the mass percent of sugar in this solution is 46%.
Answer:
Solution given:
mass of solute=34.5g
mass of solvent=75g
mass percent=[tex]\frac{mass\:of\:solute}{mass\:of \:solvent}*100\%[/tex]
=[tex]\bold{\frac{34.5}{75.0}*100\%=46\%}[/tex]
We know that,
[tex]mass \: solute = \frac{mass \: of \: solute}{mass \: of \: solvent} [/tex]×100
[tex] \frac{34.5}{75.0} = 0.46[/tex]
hope it helps..
Which acid or base (along with its corresponding salt) should be used to generate a buffer solution with pH around 3.5
Answer:
Formic acid
Sodium formiate
Explanation:
To determine acid or base that can generate a buffer solution with pH around 3.5, we have to think in the acid whose pKa = pH.
Although we have to also think in buffer capacity, a measure which can cause a change of 1 pH unit in 1 L of solution.
Buffer capacity does not only depend on the concentration of its components but also of the relationship between that concentrations.
When pH = pKa, buffer capacity is maximum which means that the concentration of conjugated species is the same and the ability to oppose pH changes is maximum.
One example with pH = pKa or nearly if:
COOH⁻ + Na⁺ → NaCOOH
HCOOH + H₂O → COOH⁻ + H₃O⁺ Ka: 1.8×10⁻⁴
pKa = 3.74
0.28 M Ca(NO3)2
Express your answer using two significant figures.
Answer:
Mass=Moles × RFM
Mass= 0.28M× 164
Mass= 45.92 grammes
3. At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
be if the temperature were changed to 22 C and the pressure to 1.25 atmospheres
Answer:
The volume will be 185.83 mL.
Explanation:
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T} =k[/tex]
Where P = pressure, T = temperature, k = Constant
Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. Boyle's law is expressed mathematically as:
P*V=k
Where P = pressure, V = volume, k = Constant
Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is kept at a constant pressure. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement of the gas molecules. .
In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
[tex]\frac{V}{T} =k[/tex]
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
[tex]\frac{P*V}{T} =k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
In this case:
P1= 720 torr V1= 256 mLT1= 35 C= 308 K (being 0 C= 273 K)P2= 1.25 atm= 950 torr (being 1 atm= 760 torr)V2= ?T2= 22 C= 295 KReplacing:
[tex]\frac{720 torr*256 mL}{308 K} =\frac{950 torr*V2}{295 K}[/tex]
Solving:
[tex]V2= \frac{295K}{950 torr} *\frac{720 torr*256 mL}{308 K}[/tex]
V2= 185.83 mL
The volume will be 185.83 mL.
the pressure on a 205 mL volume of gas is decreased from 985 mm hg to 615 mm hg while constant temperature is maintained, What is the new volume of the gas?
Answer:
[tex]\boxed {\boxed {\sf 328 \ mL}}[/tex]
Explanation:
The pressure on the gas changes, while the temperature remains constant, and we want to find the new volume. So, we will use Boyle's Law, which states there is an inverse relationship between the pressure on a gas and the volume of the gas. The formula is:
[tex]{P_1V_1}= P_2V_2[/tex]
The pressure of the gas is originally 985 mm Hg and the volume is 205 milliliters.
[tex]985 \ mm \ Hg * 205 \ mL = P_2V_2[/tex]
The pressure is decreased to 615 mm Hg, but the new volume is unknown.
[tex]985 \ mm \ Hg * 205 \ mL = 615 \ mm \ Hg * V_2[/tex]
We are solving for the new volume, so we must isolate the variable V₂. It is being multiplied by 615 millimeters of mercury.. The inverse of multiplication is division, so we divide both sides of the equation by 615 mm Hg.
[tex]\frac {985 \ mm \ Hg * 205 \ mL}{615 \ mm \ Hg} = \frac{615 \ mm \ Hg * V_2}{615 \ mm \ Hg}[/tex]
[tex]\frac {985 \ mm \ Hg * 205 \ mL}{615 \ mm \ Hg} = V_2[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]\frac {985 \ * 205 \ mL}{615 } = V_2[/tex]
[tex]\frac{201925 }{ 615} \ mL = V_2[/tex]
[tex]328.3333333 \ mL = V_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.
The 3 to the right in the tenths place tells us to leave the 8 in the ones place.
[tex]328 \ mL \approx V_2[/tex]
The new volume of the gas is approximately 328 milliliters.
