Answer:
0.41 atm
Explanation:
A reaction between liquid reactants takes place at 10.0 °C in a sealed, evacuated vessel with a measured volume of 5.0 L. Measurements show that the reaction produced 13. g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.
Step 1: Given data
Temperature (T): 10.0 °CVolume of the vessel (V): 5.0 LMass of sulfur hexafluoride gas (m): 13. gStep 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 10.0 °C + 273.15 = 283.2 K
Step 3: Calculate the moles (n) of SF₆
The molar mass of SF₆ is 146.06 g/mol.
13. g × 1 mol/146.06 g = 0.089 mol
Step 4: Calculate the pressure (P) of SF₆
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/V
P = 0.089 mol × (0.0821 atm.L/mol.K) × 283.2 K/5.0 L = 0.41 atm
How does the government rely on scientists?
What is the oxidation number of calcium
Answer:
0 for the elemental form, +2 in its compounds.
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1. Jimmy picks up a 5kg box and places it on a shelf 1 meter from the ground. What is the
gravitational potential energy of the box?
Answer: 49 joules
Explanation: gravitational potential energy = mgh
m= mass kg, g= acceleration due to gravity 9.8 m/sec/sec, h= height m
=5*9.8.1 joules = 49 joules
Why are the non-living things important to an ecosystem?
Answer:
well just like grass that helps animals eat rocks may give an animal a home
Find the mass in grams of 1.38 moles of Sr
Answer:
116.78 grams.
Explanation:
1 mol of Strontium (Sr) = 87.62 grams
1.38 mol of Strontium = x
Cross Multiply
1 * x = 1.38 * 87.62
x = 116.78 grams
How many moles are there in 24.0 grams of H2O
Answer:So, one mole of water has a mass of 16 +1+1 = 18 grams. So, if one mole has a mass of 18 grams, 25 grams would have a mass of 25 grams/ 18 grams per mole or 1.39 moles
Answer:
The answer would be 1.33
Explanation:
do you need an explanation?
If you have 3.5 L of He to blow up balloons... at STP...
a) How many moles of He do you have?
b) How many grams of He do you have?
Answer:
B
Explanation:
What is the mass of 1.75 moles of Ca(H2C302)2?
Explanation:
first we have to find molar mass of ca(H2c3o2)2
40+(1*2)2+(12*3)2+(16*2)2
40+4+72+64=180g/mole
m=n*Mm
m=1.75mole*180g/mole
m=315g
A compound is made of 6.00 grams of oxygen, 7.00 grams of nitrogen, and 20.00grams of hydrogen. Find the percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
B O-11.18%, N-22.21%, H-69.60%
C O-20%, N-30%, H-50%
D O-60.60%, N-21.21%, H-18.18%
The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
A state of matter where the particles that make up a substance start to break apart
Answer:
Liquid
Explanation:
When she introduced CO(g) and Cl2(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 1.86 atm and the initial partial pressure of Cl2 was 1.27 atm, she found that the equilibrium partial pressure of COCl2 was 0.823 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction.
Answer:
Kp is 0.00177
Explanation:
We state the equilibrium:
CO(g) + Cl₂(g) ⇆ COCl₂(g)
Initially we have these partial pressures
1.86 atm for CO and 1.27 for chlorine.
During the reaction, x pressure has been converted. As we have 0.823 atm as final pressure in the equilibrium for COCl₂, pressure at equilibrium for CO and chlorine will be:
1.86 - x for CO and 1.27 - x for Cl₂.
And x is the pressure generated for the product, because initially we don't have anything from it. So pressure in equilibrium for the reactants will be:
1.86 - 0.823 = 1.037 atm for CO
1.27 - 0.823 = 0.447 atm for Cl₂
Let's make, expression for Kp:
Partial pressure in eq. for COCl₂ / P. pressure in eq. for CO . P pressure in eq. for Cl₂
Kp = 0.823 / (1.037 . 0.447) → 0.00177
Explain the differences between an ideal gas and a real gas.
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
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A 5.5 gg sample of a substance contains only carbon and oxygen. Carbon makes up 35%% of the mass of the substance. The rest is made of oxygen. You are asked to determine the mass of oxygen in the sample. Which of the following expressions demonstrates a mathematical procedure to solve this problem using the proper order of operations?
a. ((100 - 35)/100) times 5.5 grams =
b. 100 - 35/100 times 5.5 grams =
c. 100 - (35/100) times 5.5 grams=
Answer:
a. ((100 - 35)/100) times 5.5 grams = 3.575 g
Explanation:
Given that:
The sample of carbon and oxygen = 5.5g
where carbon makes 35% of the mass of the substances.
It implies that oxygen will make: (100 - 35)% = 65%
Suppose y be the mass of the oxygen;
Then:
[tex]y = \dfrac{(100-35)}{100} \times 5.5 \ g[/tex]
[tex]y = \dfrac{(65)} {100} \times 5.5 \ g[/tex]
[tex]y = 3.575 \ g[/tex]
The mass of carbon [tex]= \dfrac{35}{100} \times 5.5 \ g[/tex]
= 1.925 g
How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)
Answer:
0.229 cm³.
