Answer:
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
Explanation:
The equation for the reaction is given as:
[tex]\mathbf{ CH_{4(g)} + 4 F_{2(g)} \to CF_{4(g)} + 4 HF_{(g) }}[/tex]
At standard conditions; the bond energies are as follows;
Bond Bond Energies (kJ/mol)
C-H 413
F-F 155
C-F 485
H-F 567
[tex]\mathbf{\Delta \ H_{rxn} = \sum \Delta H ( reactant) - \sum \Delta H (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum [\Delta H \ 4( C-H) + \Delta H \ 4(F-F) ]- \sum[ \Delta H \ 4( C-F)+\Delta H \ 4( H-F)] (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum{ \Delta \ H (4*413) + \Delta \ H (4*155) - \Delta \ H (4(485)) + \Delta H (4(567) }}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({ (1652 + 620) - (1940 + 2268)})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({2272- 4208})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Answer:
21.88mL is the volume of base required for the titration.
Explanation:
For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.
If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.
In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:
23.82mL - 1.94mL
21.88mL is the volume of base required for the titration.Suppose you have a bucket of sand containing 5.5 billion grains of sand ( 5.5×109 grains). Most of the grains of sand are white, but there are some brown grains mixed evenly throughout.
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
brown grains
Explanation:
5.5 billion grains of sand ( 5.5×109 grains)
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
Brown grains = 0.06 * 5.5×10^9 = 0.33 x 10^9 = 3.3 x 10^8 grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppm = 6 / 1000000 = 0.000006
Brown grains = 0.000006 * 5.5×10^9 = 3.3 x 10^4 grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppb = 6 / 1000000000 = 0.000000006
Brown grains = 0.000000006 * 5.5×10^9 = 3.3 x 10^1 = 33 grains
What are the correct formulas and coefficients for the products of the following double-replacement reaction? RbOH + H3PO4→
Answer:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Explanation:
Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Identify a reaction of sugars, starch, or cellulose to produce ethanol and carbon dioxide halogenation polymerization
Answer:
Fermentation
Explanation:
Fermentation is the general term used to describe the process by which sugars such as glucose, starch or cellulose are converted to ethanol and carbon (iv) oxide. It is anaerobic process meaning that it occurs in the absence of air or in very low oxygen concentrations.
Yeast and other microorganisms ferment glucose into ethanol and carbon (iv) oxide with the help of the enzyme zymase. Polysaccharides such as starch and cellulose are first broken down into glucose by enzymes such as diastases, maltase and cellulase, before it is then converted into ethanol and carbon (iv) oxide.
The equation for the conversion of glucose to ethanol and carbon (iv) oxide is as follows:
C₆H₁₂O₆(aq) -----> 2C₂H₅OH(aq) + 2CO₂(g)
Given that the molar mass of NAOH is 40.00G/MOL, what mass of NAOH is needed to make 2.500 L of a 2.000 M NaOH Solution?
Answer:
200g
Explanation:
n = CV
n = mass/molar mass
mass/molar mass = CV
mass/40 = 2 x 2.5
mass/40 = 5
mass = 5x 40
mass = 200g
What element forms an ion with an electronic configuration of 1s22s22p6 (or [Ne] ) and a −2 charge? Give the symbol for the element. g
Answer:
Mg²⁺
Explanation:
Electronic configuration = 1s22s22p6 (or [Ne] )
Charge = -2
This means the element has two extra electrons. So total electrons = 12.
The lement is Magnesium and the ion is Mg²⁺
Kinetic energy and gravitational potential energy are both forms of which type
of energy?
A. Internal energy
B. Mechanical energy
C. Potential energy
D. Thermal energy
Answer:
C. Potential energy
Explanation:
Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.
Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs to be done in 30 mins
Answer:
E) Al³⁺
Explanation:
A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.
In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;
A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.
K + 2e⁻ ----> K⁻²
B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.
Fe⁺ + 2e⁻ ----> Fe
C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.
O²⁻ + 2e⁻ ----> O⁴⁻
D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.
Ne + 2e⁻ ----> Ne²⁻
E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.
Al³⁺ + 2e⁻ ----> Al⁺
What concentration unit is measured in moles of solute per kilogram of
solvent?
Answer:
molality
Explanation:
The SI unit for molality is moles per kilogram of solvent. A solution with a molality of 3 mol/kg is often described as "3 molal", "3 m" or "3 m". hope this helps you :)
Observing Gas Properties - Flammability
Chemicals Observations
HCl
Zn
HCl + Zn Reaction
Gas + Flame
Conclusion:
A. When performing this experiment, when is the tetradecanol in a condensed phase? Explain your answer.
B. The published melting point of H2O is 0°C, and the published boiling point is 100°C. Why would you have found slightly different values?