A 2.584 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 5.874 g CO2 and 2.404 g H2O. What mass of oxygen is contained in the original sample?a. 0.7119 g.b. 0.8463 g.c. 0.29168 g.d. 0.1793 g.e. 0.6230 g.
Answer:
a. 0.7119 g
Explanation:
To solve this question we need to know that all carbon of the compound will react producing CO2 and all Hydrogen producing H2O.
Thus, we can find the mass of C and the mass of H and by difference regard to the 2.584g of the compound we can find the mass of oxygen as follows:
Moles CO2 = Moles C -Molar mass: 44.01g/mol-
5.874g CO2 * (1mol/44.01g) = 0.1335 moles CO2 = 0.1335 moles C
Mass C -Molar mass: 12.01g/mol-:
0.1335 moles C * (12.01g /mol) = 1.6030g C
Moles H2O -Molar mass: 18.01g/mol-
2.404gH2O * (1mol / 18.01g) = 0.1335 moles H2O * (2mol H / 1mol H2O) = 0.267 moles H
Mass H -Molar mass: 1g/mol-
0.267 moles H * (1g/mol) = 0.2670g H
Mass Oxygen =
Mass O = 2.584g compound - 1.6030g C - 0.2670g H
Mass O = 0.714g O ≈
a. 0.7119 gWhich technique is best suited to each application?
a. In the second week of a four week biochemistry experiement, you have 50 fractions collected from a gel filtration column to determine which fractions contain lactate dehydrogenase. You are given only 400 uL of 0.100 mg/mL lactate dehydrogenase to prepare your calibration curve. 96-well microplate
b. Your environmental lab has 2000 samples to be analyzed for trace ammonia by next week. discrete analyzer.
c. Twenty water samples must be analyzed for Cl-, NH3, PO3-, and So during each work shift. flow injection analysis colorimeter.
d. Your professor heard you will be hiking the Appalachian Trail next summer. She asks you to collect 100-mL water samples from the ten streams with the highest concentration of phosphate.
Answer:
a. discrete analyzer
b. 96 well microplate
c. flow injection analysis
d. colorimeter
Explanation:
96 well microplates are instruments designed for sample collection and throughput screening. If an environment lab has collected 2000 samples then 96 well microplate is best suited application. Discrete analyzer is automated chemical analyzer which performs test on samples kept in discrete cells. Flow injection analysis is approach used for chemical analysis. It injects a plug of sample into a flowing carrier stream. Colorimeter is a device which measures absorbance of wavelength of light by a specific solution.
What kind of element is Phosphorus is
Answer:
NON-METAL
Explanation:
Phosphorus is a non-metal that sits just below nitrogen in group 15 of the periodic table. This element exists in several forms, of which white and red are the best known.
You prepare a solution by dissolving 25.3 g sucrose (C12H22O11) 705 mL of water. Calculate the molarity of the solution.
Answer:
0.105 M
Explanation:
First we convert 25.3 grams of sucrose into moles, using sucrose's molar mass:
Molar Mass of C₁₂H₂₂O₁₁ = 342.3 g/mol25.3 g C₁₂H₂₂O₁₁ ÷ 342.3 g/mol = 0.0739 mol C₁₂H₂₂O₁₁Now we calculate the molarity of the solution, using the given volume and the calculated number of moles:
Converting 705 mL ⇒ 705 mL / 1000 = 0.705 LMolarity = 0.0739 mol / 0.705 L = 0.105 Mg Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of of acetylene. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
3.0 mol CO₂
Explanation:
Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of 1.5 moles of acetylene. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Step 1: Write the balanced combustion reaction
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of C₂H₂ to CO₂ is 1:2.
Step 3: Calculate the moles of CO₂ produced from 1.5 moles of C₂H₂
1.5 mol C₂H₂ × 2 mol CO₂/1 mol C₂H₂ = 3.0 mol CO₂
2) If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm:
a) What is the volume of the brick?
b) If the brick has a mass of 895.3 g, what is its density?
Answer:
a. 599 cm³
b. 1.49 g/cm³
Explanation:
A. VolumeVolume is the amount of space an object occupies. Since this is a brick, the object is a rectangular prism. The formula for the volume of a rectangular prism is the product of length, width, and height.
[tex]V= l *w*h[/tex]
The brick's length (l) is 13.77 centimeters, the width (w) is 8.50 centimeters, and the height (h) is 5.12 centimeters. Substitute these values into the formula.