Explanation:
The following data were obtained from the question:
Volume (in in³) = 0.014 in³
Volume (in cm³) =?
1 in = 2.54 cm
Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:
1 in = 2.54 cm
Therefore,
1 in³ = 2.54³ cm³
1 in³ = 16.387 cm³
Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:
1 in³ = 16.387 cm³
Therefore,
0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³
0.014 in³ = 0.229 cm³
Thus, 0.014 in³ is equivalent to 0.229 cm³.
Derive the isentropic and isothermal compressibility terms (for an ideal gas) in the most simplified form and compare them.
Answer:
Isothermal compressibility is different from isentropic compressibility by temperature instead of entropy.
Explanation:
Isothermal compressibility refers to that type of compressibility where volume change takes place at constant temperature whereas isentropic compressibility refers to that in which volume change takes place at constant entropy. Entropy is the measure of a thermal energy of the system per unit temperature that is unavailable for doing useful work.
You can use solid, wedged, and dashed lines to better represent the 3D structure of molecules. The solid lines represent bonds in the plane of the paper, wedged lines represent a bond coming out of the plane, and the dashed line represents a bond going back behind the plane of the paper. Draw a representation of methane that better depicts its 3D structure.
Answer:
See explanation and image attached
Explanation:
Often times, there is a need for a three dimensional representation of a molecule on paper. These three dimensional representations give us an idea of what the molecule really looks like if we were to be looking at it physically.
In order to make a three dimensional representation, we use wedged and dashed bonds. The wedged bonds are coming out of the plane of the paper towards you while the dashed bonds are going into the plane of the paper away from you.
In the image attached, you will find the three dimensional representation of the methane molecule.
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Answer:
no way
Explanation:
more free points
How many mols in 2.25x10^25 atoms of Zinc
Answer:
37.4 mol.
Explanation:
Hello!
In this case, since the Avogadro's number help us to realize that one mole of any substance contains 6.022x10²³ formula units, in this case atoms of zinc, the following dimensional analysis provides the correct answer:
[tex]=2.25x10^{25} atoms*\frac{1mol}{6.022x10^{23}atoms}\\\\= 37.4mol[/tex]
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If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at
the end of the series of reactions, what was the percent recovery? Briefly explain how you found your
answer.
Answer: Percent recovery is 47.34 %
Explanation:
Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.
[tex]{\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100[/tex]
Putting in the values we get:
[tex]{\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%[/tex]
Therefore, the percent recovery is 47.34 %
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A sample is 54.7g CaCl2 and 53.64 g H20. How
many water molecules are associated with this
hydrate?
The molecular formula of hydrate : CaCl₂.6 H₂O
So there are 6 molecules of H₂O
Further explanationGiven
54.7g CaCl₂ and 53.64 g H₂O
Required
The number of molecules H₂O
Solution
mol CaCl₂ :
= mass : MW
= 54.7 : 111 g/mol
= 0.493
mol H₂O :
= 53.64 : 18 g/mol
= 2.98
mol ratio H₂O : CaCl₂ :
= 2.98/0.493 : 0.493/0.493
= 6 : 1
The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of H2OH2O is ________% by mass. The molar mass of urea is 60.0 g/mol. The concentration of urea in a solution prepared by dissolving 16 g of urea in 20 g of is ________% by mass. The molar mass of urea is 60.0 g/mol. 44 80 0.80 0.48 0.44
Answer:
44
Explanation:
Given that :
Mass of solute = Mass of urea = 16g
Mass of water = 20g
Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g
Percentage Mass = (mass of solute / mass of solution) * 100%
Percentage Mass = (16 / 36) * 100%
Percentage Mass = 0.4444444 x 100%
Percentage Mass = 44.44%
Percentage Mass = 44%
An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically, but it can also be calculated using calculus. The reaction of compound A forming compound B was studied and the following data were collected:
Time (s) [A] (M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016
Required:
a. What is the average reaction rate between 0 and 1500s?
b. What is the average reaction rate between 200. and 1200s ?
c. What is the instantaneous rate of the reaction at t=800s?
Answer:
Explanation:
a )
Average reaction rate between 0 and 1500s
Time duration = 1500 s
moles reacted = .184 - .016 = .168 moles
Moles reacted per second = .168 / 1500
= 112 x 10⁻⁶ moles /s
b )
Average reaction rate between 200 and 1200s
Time duration = 1000 s
moles reacted = .129 - .019 = .11 moles
Moles reacted per second = .11 / 1000
= 110 x 10⁻⁶ moles /s
c )
the instantaneous rate of the reaction at t=800s
We shall assume that between 500 s and 1200 s , rate of reaction is uniform
rate between 500 and 1200
Time duration = 700 s
moles consumed = .069 - .019 = .05 moles
Rate of reaction = .05 / 700
= 71 .4 x 10⁻⁶ moles / s
This will also be instantaneous rate of reaction at t = 800 s .
