Answer: Hợp chất CTHH 0 °C 10 °C 20 °C 30 °C 40 °C 50 °C 70 °C
Actini(III) hydroxide Ac(OH)3 0,0022
Amonia NH3 1176 900 702 565 428 333 188
Amoni azua NH4N3 16 25,3 37,1
View 42 more rows
hehe
What happens if we put raw eggs in a pot full of hot oil?
Which two layers are part of the thermosphere?
O exosphere and ionosphere
O ionosphere and mesosphere
mesosphere and stratosphere
O stratosphere and troposphere
The two layers are part of the thermosphere are exosphere and ionosphere.
What is the exosphere layer?The Exosphere is the topmost layer of the Earth's atmosphere.
and its gradually disappear into the vacuum of space.
It consist two parts that are:
exosphere and ionosphere.
Thus, option "A" is correct, the rest of the option is not a part of thermosphere.
To learn more about atmospheric layers click here:
https://brainly.com/question/25881294
#SPJ2
Answer:
hi hope your doing great the answer is A
Explanation:
its on Edge 2020
hope i helped :)
How many moles of Cl− are in 5.76 mg of FeCl3?
Answer:
0.0061650760770388 mole
What are the conjugate acid-base pairs in the following chemical reaction? HBr(aq)+ CH3COOH(aq) ⇌ CH3C(OH)2+(aq) + Br-(aq)
Answer:
HBr, CH3C(OH)2 and CH3COOH, Br-
Explanation:
The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.
Conjugate acid is formed when a bases receives a proton (H+) and a conjugate base is formed when an acid losses a proton (H+).
From the given equation:
HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.
How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.0135
Answer:
C) 0.027
Explanation:
In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:
[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]
We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).
Now we can balance the reaction:
[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]
If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]
[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]
In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:
[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]
I hope it helps!
Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original valueH2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.
Answer: [tex]\Delta S[/tex] = 473.92J/K.mol
Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:
[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]
The balanced reaction:
[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]
gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:
3.2 moles of water is used;
4.8 moles of hydrogen gas is formed;
1.6 moles of sulfur dioxide is also formed;
Calculating entropy change:
[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]
[tex]\Delta S=628.8+398.08-328.96-224[/tex]
[tex]\Delta S[/tex] = 473.92J/K.mol
Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol
Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction
Answer:
d. Oxidation and reduction
Explanation:
For this question we have to remember the definition of each type of reaction:
-) Hydrogenation
In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.
-) Alkylation
In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.
-) Hydrolysis
In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.
-) Halogenation
In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.
-) Ammoniation
In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.
-) Oxidation and reduction
In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:
[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction
[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation
03/08/2020
Question
1. (a) State the definition of a chemical formula.
(b) What does it tell about a compound?
(c) What information is conveyed by the formulation H2SO4?
Explanation:
According to your question,
no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.
b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.
c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).
Hope it helps...
The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
A laboratory assistant needs to prepare 217 mL of 0.246 M solution. How many grams of calcium chloride will she need
Answer:
5.92 g
Explanation:
Convert milliliters to liters.
217 mL = 0.217 L
Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.
0.217 L × 0.246 M = 0.05338 mol
Now, convert the moles to grams using the molar mass. The molar mass of calcium chloride is 110.98 g/mol.
0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g
You will need 5.92 g of calcium chloride.
If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water. a. What is the molality of sodium hydroxide in this solution
Answer:
2.77 mol/kg
Explanation:
Molality is a sort of concentration that indicates the moles of solute in 1kg of solvent. In this case our solvent is water and, if we consider water's density as 1g/mL, we determine that the mass of solvent is 750 g.
We convert the mass to kg → 750 g . 1kg /1000g = 0.750 kg
Our solute is the NaOH → 83 g.
We convert the mass to moles → 83 g . 1mol /40g = 2.075 mol
Molality (mol/kg) = 2.075 mol / 0.75kg = 2.77 m
Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significant digits.
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
The mole is a counting number that allows scientists to describe how individual molecules and atoms react. If one mole of atoms or molecules is equal to 6.022 x 10^32 atoms or molecules, how many molecules are in 23.45 g sample of copper (II) hydroxide, Cu(OH)2? Express your answer to the correct number of significant figures. (MM of Cu(OH)2 is 97.562g/mol. Be sure to show all steps completed to arrive at the answer.
Answer:
[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]
Explanation:
You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.
1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]
2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].