[tex]V= 13.77 \ cm * 8.50 \ cm * 5.12 \ cm[/tex]
Multiply the numbers together.
[tex]V= 117.045 \ cm^ 2* 5.12 \ cm[/tex]
[tex]V= 599.2704 \ cm^3[/tex]
The original measurements have at least 3 significant figures, so our answer must have 3. For the number we calculated, that is the ones place. The 2 in the tenths place tells us to leave the 9 in the ones place.
[tex]V \approx 599 \ cm^3[/tex]
[tex]\bold {The \ volume \ of \ the \ brick \ is \ approximately \ 599 \ cubic \ centimeters}}[/tex]
2. DensityDensity is the amount of matter in a specified space. The formula for density is mass over volume.
[tex]d= \frac{m}{v}[/tex]
The mass of the brick is 895.3 grams and we just found the volume to be 599.2704 cubic centimeters. Substitute the values into the formula.
[tex]d= \frac{895.3 \ g}{599 \ cm^3}[/tex]
Divide.
[tex]d= 1.494657763 \ g/cm^3[/tex]
Round to three significant figures. For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 9 in the hundredth place.
[tex]d \approx 1.49 \ g/cm^3[/tex]
[tex]\bold {The \ density\ of \ the \ brick \ is \ approximately \ 1.49 \ grams /cubic \ centimeters}}[/tex]
A mixture of 1-heptyne, 2-heptyne, and 3-heptyne was hydrogenated in the presence of a palladium catalyst until hydrogen uptake ceased. How many seven-carbon hydrocarbons were produced
Answer:
four
Explanation:
Recall that the hydrogenation of 1-heptyne, 2-heptyne, and 3-heptyne is an addition reaction. The hydrogen is added across the triple bond in the presence of palladium serving as the catalyst.
Note that 1-heptyne and 2-heptyne do not exist as geometric isomers hence they yield one corresponding hydrocarbon each. However, 3-heptyne exhibits geometric isomerism. As a result of this, 3-heptyne yields cis and trans products making a total of four seven-carbon hydrocarbons produced.
So, the number of seven carbon hydrocarbons produced is one.
1-heptyne, 2-heptyne, and 3-heptyne
Note that if there is excess hydrogen, then all triple bonds will be substituted by hydrogen so only heptane is formed, double bonds and tirple bonds disappear to
[tex]CH_3-(CH_2)_5-CH_3[/tex]
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The tabulated data show the rate constant of a reaction measured at several different temperatures. Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction.
Temperature (K) Rate Constant (1/s)
300 0.0134
310 0.0407
320 0.114
330 0.303
340 0.757
Required:
Use an Arrhenius plot to determine the activation barrier for the reaction.
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Can someone help me with problem 21.105? Many thanks!
What is the molecular geometry of CIO3F as predicted by the VSEPR model?
Multiple Choice
trigonal pyramidal
square planar
square pyramidal
tetrahedral
octahedral
Explanation:
since there are no lone pairs on the central atom, the shape will be tetrahedral
How must the hydrogen in an acid be attached (ionically or covalently) to be an acid?
Answer:
Search it online it will help
What element
is a basic element of life
Answer:
Oxygen is the main element of life.
Explanation:
Because Oxygen allows us to breathe..
I hope this helps you!
XoXo Makayla
Calculate the boiling point of a 3.5 % solution (by weight) of sodium chloride in water.
Kb of H2O = 0.512 oC/M
Answer: The boiling point of the solution is [tex]101.02^oC[/tex]
Explanation:
We are given:
3.5 % (by weight) NaCl
This means that 3.5 g of NaCl is present in 100 g of solution
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent (water) = (100 - 3.5) g = 96.5 g
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (water) = [tex]100^oC[/tex]
Boiling point of solution = ?
i = Vant Hoff factor = 2 (for NaCl)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]0.512^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 3.5 g
[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 36.5 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent (water) = 96.5 g
Putting values in equation 1, we get:
[tex]\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC[/tex]
Hence, the boiling point of the solution is [tex]101.02^oC[/tex]
What is the molarity of a solution prepared by dissolving 12.0 g of potassium permanganate, KMnO4, in water to make 250.0 mL of solution
Answer:
Explanation:
Molecular weight of potassium permanganate is 158 g .
12 g = 12 g / 158 g moles
= .076 moles .
250 mL = .25 L .
