂ produced
mole of Na₃PO₄.12H₂O = mass / molar mass
= 0.629 / 380.12 = 0.001654 mol
molar mass of BaCl₂.2H₂O = 0.527 / 244.26 = 0.002157 mol
After mixture : Determine the limiting reactant by lower mole ratio
mole ratio of BaCl₂.2H₂O = 0.002157 / 3
= 0.0007191
mole ratio of Na₃PO₄.12H₂O = 0.001654 / 2
= 0.000827
therefore the limiting reactant is : BaCl₂.2H₂O
Final step : Determine the mass of BaCl₂.2H₂O
molar mass of BaCl₂.2H₂O = 601.93 g/mol
Therefore :
mass of BaCl₂.2H₂O = moles * molar mass
= 0.0007191 * 601.93
= 0.4328 gram
Hence we can conclude that The number of grams of Ba₃(PO₄)₂ that is produced is : 0.4328 gram
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1 = 2 Amps, R = 3 Ohms, V = ? Volts
Answer:
6 volts
Explanation:
Use Ohm's Law:
V = IR
V = (2 amps)(3 ohms) = 6 volts
What type of cloud produces severe weather like thunderstorms, thunder
and lighting, occasional hail, and possibly tornadoes?
Answer:
The cumulonimbus cloud
Explanation:
The cumulonimbus cloud, or thunderstorm, is a convective cloud or cloud system that produces rainfall and lightning. It often produces large hail, severe wind gusts, tornadoes, and heavy rainfall.
pls mark brainliest
Answer:
cumulonimbus clouds
Explanation:
that's the answer.
A balloon is filled with 30.0 L of helium in the morning when the temperature is 25.00 ℃. By noon the temperature has risen to 55.00 ℃. Determine the new volume of the balloon. *
Answer:
33.0 L
Explanation:
Step 1: Given data
Initial volume of the balloon (V₁): 30.0 LInitial temperature (T₁): 25.00 °CFinal volume of the balloon (V₂): ?Final temperature (T₂): 55.00 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
T₁: K = 25.00°C + 273.15 = 298.15 K
T₂: K = 55.00°C + 273.15 = 328.15 K
Step 3: Calculate the final volume of the balloon
If we assume ideal behavior, we can calculate the final volume of the balloon using Charles' law.
V₁/T₁ = V₂/T₂
V₂ = V₁ × T₂/T₁
V₂ = 30.0 L × 328.15 K/298.15 K = 33.0 L
Which of the following has the highest pH?
A. 1 M HCl
B. 0.01 M HCl
C. 0.1 M HCl
D. 0.001 M HCl
Answer: D
Explanation: pH = - log( c). If c= 1.0 M. PH = 0, if c=0.001 M. PH = 3
Answer:b
Explanation:
ik
5. 1.250 g of an alloy which is 90.00 % Ag and 10.00 % Cu is reacted with nitric acid producing AgNO3 and Cu(NO3)2 dissolved in solution. To the solution of the nitrate compounds is added a piece of copper wire that reacts only with the AgNO3 in solution to precipitate all of the silver, and produce additional Cu(NO3)2 (aq) . (The equation for this single-replacement reaction is: AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2 Ag(s) ) The resulting solution containing all of the Cu(NO3)2 produced by the above processes is treated with excess zinc to precipitate the entire amount of copper (ll) ion present. ( The equation for this step is Cu(NO3)2(aq) + Zn(s) Zn(NO3)2(aq) + Cu(s) ) What mass of copper will be precipitated?
Answer:
See explanation
Explanation:
According to the question Copper is obtained from the original alloy and the copper wire used to displace the silver.
So,
Mass of the copper from the original alloy: (10.00 % * 1.250 g) = 0.125 g
We now need to obtain the mass of the copper from the wire, then we consider how much silver is precipitated:
Mass of silver in the alloy; 90.00 % * 1.250 g = 1.125 g
So,
Amount of silver precipitated =
1.125 / 107.9 = 0.01043 mole
From the reaction equation, it takes two moles of silver to displace one mole of copper.
Hence, the amount of copper from the wire is: 0.01043 / 2 = 0.005213 mol
And a mass of 0.005213(63.55) = 0.3313 g
Total mass of copper= 0.125 + 0.3313 = 0.456 g
Describe how to prepare 100.00 mL of a 0.250 M sodium chloride (NaCl) solution using sodium chloride powder.
Answer:
The answer to your question is given below.
Explanation:
We'll begin by converting 100 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 mL
100 mL = 0.1 L
Next, we shall determine the number of mole of NaCl in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity of NaCl = 0.250 M
Mole of NaCl =?
Molarity = mole / Volume
0.250 = mole of NaCl / 0.1
Cross multiply
Mole of NaCl = 0.250 × 0.1
Mole of NaCl = 0.025 mole
Finally, we shall determine the mass of 0.025 mole of NaCl. This can be obtained as follow:
Mole of NaCl = 0.025 mole
Molar mass of NaCl = 23 + 35.5
= 58.5 g/mol
Mass of NaCl =?
Mole = mass / Molar mass
0.025 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.025 × 58.5
Mass of NaCl = 1.46 g
Thus, to prepare the solution, weigh out 1.46 g of NaCl and dissolve in 1 L of water.
