Answer:
One of the organelles in eukaryotic cells that carry out digestion and waste removal.
Answer:
It's X
Explanation:
Select the correct answer.
What effect does an increase in products have on the reaction rate of a mixture at equilibrium?
A.
The forward reaction rate increases.
B.
Both the forward and the reverse reaction rates decrease.
Both the forward and the reverse reaction rates increase.
D.
The reverse reaction rate increases.
Reset
Next
Answer:
At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.
When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.
Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.
Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.
Explanation:
Hope This Helps Amigo!
Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.
a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene
If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?
Answer:
572. 3 g of NH3
Explanation:
Equation of the reaction: 3H2 + N2 ----> 2NH3
From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.
Since N2 is in excess in the given reaction, H2 is the limiting reactant.
Molar mass of H2 = 2 g/mol
Molar mass of NH3 = 17 g/mol
Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3
6 g of H2 produces 34 g of NH3
101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3
Therefore, 572.3 g of NH3 are produced
Answer:
572.33g of NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
Molar Mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g.
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.
Therefore, 572.33g of NH3 is produced from the reaction.
You are trying to recrystallize compound X. You consider using ethyl acetate as your recrystallizing solvent and test a small amount of compound X with ethyl acetate. You find that compound X is soluble in ethyl acetate at room temperature and at boiling. Is ethyl acetate a good recrystallization solvent? No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath. Yes, you want the sample to fully dissolve at room temperature and boiling so that it will crystallize in the ice bath. Yes, you can only be sure that all the impurities dissolved if the sample is soluble at room temperature
Answer:
No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath.
Explanation:
For a solvent to be adequate it must completely dissolve the substance to be purified when it is hot, that is, at boiling temperature only. It should be practically insoluble when the solvent is cold or at room temperature. This must occur in this way since impurities must be removed by hot filtering or dissolved in the mother liquor.
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
The modern view of an electron orbital in an atom can best be described as
Answer:
An orbital is a region in space where there is a high probability of finding an electron.
Explanation:
The orbital is a concept that developed in quantum mechanics. Recall that Neils Bohr postulated that the electron occupied stationary states which he called energy levels. Electrons emit radiation when the move from a higher to a lower energy level. Similarly, energy is absorbed by an electron to move from a lower to a higher orbit.
This idea was upturned by the Heisenberg uncertainty principle. This principle state that the momentum and position of a particle can not be simultaneously measured with precision.
Instead of defining a 'fixed position' for the electron, we define a region in space where there is a possibility of finding an electron with a certain amount of energy. This orbital is identified by a set of quantum numbers.
Answer:
three - dimensional space that shows the probability where an electron is most likely to be found
When pressure is increased on the following equilibrium, where will the shift be? 3H2 + N2 2NH3
Answer:
Explanation:
it is based on le chatliers principles
the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,
so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Answer:
−153.1 J / (K mol)
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Hg(liq) + Cl2(g) → HgCl2(s)
Given that;
The standard molar entropies of the species at that temperature are:
Sºm (Hg,liq) = 76.02 J / (K mol) ;
Sºm (Cl2,g) = 223.07 J / (K mol) ;
Sºm (HgCl2,s) = 146.0 J / (K mol)
The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]
= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]
= -153.09 J / (K mol)
= or -153.1 J / (K mol)
Hence the answer is −153.1 J / (K mol)
If complications arise after cataract surgery, the ophthalmologist will use a Nd:YAG laser to perform a posterior capsulotomy. If the wavelength of the laser used is 1064 nm (infrared), and the pulse duration is 2.00 x 10–6 s whose energy is 0.245 J per pulse, how many photons are produced in each pulse?
Answer: 1.311 × 10^18 photons are produced in each pulse
Explanation: Please see the attachments below
Amylase is the enzyme that controls the breakdown of starch to glucose. Describe how the student could investigate the effect of pH on the breakdown of starch by amylase.
Answer:
Explanation:
You will investigate the breakdown of starch by amylase at different pHs.