Molarity of solution = moles dissolved / volume of solution
= .076 / .25 L
= .304 M .
Molarity of solution is .304 M.
The molarity of a solution prepared by dissolving 12.0 g of potassium permanganate, KMnO4, in water to make 250.0 mL of solution will be 0.304 M. It can be calculated by using mole formula.
What is molarity ?The amount of material in a given volume of the given solution has been measured in molarity (M). The moles of a particular solute per liter of a solution can be known as molarity. The molar solution of known concentration is indeed known as molarity.
What is mole?A mole is just the quantity of a material that includes exactly 6.022 × [tex]10^{23}[/tex] of the particular chemical elementary entities.
Molecular weight potassium potassium permanganate = 12 g
Volume of the solution = 250mL = 25 L.
Calculation of mole:
Mole can be calculated by using the formula.
Mole = mass / molar mass
Mole = 12 g / 158 g
Mole = 0.076 mole.
Calculation of molarity:
Molarity can be calculated by using the molarity formula:
Molarity = moles of dissolved / volume of solution
Molarity = 0.076 / 0.25
Molarity = 0.304 M
Therefore, the molarity of the solution will be 0.304 M.
To know more about molarity and mole
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Carbon monoxide, a product of combustion, is a toxic gas that has an extremely high affinity for hemoglobin (much higher than that of oxygen for hemoglobin); consequently, as soon as it dissolves in the liquid part of blood at low partial pressure, it diffuses quickly into red blood cells and binds to hemoglobin. In carbon monoxide (CO) poisoning, even with very low partial pressure of inspired CO, CO rapidly binds to hemoglobin (Hgb), leaving a lower fraction of oxygen binding sites on Hgb available to be occupied by oxygen. What would you expect to find if you measure the arterial PO2 of a person with CO poisoning
Answer:
Decreases.
Explanation:
The arterial PO2 of a person is decreasing because of carbonmonoxide poisoning due to higher affinity to hemoglobin than oxygen. The PO2 in the blood of arteries are used up by the cells and then there is higher concentration of carbonmonoxide in the blood as compared to oxygen. Due to this higher amount of carbonmonoxide in the blood, many complications occur such as headache, dizziness, weakness, upset stomach, vomiting, chest pain, and confusion.
If a 520 mg sample of technetium-99 is used for diagnostic procedure, how much of Tc-99 remains after 30.0h? Half life of Tc-99 is 6.0 hours.
Carbonic anhydrase is strongly inhibited by the drug acetazolamide, which is used as a diuretic (i.e., to increase the production of urine) and to lower excessively high pressure in the eye (due to accumulation of intraocular fluid) in glaucoma.
a. True
b. False
Answer:
a. True
Explanation:
There is strong inhibition of Carbon Anhydrase by Aceta-zolamide Carbonic Anhydrase. The drug acetazolamide is used as diuretic which increase the urine production in human body. It lowers pressure in eye in glaucoma.
from kinatic point of view explain the change from solid to liquied based on the effect of change of tempreture.
Answer:
Temperature affects the kinetic energy in a gas the most, followed by a comparable liquid, and then a comparable solid. The higher the temperature, the higher the average kinetic energy, but the magnitude of this difference depends on the amount of motion intrinsically present within these phases.
Explanation:
Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.
______ drives the aminoacylation reaction hydrolysis of gtp hydrolysis of phosphate phosphorylation hydrolysis of pyrophosphate
Answer:
Hydrolysis of pyrophosphate
Explanation:
The hydrolysis of pyrophosphate into two molecules of inorganic phosphate (Pi) reaction which is highly energetic that drives the other two reactions. These two reactions highly exergonic reactions that take place inside the aminoacyl-tRNA synthetase which is specific for that amino acid. Exergonic reactions are the reactions that produce free energy that can be used in the next phases of reactions so we can say that hydrolysis of pyrophosphate is responsible for the happening of aminoacylation reaction.
Write the precipitation reaction for silver sulfate in aqueous solution: (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
The balanced chemical reaction is :
[tex]2Ag^+(aq)+SO_4^{2-}(aq)\rightarrow Ag_2SO_4(s)[/tex]
Explanation:
The precipitation reaction is defined as a chemical reaction in which a solid substance is formed when two aqueous solutions of different compounds are allowed to react with each other.
When an aqueous solution of silver ion and sulfate ions are mixed together solid white precipitate of silver sulfate is formed.
