Answer:
Glucose: (a) moves into sac
Water: (b) moves out of sac
Albumin: (c) does not move
NaCl: (a) moves into sac
Explanation:
A semi-permeable membrane is a membrane that allow certain molecules or ions to pass through by diffusion.
Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration until equilibrium is attained. A special form of diffusion known as osmosis transports water molecules across a semi-permeable membrane.
Osmosis is the movement of water molecules from a region lower solute concentration (high water molecules concentration) to a region of higher solute concentration (low water molecules concentration) until equilibrium is attained.
From the above definitions and the given data;
Glucose concentration is higher in solution outside the sac, thus, glucose molecules will move into the sac.
Water molecules are higher in the sac as a result of the lower concentration of solutes, therefore, water molecules will move out of the sac into the solution outside.
Since the sac is impermeable to albumin, it does not move.
NaCl concentration is lower in the sac, therefore, it will move ro the solution outside into the sac.
Glucose will move into the sac, water will move out of the sac, albumin will neither move in nor out, and NaCl will move into the sac.
The molecules of each of the substances will move by diffusion from the region of higher concentration to the region of lower concentration of each substance as long as there is a permeable membrane. Water, on the other hand, will move by osmosis from the region of high to low water potential through a permeable membrane. Regions with higher concentrations of substances usually have low water potentials and vice versa.Thus, both glucose and NaCl molecules will diffuse from the solution into the sac, and water molecules will move from the sac into the surrounding solution. Since the sac is not permeable to albumin, then the movement in or out is inhibited.
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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.
Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=39.61g[/tex]
Mass of [tex]H_2O=9.01g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.
For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]
For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]
For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 9 : 10 : 1
Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
We are given:
Mass of molecular formula = 268.34 g/mol
Mass of empirical formula = 134 g/mol
Putting values in above equation, we get:
[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]
Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].
What can be known about the salt sample that Gerry is looking at?
Answer:
That its small pointed. Pink(Himalayan salt)or white(normal salt)
Explanation:
Summa dees questions are so stupid, deys makin me salty.
One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the CO2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 107 CO2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?
Answer:
Enzyme is carbonic anhydrase
Substrate is [tex]CO_2[/tex]
Turnover number is [tex]10^{7}[/tex]
Explanation:
An enzyme is used by a living organism as a catalyst to perform a specific biochemical reaction.
A substrate is a molecule upon which an enzyme acts.
Turnover number refers to the number of substrate molecules transformed by a single enzyme molecule per minute. Here, the enzyme is the rate-limiting factor.
Here,
Enzyme is carbonic anhydrase
Substrate is [tex]CO_2[/tex]
Turnover number is [tex]10^{7}[/tex]
A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute HCl, step 1? Group of answer choices
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is ammonia
Explanation:
The mixture contains two base compound which are
ammonia,
and diethylamine
Now the addition of HCl which is a strong acid in step 1 will cause the protonation of the two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below
[tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]
and
[tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]
Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
1. ______The force that keeps the nucleons bound inside the nucleus of an atom
A. Strong electrostatic force
B. Strong nuclear force
C. Strong centripetal force
D. Gravitational attraction
2._____The amount of energy needed to split the nucleus into individual protons and neutrons
A. Nuclide transfer energy
B. Nuclear binding energy
C. Mass energy equivalence
D. Nuclear energy
3._______ The difference between the mass of the nucleons and the mass of an Atom
A. Mass of nucleus
B. Mass defect
C. Atomic mass
D. Isotopic mass
Answer:
1). strong nuclear force 2). nuclear binding energy 3), mass defect
Explanation:
Right on Edge
1. Strong nuclear force the force that keeps the nucleons bound inside the nucleus of an atom.
2. Nuclear binding energy the amount of energy needed to split the nucleus into individual protons and neutrons.
3. Mass defect the difference between the mass of the nucleons and the mass of an Atom.
What is strong nuclear force ?The term strong nuclear force is defined as the force that binds protons and neutrons together. It also binds them all together in a nucleus and is responsible for the energy released in nuclear reactions.
The examples of strong nuclear force are the force that hold protons and neutrons in nuclei of atoms. The elements' greater than the hydrogen atom. The fusion of hydrogen into helium in the sun's core.
Thus, 1. option B, 2. option B and 3. option B is correct.
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Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3857 kJ/mol Given that ΔH°f[CO2(g)] = –393.5 kJ/mol and ΔH°f[H2O(l)] = –285.8 kJ/mol, calculate the enthalpy of formation of glycine.
Answer:
ΔH°f C₂H₅O₂N(s) = -537.2kJ
Explanation:
Based on the reaction:
4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)
ΔHrxn = ΔH°f products - ΔH°f reactants.