The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.
The amylase will break down the starch.
A series of test tubes containing a mixture of starch and amylase is set up at different pHs.
A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.
A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.
When the starch solution is added:
Start timing immediately.Remove a sample immediately and test it with iodine solution.Sample the starch-amylase mixture continuously, for example every 10 seconds.For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.
The time taken for the disappearance of starch is not the rate of reaction.
It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.
We can calculate the rate of the reaction by calculating \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.
A similar experiment can be carried out to investigate the effect of temperature on amylase activity.
Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.
Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.
good luck :)
A chemistry student weighs out 0.306 g of citric acid (H3C6H5O7) , a triprotic acid, into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits
Answer:
THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L
Explanation:
The titration is between citric acid (H3C6H507) and NaOH
mass of citric acid = 0.306 g
Volume of citric acid = 250 mL = 250 /1000 = 0.25 L
Concentration of NaOH = 0.1000 M
Volume = unknown
First calculate the molar mass of citric acid
( 1 * 3 + 12* 6 + 1*5 + 16*7) = (4 + 72 + 5 + 112) = 193 g/mol
Since,
Concentration in moles/dm3 = concentration in g/dm3 / RMM
So the molarity of citric acid is:
Molarity = 0.306g / 0.25dm3 / Rmm
Molarity = 1.224g/dm3 / 193 g/mol
Molarity = 0.0063 M
Equation for the reaction is:
C3H5O(COOH)3 + 3NaOH → Na3C3H5O(COO)3 + 3H2O
Using the formula:
CaVa / CbVb = na/ nb
Ca = 0.0063 M
Cb = 0.1000 M
Va = 0.25 L
Vb = unknown
na = 1
nb = 3
Vb = Ca Va nb/ Cb na
Vb = 0.0063 * 0.25 * 3 / 0.1000 * 1
Vb = 0.4725 / 0.1000
Vb = 4.725 L
The volume of NaOH needed to reach the equivalent point is therefore 4.725 L
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
Best regards.
Answer:
B. Mixing a solute and a solvent
Explanation:
ap3x
What is the temperature at which the substance can be both in the solid and the liquid phase?
Answer: Gas–liquid–solid triple point
The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).
Explanation:
It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.
Identify the state(s) of matter that each property describes.
Answer:solid,liquid,gas,plasma
Explanation:
This question seems incomplete. I believe the full question is as followed:
Identify the state(s) of matter that each property describes.
1.) takes the shape of its container:
O gas
O liquid
O solid
2.) fills all available space:
O gas
O liquid
O solid
3.) maintains its shape:
O gas
O liquid
O solid
4.) can be poured:
O gas
O liquid
O solid
5.) is compressible:
O gas
O liquid
O solid
6.) has a fixed volume:
O gas
O liquid
O solid
The answers to the 1st are gas and liquid.
The answer to the 2nd is gas.
The answer to the 3rd is solid.
The answer to the 4th is liquid.
The answer to the 5th is gas.
The answers to the 6th are liquid and solid.
consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.
Answer:
[tex]pH=10.5[/tex]
Explanation:
Hello,
In this case, the dissociation of the given weak acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the law of mass action for it turns out:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent is:
[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]
Thus, by solving with the quadratic equation or solver, we obtain:
[tex]x=31.6x10^{-12}M[/tex]
Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:
[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]
Regards.
Please help! (:
question above — how much money would you need to buy 7.0 lb of arugula? If 27lb of arugula cost $16
Answer:
$11.81
Explanation:
27 lb cost $16
27/16=$1.69 per pound
$1.69*7=$11.81 for 7 lbs
Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of water produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
1.7 moles of ethanol would be needed.
Explanation:
* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
First off, we have to state the equation for the reaction.
So we know that;
ethanol + oxygen → acetic acid + water
This leads us to;
C2H5OH + O2 → CH3COOH + H2O
1 1 1 1
To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.