The balanced chemical reaction is written as:
[tex]2Ag^+(aq)+SO_4^{2-}(aq)\rightarrow Ag_2SO_4(s)[/tex]
what would happen if the number of protons and electrons in the Atom did change
To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?
Answer:
4552 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 55 mL
Molarity of stock solution (M₁) = 12 M
Molarity of diluted solution (M₂) = 0.145 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
12 × 55 = 0.145 × V₂
660 = 0.145 × V₂
Divide both side by 0.145
V₂ = 660 / 0.145
V₂ ≈ 4552 mL
Thus, the volume of the diluted solution is 4552 mL
What’s the IUPAC name
O
||
CH3CH2CH2CH2COCH3
Answer:
Methyl pentanoate.
Explanation:
Hello there!
In this case, according to the given information, we can see the correct structure will be:
O
||
CH3CH2CH2CH2COCH3
Which matches with the structure of an ester due to the -COO- functional group. In such a case, the first part of the name is in function of the right side of the ester, in this case, methyl, followed by the left side, pentanoate, as it has five carbon atoms and is an ester (similar to an inorganic salt, but organic) and therefore, the name will be methyl pentanoate.
Regards!
A mixture of hydrocarbons contains 38.3% hexane, C6H14, 13.9% octane, C8H18, and 47.8% decane, C10H22. The mixture is combusted in an excess of oxygen. What mass of CO2 is generated from the combustion of 19.3 kg of the mixture
Answer:
52.206 kg
Explanation:
From the given information:
Mass of hexane C6H14 = [tex]19.3*10^3 \ g \times \dfrac{38.3}{100}[/tex]
= 7391.9 g
Mass of octane C8H18 = [tex]19.3*10^3 \ g \times \dfrac{13.9}{100}[/tex]
= 2682.7 g
Mass of decane C10H22 = [tex]19.3*10^3 \ g \times \dfrac{47.8}{100}[/tex]
= 9225.4 g
However, recall that:
number of moles of an atom = mass/molar mass
∴
For hexane, no of moles = 7391.9 g/86.18 g/mol
= 85.77 moles
For octane, no of moles = 2682.7 g/114.23 g/mol
= 23.49 moles
For decane, no of moles = 9225.4 g/142.29 g/mol
= 64.84 moles
Therefore:
number of moles of CO2 produced = (6 × 85.77)+(23.49)+(10×64.84) moles
= 1186.51 moles
Finally, the mass of CO2 produced is:
= 1186.51 mol × 44 g/mol
= 52206.44 g
= 52.206 kg
The following list of properties is most descriptive of a(n) ______________. Low melting point, non-conductor of electric current. Group of answer choices
Answer:
This question is incomplete
Explanation:
This question is incomplete but the completed question is below;
"The following list of properties is most descriptive of a(n) ______
1. High melting point, conductor of electricity when dissolved in water
2. Low melting point, non-conductor of electric current
3. Malleable, ductile, insoluble in water.
The choices for all 3 are: a) metallic solid b) molecular solid c) ionic solid d) all of these e) none of these f) more than one of these"
1. Ionic solid: Ionic solids are solids that have ionic/electrovalent bonds holding it's constituent molecules together. These bonds are strong bonds that involve the transfer of electrons from one constituent atom (the metal) to another constituent atom (the nonmetal). This strong bond causes the solid to have a high melting and boiling point. Also, when dissolved in water, the constituent atoms (involved in the electron transfer) dissociate to form ions (become charged) and thus easily carry electric charges (i.e conduct electricity).
Examples are Sodium Chloride and Potassium Iodide
2. Molecular solids: These are solids whose constituent molecules are held together by a weak bond/force known as Van der Waal forces. This forces are easily broken down when subjected to heat and thus the molecular solids have a low melting point. Also, these solids do not have carriers of heat or electric charges in them and are thus non-conductors of electric current.
Examples are Ice (frozen water) and sucrose
3. Metallic solids: These of solids made from constituent metal atoms only. The nuclei of these constituent metal atoms have the ability to move past one another without disrupting there metallic bonding; it is for this reason they are malleable and ductile. There constituent atoms however do not dissociate in water and are thus insoluble in water.
Examples are aluminium and copper crystal.
Write a balanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
Explanation:
Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.
H₂O(l) ⇒ H₂(g)
First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.
2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)
Then, we perform the charge balance by adding 2 electrons to the left side.
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