As:
ΔH°fO₂(g) = 0
ΔH°fCO₂(g) = -393.5kJ/mol
ΔH°fH₂O(l) = -285.8kJ/mol
ΔH°fN₂(g) = 0
The ΔHrxn is:
ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol
ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4
ΔH°f C₂H₅O₂N(s) = -537.2kJiron oxide + oxygen equals to ?
Answer:
It's ferric oxide Fe2O3
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...
Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of the following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.
Answer: (e) The pressure in the container increases but does not double.
Explanation:
To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant
The pressure in the container increases but does not double.
At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.
Number of moles of He = 10 g/4 g/mol = 2.5 moles
Number of moles of Ne = 10 g/20 g/mol = 0.5 moles
We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.
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What is the systematic name of the following compound?
Mn3(PO4)2
The polyatomic ion phosphate has the formula PO
Answer:
Manganese(II) phosphate | Mn3(PO4)2 - PubChem
Answer:
Magnese(ll) posphate M23 (p042) Molecular weight.
That is what the leters stand for!
IF THIS HELPED AND IF YOU DON'T MIND CAN YOU PLEASE MARK ME BRAINLIEST?8) What is the molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22011) in 35.5 mL of solution?
A) 3.52 M
B) 1.85 x 10-2M
C) 0.104 M
D) 0.0657 M
E) 1.85 M
Answer:
E) 1.85 M
Explanation:
M(C12H22O11) = 342.3 g/mol
22.5 g * 1mol/342.3 g = 0.0657 mol
35.5 mL = 0.0355 L
Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M
The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
From the question,
We are to determine the molarity (that is, concentration) of the given sucrose solution
First, we will determine the number of moles present in the given mass of sucrose
Mass of sucrose = 22.5 g
Using the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of sucrose = 342.2965 g/mol
∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]
Number of moles of sucrose present = 0.0657325 moles
Now, for the molarity (concentration) of the sucrose solution
From the formula
Number of moles = Concentration × Volume
Then,
[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]
From the question,
Volume = 35.5 mL = 0.0355 L
∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]
Concentration = 1.85 M
Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
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Write the limiting forms (or Canonical forms) of the following ions:
i. H3O+
, ii. CO3
2-
, iii. NO3-
Answer:
Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.
Explanation:
Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.
There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.
When you turn on the air conditioner during a hot summer day the cooler air will sink to the floor, while warmer air rises to the
ceiling
Which type of heat transfer is this an example of?
(A) conduction
(B) convection
(C) radiation
(D)
kinetic
tertbutylamine and ammonia. Which is more basic
Answer:
ammonia
Explanation:
Nitroglycerin, an explosive, decomposes according to the following equation 4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) Calculate the total volume of gases produced when collected at 1.45 atm, and 18.0°C from 2.70 × 102 g of nitroglycerin.
Answer:
6.65dm³
Explanation:
Equation of reaction,
4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)
From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.
Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g
Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.
908g of C₃H₅(NO₃)₃ = 908g of products
2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products
Number of moles = mass / molar mass
Molar mass of C₃H₅(NO₃)₃ = 908g/mol
Number of moles = 2.70×10² / 908
Number of moles = 0.297 moles
But 1 mole = 22.4dm³
0.297mole = x dm³
x = (0.297 × 22.4) / 1
x = 6.65dm³
The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³
A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of diethylamine, tetrahydrofuran, chloroform, ethanolamine, and acetone. First, from her collection of Material Safety Data Sheets (MSOS), the chemist finds the following information:
liquid density
diethylamine 1.1 gcm-3
tetrahydrofuran 0.7 9gcm-3
chloroform 0.71 gcm-3
ethanolamine 0.89 gcm-3
acetone 1.6 gcm-3
Next, the chemist measures the volume of the unknown liquid as 0.767 L and the mass of the unknown liquid as 682 g.
1. Calculate the density of the liquid.
2. Given the data above, is it possible to identify the liquid?
3. If it is possible to identify the liquid, do so.
a. dimethyl sulfoxide.
b. acetone.
c. diethylamine.
d. tetrahydrofuran .
e. carbon tetrachloride
Answer:
1. density = 0.89 g/cm3
2. Yes is possible to identify the liquid
3. ethanolamine
Explanation:
Data:
mass = 682 g
volume = 0.767 L = 767 mL or cm3
1.
To calculate the density of the liquid it is necessary to know that the density formula is:
[tex]density=\frac{mass(g)}{volume(cm^{3}) }[/tex]
The data obtained is replaced in the formula:
[tex]density=\frac{682g)}{767(cm^{3}) }=0.89\frac{g}{cm^{3} }[/tex]
2.
With the given data it is possible to identify the liquid, this because the density value is a basic property of each liquid.
3.
It is possible to determine what liquid it is, since when comparing the value obtained with those reported in the collection of Material Safety Data Sheets (MSOS), the value that agrees is that of ethanolamine.