1 mol of ethanol produces 1 mole of water
x mol of thanol would produce 1.7 mol of water
Thus we have;
1 = 1
x = 1.7
x = 1.7 moles of ethanol would be needed.
According to the ideal gas law, what happens to the volume of a gas when the
temperature doubles (all else held constant)?
A. The volume stays constant.
B. The volume doubles.
OOO
C. It cannot be determined
D. The volume is halved
According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).
What does the ideal law state?The ideal gas law relates the pressure, volume, number of moles and temperature of an ideal gas.
Let's consider the equation of the ideal gas law.
P . V = n . R .T
V = n . R . T / P
As we can see, there is a direct relationship between the volume and the temperature. Thus, if the temperature doubles, the volume will double as well.
According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).
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A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
50.0 g N204 (92.02 g/mol) react with 45.0 g N2H4 (32.05 g/mol) forming nitrogen gas, N2
(28.01 g/mol) and water, H20 (18.02 g/mol). What mass in grams of excess-reactant is
left over?
Answer:
The excess reactant is N2H4 and the leftover mass is 10.17g.
Explanation:
Step 1:
The balanced equation for the reaction.
N2O4 + 2N2H4 —> 3N2 + 4H2O
Step 2
Determination of the masses of N2O4 and N2H4 that reacted from the balanced equation:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.
Step 3:
Determination of the excess reactant. This is illustrated below:
From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g of N2H4 reacted out of 45g that was given. Therefore, N2H4 is the excess reactant.
Step 4:
Determination of the mass of excess reactant that is leftover.
The excess reactant is N2H4 and the leftover mass can be obtained as follow:
Mass of N2H4 given = 45g
Mass of N2H4 that reacted = 34.83g
Leftover mass of N2H4 =..?
Leftover mass of N2H4 = (Mass of N2H4 given) – (Mass of N2H4 that reacted)
Leftover mass of N2H4 = 45 – 34.83
Leftover mass of N2H4 = = 10.17g.
A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution
Answer:
[tex]0.745~M[/tex]
Explanation:
In this case, we have a dilution problem. So, we have to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
Now, we have to identify the variables:
[tex]C_1~=~14.9~M[/tex]
[tex]V_1~=~25~mL[/tex]
[tex]C_2~=~?[/tex]
[tex]V_2~=~0.5~L[/tex]
Now, we have different units for the volume, so we have to do the conversion:
[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]
Now we can plug the values into the equation:
[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]
I hope it helps!
How would I find the quantity of heat absorbed or released when 2.0g of LiOH is dissolved in 100g of H₂0 when the enthalpy of the solution is -23.6KJ/mol?
Answer:
1.97kJ of energy are released.
Explanation:
The dissolution of LiOH in water is:
LiOH(s) → Li⁺(aq) + OH⁻(aq) ΔH = -23.6kJ
That means, when 1 mole of LiOH is dissolved, there are released (Because of the - in the enthalpy) 23.6kJ
2.0 g of LiOH (Molar mass: 23.95g/mol) are:
2.0g LiOH × (1 mol / 23.95g) =0.0835 moles of LiOH.
As 1 mole of LiOH release 23.6kJ, 0.0835moles release:
0.0835moles × (-23.6kJ / 1mole) = 1.97kJ of energy are released
Write a Lewis structure for each atom or ion. Draw the particle by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons and non-bonding electrons. Show the charge of the atom. Particles: S2-, Mg, Mg+2, P.
Answer:
The Lewis structure to this question can be described as follows:
Explanation:
Structure of Lewis for [tex]S^{2-}[/tex]:
The maximum number of electrons from valence in [tex]S^{2-}[/tex] is 8 (6 from S as well as 2 from negative change).
The valence electrons in the Lewis structure are placed on four sides of the atom.
Thus the structure of Lewis for [tex]S^{2-}[/tex] is as follows:
[tex]\left[\begin{array}{ccc} &. .&\\: &S&:\\&. .&\end{array}\right] ^{2-}[/tex]
Lewis Mg Structure:
Complete valence electrons are 2 in Mg.