The moles of a compound has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]
The moles in 23.45 g copper (II) hydroxide has been:
[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]
The moles of copper (II) hydroxide has been 0.24 mol.
The number of molecules in 0.24 mol sample has been driven by:
[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
For more information about molecules in a mole of sample, refer to the link:
https://brainly.com/question/24577356
Identify four general properties that make an NSAID unique as compared to the NSAID aspirin. List specific properties that make aspirin, naproxen, and ibuprofen unique from one another
Answer:
NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).
These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.
The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.
On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.
NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.
In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.
That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.
Explanation:
Both types of drugs generate the same effect but by different mechanisms.
Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.
NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.
Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.
If we want to change a gas to its liquid state, should we add or remove energy from the gas?
In which ONE of the following compounds would the bonding be expected to have the highest percentage of ionic character? A) LiBr B) CsCl C) BaBr2 D) NaCl E) KI
Answer:
B) CsCl
Explanation:
The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.
In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).
The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.
HELP!!!!! Substance A and substance B are mixed together. To separate the mixture, water is added, and substance A is filtered out. Then, the remaining liquid is heated to remove the water, leaving a residue of substance B. Which statement about substance A and substance B could be correct?
A. Substance A is rice, and substance B is sugar.
B. Substance A is alcohol, and substance B is salt.
C. Substance A is sand, and substance B is alcohol.
D. Substance A is sugar, and substance B is instant coffee.
The statement that "substance A is rice, and substance B is sugar" is correct.
WHAT IS A MIXTURE?
A mixture in chemistry is a substance that contains two or more different substances. One notable characteristics of a mixture is that it can easily be separated using physical means.According to this question, a mixture contains two substances A and B. Water is added to this mixture for easy separation.
Substance A is filtered out. This means that substance A is a solid that is insoluble in water. This substance is RICE. The remaining liquid is heated to remove the water content leaving a residue of substance B. This shows that substance B is also a solid but soluble in water. SUGAR best fits this description.Therefore, in the mixture of two substances A and B, substance A is rice, and substance B is sugar.
Learn more at: https://brainly.com/question/6594631
Question 14 (5 points)
What's the acid ionization constant for an acid with a pH of 2.11 and an equilibrium
concentration of 0.30 M?
O A) 4.87x10-8
B) 1.99x10-6
C) 3.32x10-4
OD) 2.01x10-4
Answer:
D) 2.01 x 10⁻⁴ .
Explanation:
pH = 2.11
[ H⁺ ] = [tex]10^{-2.11}[/tex]
Let the acid be HA
It will ionise as follows .
HA ⇄ H⁺ + A⁻
in equilibrium .30 [tex]10^{-2.11}[/tex] [tex]10^{-2.11}[/tex]
Acid ionisation constant Ka = [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]
= 2 x 10⁻⁴
Answer:
D) 2.01 x 10⁻⁴ is correct!
Explanation:
I got it in class!
Hope this Helps!! :))
How many grams of PtBr4 will dissolve in 250.0 mL of water that has 1.00 grams of KBr dissolved in it
Answer:
[tex]m_{PtBr_4}=0.306gPtBr_4[/tex]
Explanation:
Hello,
In this case, since the solubility product of platinum (IV) bromide is 8.21x10⁻⁹, and the dissociation is:
[tex]PtBr_4(s)\rightleftharpoons Pt^{4+}(aq)+4Br^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Pt^{4+}][Br^-]^4[/tex]
Thus, since the salt is added to a solution initially containing 1.00 grams of potassium bromide, there is an initial concentration of bromide ions:
[tex][Br^-]_0=\frac{1.00gKBr*\frac{1molKBr}{119gKBr}*\frac{1molBr^-}{1molKBr} }{0.250L}=0.0336M[/tex]
Hence, in terms of the molar solubility [tex]x[/tex], we can write:
[tex]8.21x10^{-9}=(x)(0.0336+4x)^4[/tex]
In such a way, solving for [tex]x[/tex], we obtain:
[tex]x=0.00238M[/tex]
Which is the molar solubility of platinum (IV) bromide. Then, since its molar mass is 514.7 g/mol, we can compute the grams that get dissolved in the 250.0-mL solution:
[tex]m_{PtBr_4}=0.00238\frac{molPtBr_4}{1L}*0.250L *\frac{514.7gPtBr_4}{1molPtBr_4} \\\\m_{PtBr_4}=0.306gPtBr_4[/tex]
Best regards.