A base has a molarity of 1.5 M with respect to the hydroxyl ion (OH-) concentration. If 7.35 cm³ of this base is taken and diluted to 147 cm³, then what is the concentration of the hydroxyl ion. How many moles of hydroxyl ion are there in the 7.35 cm³? In the 147 cm³?
Answer:
0.077M is the concentration of the hydroxyl ion
Explanation:
If 7.35 cm3 of this base is take and diluted to 147 cm3, then what is the concentration of the hydroxyl ion?
Use the dilution equation:
M1V1 = M2V2
M1 * 147cm³ = 1.5 M * 7.35 cm³
M1 = 1.5 M * 7.35 cm³ / 147 cm³
M1 = 0.077 M
0.077M is the concentration of the hydroxyl ionHow many moles of hydroxyl ion are there in the 7.35 cm3?
1000 cm³ contains 1.5 mol OH- ions
7.35 cm³ contains : 7.35 cm³ / 1000 cm³ *1.5 mol
= 0.011025 mol
Answer correct to 2 significant digits = 0.011 mol OH- ions.Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?
Answer:
ΔH = - 44.0kJ
Explanation:
H2O(l)→H2O(g), ΔH =44.0kJ
In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH is positive.
If he reaction is reversed, we have;
H2O(g)→H2O(l)
In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still have the same value.
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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How many moles of H2 are needed to produce 34.8 moles of NH3?
2 i hope this helps
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According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol.
Answer:
Mass of Al2S3 remaining is 17.212 g
Explanation:
Equation of the reaction is given below:
Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S
From the balanced equation above
6 mole of H20 reacts with 1 mole of Al2S3
i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3
= 108.12 g of H2O reacts with 150.71 g of Al2S3
Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3
= 2.788 g of Al2S3
Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g
According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.
Given the following data:
Mass of [tex]AL_2S_3[/tex] = 20.00 gramsMass of [tex]H_2O[/tex] = 2.00 gramsMolar mass of [tex]AL_2S_3[/tex] = 150.17 g/molMolar mass of [tex]H_2O[/tex] = 18.02 g/mol.To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:
First of all, we would write a properly balanced chemical equation for this chemical reaction.
[tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]
By stoichiometry:
1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]
Next, we would calculate the mass of each compound.
For [tex]AL_2S_3[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]
Mass = 150.17 grams
For [tex]H_2O[/tex]:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]
Mass = 108.12 grams
108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]
2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]
Cross-multiplying, we have:
[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]
X = 2.78 grams of [tex]AL_2S_3[/tex]
Remaining mass = [tex]20.00 - 2.78[/tex]
Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]
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please help!!!! Chem question
Answer : The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The given balanced ionic equation will be,
[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]
In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?
Rate = K[A]2[B]2
A. 1.6 (mol/L)/s
B. 0.8 (mol/L)/S
C. 0.2 (mol/L)/S
D. 0.4 (mol/L)/S
Answer:
D. 0.4 (mol/L)/S
Explanation:
You simply have to plug in the given values into the rate law.
Rate = k[A][B]
Rate = (0.1)(1)²(2)²
Rate = (0.1)(1)²(4)²
Rate = 0.4
The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion
Answer:
[Cd²⁺] = 2.459x10⁻⁶M
Kf = 9.96x10⁶
Explanation:
Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸
The ksp of CdC₂O₄ is:
CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)
ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸
As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:
[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]
[Cd²⁺] = 2.459x10⁻⁶MAll Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:
[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M
[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M
[Cd(NH₃)₄²⁺] = 6.098x10⁻³MIn the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:
[0.150M] = [NH₃] + 0.024M
0.1256M = [NH₃]The equilibrium of the complex formation is:
Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺
The kf, formation constant, is defined as:
Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴
Replacing:
Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴
Kf = 9.96x10⁶At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)
Answer:
The correct answer is 2.016 x 10⁻¹⁷
Explanation:
We have the following chemical reactions and their equilibrium constants (K):
(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
And we have to obtain K for the following reaction:
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)
If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.
H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
+
HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
-------------------------------------------------------------
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq) K= K₁ x K₂
K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷
What is the atomic mass of AlNO2?
Answer:
I am not sure, but I think this is the answer 72.987 g/mol
Describe why some acids are strong while other acids are weak
Answer:
I hope this help you. Mark me as brainliest and rate pleaseExplanation:
the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.
As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.
It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.
An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction
Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]
Now, we can balance the reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]
Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:
1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).
2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).
3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).
[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]
Finally, we can calculate the yield percent:
[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]
I hope it helps!
The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%
We'll begin by writing the balanced equation for the reaction. This is given below: [tex]Na_{2}CO_{3} + CaCl_{2} - > CaCO_{3} + 2NaCl[/tex]Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
SUMMARY
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]
= 93.2%Therefore, the percentage yield of the reaction is 93.2%
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