For each of the following acid-base reactions,
calculate the mass (in grams) of each acid necessary
to completely react with and neutralize 3.65 g of the
base.
Part A
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Part B
2 HNO3(aq) + Ca(OH)2(aq)
2 H2O(1) + C
Answer: A. 3.28 g of HCl
B. 6.30 g of [tex]HNO_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} NaOH=\frac{3.65g}{40g/mol}=0.09moles[/tex]
[tex]\text{Moles of} Ca(OH)_2=\frac{3.65g}{74g/mol}=0.05moles[/tex]
a) [tex]HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)[/tex]
According to stoichiometry :
1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]
Thus 0.09 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.09=0.09moles[/tex] of [tex]HCl[/tex]
Mass of [tex]HCl=moles\times {\text {Molar mass}}=0.09moles\times 36.5g/mol=3.28g[/tex]
b) [tex]2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)[/tex]
According to stoichiometry :
1 mole of [tex]Ca(OH)_2[/tex] require = 2 moles of [tex]HNO_3[/tex]
Thus 0.05 moles of [tex]Ca(OH)_2[/tex] will require=[tex]\frac{2}{1}\times 0.05=0.1moles[/tex] of [tex]HNO_3[/tex]
Mass of [tex]HNO_3=moles\times {\text {Molar mass}}=0.1moles\times 63g/mol=6.3g[/tex]
What is the pH of this solution?
Answer:
pH = 11.216.
Explanation:
Hello there!
In this case, according to the ionization of ammonia in aqueous solution:
[tex]NH_3+H_2O\rightleftharpoons NH_4^++OH^-[/tex]
We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:
[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}[/tex]
However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:
[tex]1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M[/tex]
Which is also:
[tex][OH^-]=1.643x10^{-3}M[/tex]
Thereafter we can compute the pOH first:
[tex]pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784[/tex]
Finally, the pH turns out:
[tex]pH=14-2.784\\\\pH=11.216[/tex]
Regards!
Compare the weight of a 100-lb person on Mercury with the weight of that same person on Jupiter. What can you conclude about the mass and force of gravity of these planets?
A
Mercury has much less mass than Jupiter, so the force of gravity is weaker on Mercury and this causes the person to weigh less.
B
Mercury is closer to the sun than Jupiter so this causes the person to weigh less.
C
The mass of Mercury and Jupiter has no impact on the force of gravity.
D
Mercury has more mass than Jupiter, so the force of gravity is greater on Mercury and this causes the person to weigh less.
Answer:
if you weigh 100 pounds on Earth, you would weigh only 38 pounds on Mercury. That's because Mercury weighs less than Earth, and therefore its gravity would pull less on your body. If, on the other hand, you were on heavy Jupiter, you would weigh a whopping 253 pounds!
Most cooking utensils are made up of aluminum because aluminum is _____
Explanation:
Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.
A heat engine with 0.500 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle: - Isothermal expansion to 4000 cm3. - Isochoric cooling to 300 K. - Isothermal compression to 3000 cm3. - Isochoric heating to 800 K.
Answer:
Explanation:
[tex]\text{For isothermal expansion:}[/tex]
[tex]W_1 = nRT_1 In (\dfrac{V_2}{V_1}) \\ \\ W_1 = 0.5 \times 8.314 \times 800 \times In (\dfrac{4000}{3000})\\ \\ W _1 = 956.72 \ J \\ \\ Q_1 = W_1 = 956.72 \ J \ since \ (dU=0) \\ \\ \\ \text{For isochoric cooling,} W_2 = 0} \\ \\ Q_2 = nCr \Delta T \\ \\ Q_2 = 0.5 (\dfrac{3R}{2})(T_2-T_1) \\ \\ Q_2 = 0.5 \times \dfrac{3\times 8.314}{2}(-500)= -3117.75[/tex]
[tex]\text{For Isothermal compression:}\\\\ W_3 = nRT_2 \ In (\dfrac{V_4}{V_3}) \\ \\ 0.5 \times 8.314 \times 300 \times In (\dfrac{3000}{4000}) \\ \\ W_3 =- 358.77 \ J \\ \\ Q_3=W_3= - 358.77 \ J[/tex]
[tex]\text{For isochoric heating; }W_4 =0} \\ \\ Q_4 = nC_v\Delta T \\ \\ = 0.5 \times \dfrac{3}{2}\times 8.314 \times 500 \\ \\ Q_4 = 3117.5 \ J[/tex]
[tex]\text{Total workdone W}= W_1 + W_2+W_3+W_4 \\ \\ W = 956.71 \ J + 0 + (-358.77 \ J) +0 \\ \\ \mathbf{W = 597.94 J} \\ \\ \\ \eta = \dfrac{Work \ done}{heat \ taken } \\ \\ \eta = \dfrac{W}{Q_1+Q_4} \\ \\ \eta = \dfrac{597.94 \ J}{956.71 \ J + 3117.5 \ J} \\ \\ \eta = 0.1468 \\ \\ \mathbf{\eta = 14.68\%}[/tex]
Type the CONVERSION factor (fraction) to calculate the moles in 15.0 g
of potassium sulfide, K2S. *
10 points
Explanation:
5g C7H8 / 92g/mole = 0.054moles
total moles = 0.232moles
mole fraction C6H12 = 0.178moles / 0.232moles = 0.767
mole fraction C7H8 = 0.233