3. How much heat (in kl) is released when 1.65 mol NaOH(s) is dissolved in water? (The molar heat of solution of NaOH is -445.1 kJ/mol.)
a. -1.36x 10-3 kJ
b. -7.34 x 10^2 km
c. -2.69 x 10^1 km
d. -2.80 x 10^2 km
The heat released : b. -7.34 x 10² kJ
Further explanationGiven
1.65 mol NaOH
The molar heat of solution of NaOH is -445.1 kJ/mol
Required
Heat released
Solution
ΔH solution = Q : n
ΔH solution = enthalpy of solution(-=exothermic, +=endothermic)
Q = heat released/absorbed
n = moles of solute
Input the value :
Q = ΔH solution x n
Q = -445.1 kj/mol x 1.65 mol
Q = -734.415 kJ
Select one metal which will displace Sn from a compound and form metallic tin (Sn).
Cu, Cr, Ag, or Hg
Answer:
Cr
Explanation:
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It is desirable to remove calcium ion from hard water to prevent the formation of precipitates known as boiler scale that reduce heating efficiency. The calcium ion is reacted with sodium phosphate to form solid calcium phosphate, which is easier to remove than boiler scale. What volume (in liters) of 0.478 M sodium phosphate is needed to react completely with 0.225 liter of 0.279 M calcium chloride
Answer:
0.0876L of 0.478M Na₃PO₄ are needed
Explanation:
The reaction of calcium chloride, CaCl₂, with sodium phosphate, Na₃PO₄ is:
3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl
Where 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.
To solve this question we need to find moles of CaCl₂ added. Using the reaction we can find the moles of Na₃PO₄ that are needed to react completely and the volume using its concentration:
Moles CaCl₂:
0.225L * (0.279mol / L) = 0.0628moles of CaCl₂
Moles Na₃PO₄:
0.0628moles of CaCl₂ * (2mol Na₃PO₄ / 3 mol CaCl₂) =
0.0419moles of Na₃PO₄
Volume 0.478M Na₃PO₄:
0.0419moles of Na₃PO₄ * (1L / 0.478mol) =
0.0876L of 0.478M Na₃PO₄ are neededTo determine the concentration of citric acid, you will need to titrate this solution with 0.100 M NaOH. You are given a 1.00 M NaOH stock solution and will need to make enough 0.100 M NaOH to perform 3 titrations. For each titration, you will use 20.0 mL of 0.100 M NaOH solution.
Calculate the total volume (in mL) of the diluted solution you will need to prepare for the 3 titrations.
Determine the minimum volume (in mL) of 1.00 M NaOH stock solution needed to prepare the 0.100 M NaOH solution.
Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.What kind of solid tends to have the lowest melting points?
Answer:
Molecular solids
Covalent compounds
Explanation:
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Draw the organic product(s) of the following reactions, and include carbon dioxide if it is produced.
CH3CH2-C triple bond C-CH2CH3 rightarrow O3
You do not have to consider stereochemistry.
If a compound is formed more than once, add another sketcher and draw it again.
Draw carbon dioxide in its own sketcher if it is produced.
Separate multiple products using the + sign from the dropdown menu.
If no reaction occurs, draw the organic starting material.
Answer:
See picture below
Explanation:
In this case, is ocurring an ozonolysis reaction with alkines.
Alkines, unlike alkenes, when they undergo an ozonolysis reaction, the product formed is a carboxilic acid. In some cases it may produce CO₂, but that will happen only if the starting molecule has a terminal Hydrogen atom.
In other words the following:
CH₃CH₂C≡CH
In this case, it may produces the CO₂. However, it's not the case, so it will produce two molecules of carboxylic acid.
You can see the picture below for the final product.
Hope this helps.
Alkynes are subjected to ozonolysis to produce two ketones or acid anhydrides. The acid anhydride undergoes hydrolysis to produce two carboxylic acids if water is present in the process. Elastomer ozonolysis is also referred to as the ozone method.
Ozone (O₃), a reactive allotrope of oxygen, is used in ozonolysis, a process for oxidatively breaking alkenes or alkynes. The procedure enables the substitution of double or triple carbon-carbon bonds with double oxygen bonds. This reaction is frequently used to determine an unknown alkene's structure.
The products of the given reaction can be shown below:
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7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c. The temperature of the water is observed to rise by 2.316c. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound x at 25c.
Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
Answer:
[tex]\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
[tex]\Delta H_{rxn} =- m_wC_w\Delta T[/tex]
We plug in the mass of water, temperature change and specific heat to obtain:
[tex]\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ[/tex]
Now, this enthalpy of reaction corresponds to the combustion of propyne:
[tex]C_3H_4+4O_2\rightarrow 3CO_2+2H_2O[/tex]
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
[tex]\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}[/tex]
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
[tex]\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol[/tex]
Now, we solve for the enthalpy of formation of C3H4 as shown below:
[tex]\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}[/tex]
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
[tex]\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
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