The Lewis structure for Mg, therefore, is as follows:
[tex]\ . \\ Mg\\ \ .[/tex]
The Lewis structure for [tex]Mg^{2+}[/tex]
The maximum valence of electrons [tex]Mg^{2+}[/tex] in is= 0.
Thus, the structure for [tex]Mg^{2+}[/tex] is as follows:
[tex]Mg^{2+}[/tex]
Lewis structure for P :
The maximum number of valence electrons in P is = 5.
Thus, the structure for P is=
[tex]\ \ \ . \\ : P \ : \\[/tex]
Choose the slope and Y intercept that correspond with the graph
y=3x + 4
Explanation:
Where 3 is the slope and 4 is the y-intercept
How could the government enforce ethical standards of scientific
experiments?
A. The government could encourage scientists to make up their own
minds about ethics.
B. The government could take away research funds if ethical
standards are not met.
C. The government could let scientists monitor each other to
encourage ethical behavior.
D. The government could encourage the public to take a stand
against unethical scientists.
Answer: D. The Government could take away research funds if ethical standards are not met
The government enforce ethical standards of scientific experiments
B. The government could take away research funds if ethical
standards are not met.
Ethical standards are a set of principles established by the founders of the organization to communicate its underlying moral values. This code provides a framework that can be used as a reference for decision making processes.
How does the government control scientific research?Politicians and bureaucrats control scientific research and research outcomes by selectively funding projects that look for potential disasters, ideally global disasters.
What are the 8 ethical standards?This analysis focuses on whether and how the statements in these eight codes specify core moral norms (Autonomy, Beneficence, Non-Maleficence, and Justice), core behavioral norms (Veracity, Privacy, Confidentiality, and Fidelity), and other norms that are empirically derived from the code statements.
To learn more about ethical standards, refer
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RUIGA GIRLS
CHEMISTRY FORM 3. 23/06/2020
MR. GICHURU
IZ
1
Narne the elements present in
Common salt
(2 miks)
Hydrated copper (11) Sulphate.
(2 ks)
Sulphuric (VI) acid,
2 Why is a reaction between zinc metal and Nitric acid not suitable for preparing
hydrogen gae in the laboratory
(2 mi)
(1 m)
3.
What is relative atomic mass?
b)
Define 'isotopes
c)Determine the relative atomic mass of element K whose isotople misure occur in
the proportione:
(2 marks)
An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.
Answer:
U = -45.557kj
Explanation:
Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W
U=q + w .
We know that
w = PΔ V
where P is the pressure of
and V is the volume
then we can calculate the work
w = 35 atm * ( 8.20L - 2.21L)
W=35atm* 5.99L
W=209.65atmJ
But 1 atm = 101.325J
then ,
w = 209.65* 101.325 J = 21242.79 J
let us convert it to Kj
But we know that 1kJ = 10^3 J .
Then w = 21.243 kJ .
Then we can now calculate the internal energy as
U = 21.243- 66.8 kJ = -45.557kj
But we know that heat was released. Theeefore, the total internal energy change was -45.557kj
A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound
Answer:
Na₂SO₄
Explanation:
Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.
Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:
Na₂SO₄
I WILL GIVE BRAINLIEST
The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.94 grams of the compound were dissolved in 279.5 grams of diethyl ether, the solution began to boil at 35.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
Answer:
The correct answer is 179.94 g/mol.
Explanation:
Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.
The molecular weight of the compound can be determined by using the formula,
deltaTb = Kb * molality
Tb-To = Kb * molality
Tb-To = Kb*wt/mol.wt*1000/w (solvent)
35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g
0.6 = 2.02 * 53.45/ mol.wt
mol. wt = 2.02*53.45/0.6
mol. wt = 179.94 g/mol
Hence, the molecular weight of the compound is 179.94 gram per mol.