If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.
Answer:
1 L
Explanation:
ppm means parts per million. Generally the relationship between mass and litre is given as;
1 ppm = 1 mg/L
This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.
The argon atoms are excited into an excited state before emitting the 488.0 nm laser. It is known that the energy of the first ionization energy of argon is 1520 kJ mol-1. What is the energy level of the excited state (in unit eV) lies below the vacuum energy level (0 eV)
Answer:
Explanation:
Given that:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser.
the energy of the first ionization energy of argon is 1520 kJ mol-1.
SInce 1 eV = 96.49 kJ/mol
Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV
= 15.75 eV
To find where the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E =4.057 \times 10^{-19} \ J[/tex]
Converting Joules (J) to eV ; we get,
[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]
E = 2.53 eV
The energy levels of the first exited state = -13.223 eV
equal volumes 50.0 ml of 0.54897 m solution of HCL and 0.9573 M NaOH solution have an initial temperature of 18.5 degrees Celsius react in a coffee cup calorimeter the resulted Solutions records a temperature of 22.9 degrees Celsius since the solutions are mostly water the solution are assumed to have a density of 1.0 g per ml and a specific heat of 4.184 joules per gram degree celsius the heat capacity of the Colorimeter is 1.0 x 10 to power 1 joules per degrees Celsius calculate the heat and gained or released by the solution at a constant pressure in J
Answer:
Explanation:
ok
1. What are some examples of forces found in everyday life? (Give examples from the bike picture
above OR make up your own examples.)
2 Define the word "motion"...
3. What do you think the following phrase from the definition means with respect to time and in
comparison to the position of other objects used as reference points?"
4. Define the word "Yorco...
5. Give an example from everyday life of an object that is "in motion (there is no "correct answer for
this - all reasonable answers accepted).
6. What is an example from everyday life of a force acting on an object? (there is no correct answer"..)
i need help on the questions
Answer:
1.answer = pushing force
= pulling force
=gravity force
Choose the most correct answer: The endothermic (∆H > 0) reaction:
a) Cannot occur at all temperature.
b) Can occur with the positive ∆S at high temperature.
c) Can occur with the negative ∆S at low temperature.
d) Cannot occur with the positive ∆S at high temperature.
Answer:
B
Explanation:
In order to create spontaneity, an endothermic process has to occur along with positive entropy and high temperature
The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
Which of the following solutions is acidic? [H3O+] = 1.0 x 10-10 M [H3O+] < 1.0 x 10-7 M [OH-] = 1.0 x 10-10 M [OH-] = 1.0 x 10-7 M [OH-] > 1.0 x 10-7 M
Answer:
[OH-] = 1.0 x 10-10 M
Explanation:
The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.
Generally, for a neutral solution we have;
[H3O+] = [OH-] = 1.0 x 10-7 M
For an acidic solution;
[H3O+] > 1.0 x 10-7 M
[OH-] < 1.0 x 10-7 M
Comparing the options the correct option is;
[OH-] = 1.0 x 10-10 M
Suppose a small amount of a solid is added to water and, after a short time, all the solid has dissolved. Which of the following statements is most likely to be true?
A) The solution is supersaturated with solute.
B) The solution is saturated with solute.
C) The solution is unsaturated with solute.
D) The solution is either unsaturated or supersaturated with solute.
E) The solution is either saturated or supersaturated with solute.
Answer:
the option e is correct I think
The solution is either saturated or supersaturated with solute
Answer:
the option e is correct I think
The solution is either saturated or supersaturated with solute
Explanation:
2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing two similarities and two differences.
Answer:
hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
answer : It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
They have different Geometry
They differ in bond angles as well
Explanation:
The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
comparing(similarities) these two generic molecules
It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
differences between the generic molecules
They have different Geometry
They differ in bond angles as well
A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is the molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl
Answer:
mol LiCl = 4.83 m
Explanation:
GIven:
Solution of LiCl in water XLiCl = 0.0800
Mol of water in kg = 55.55 mole
Find:
Molality
Computation:
mole fraction = mol LiCl / (mol water + mol LiCl)
0.0800 = mol LiCl / (55.55 mol + mol LiCl)
0.0800 mol LiCl + 4.444 mol = mol LiCl
mol LiCl - 0.0800 mol LiCl = 4.444 mol
0.92 mol LiCl = 4.444 mol
mol LiCl = 4.83 m