C. In this experiment, there were two possible identities for the gas produced: hydrogen or chlorine, which have quite different properties. Hydrogen is a colorless, odorless, flammable gas. Chlorine gas is greenish yellow with a pungent, bleach-like odor and is non-flammable. Use your observations to determine which gas was produced in this experiment.
D. Give some examples of how it might be useful to know the melting or boiling points of a substance. Think in terms of both scientific and commercial/industrial settings.
Answer:
The answer is C love.
Explanation:
Answer:
a) It is when a solid is going into the liquid stage. The molecules are really close together.
b) The water could have impurities, altitude could effect the boiling point or the temperature could of been measured wrong.
c) The gas was colorless and odorless therefore it is Hydrogen gas.
d) In a Scientific setting finding the boiling point of a new unknown substances can help identify and organize them. In a commercial setting chemicals need to be stored a certain way according to their boiling point. Stored incorrectly could cause the substance to evaporate.
Brown algae, such as giant kelp and sargassum, are an example of ? A. macroalgae B chemosynthetic algae C Flowering plants D Phytoplankton
Answer:
The answer is option A.
I hope this helps you.
Answer:
Macro Algae
Explanation:
probz
After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Express the concentration to two significant figures and include the appropriate units.
Answer:
[tex]\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}[/tex]
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
[tex]n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}[/tex]
2. Moles of NH₃
[tex]n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}[/tex]
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
[tex]c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}[/tex]
(c) [NH₃]
[tex]c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}[/tex]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}[/tex]
Kf is large, so x ≪ 6.106×10⁻³. Then
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}[/tex]
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
A gas particle of mass 5.31 × 10^-23 kg has a velocity of 1.00 102 m/s. What is the kinetic energy of the molecule
Answer:
Kinetic energy = 1/2mv²
where m is the mass
v = velocity
m = 5.31 × 10^-23 kg
v = 1.00 × 10^2 m/s
Kinetic energy = 1/2 × 5.31 × 10^-23 × ( 1.00 × 10^2)²
= 2.655 × 10^-19 Joules
Hope this helps
Calculate how many moles of NO2 form from 3.0 mol N2O5 2N2O5(g)→4NO2(g)+O2(g)Express your answer using two significant figures.
Answer:
6.0 moles NO2(g)
Explanation:
Based on the reaction every 2 moles N2O5(g) gives reaction with 4 moles NO2(g).Then when we have 3.0 mol N2O5(g),
2 moles N2O5(g) 4 moles NO2(g)
3 moles N2O5(g) ? moles NO2(g)
______________________________________
3 *4 / 2 = 6.00 moles NO2(g)
Answer:
There are 6 mol of NO2 with respect to 3 mol of N2O5
Explanation:
Approach 1 ( dimensional analysis ) :
3 Moles of N2O5 [tex]*[/tex] ( 4 moles of NO2 / 2 Moles of N2O5 ) - moles of N2O5 cancel out, leaving you with the moles of NO2 -
3 [tex]*[/tex] 4 / 2 = 12 / 2 = 6 moles of NO2
So as you can see in the formula there are 4 moles of NO2 present per 2 Moles of N2O5 - " 4NO2 and 2N2O5. " As we wanted the moles of N2O5 to cancel out, the 2 moles of N2O5 was kept as the denominator, and hence we received the fraction we needed.
Approach 2 :
There are 3 Moles of N2O5. The ratio of Moles of N2O5 to moles of NO2 is provided by the reaction -
Moles of N2O5 : Moles of NO2,
2 : 4,
1 : 2
Therefore the moles of NO2 will be two times as much as the given moles of N2O5, or 3 [tex]*[/tex] 2 = 6 moles of NO2
need this asap , help please
Answer:
Path A-B-D involves a catalyst and is slower than A-C-D
Explanation:
The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.
The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.
The catalyzed path has a higher activation energy than the uncatalyzed path.
Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.
On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.
Draw an aldohexose.
Answer:
See explanation
Explanation:
We have to remember that theory behind the carbohydrates. Carbohydrates are molecules with several hydroxyl groups in which the main functional group can be an aldehyde or a ketone.
If we have an aldehyde as a main functional group we will have an "aldose". If we have a ketone as a main functional group we will have a "ketose".
We can also, classify the carbohydrates using the number of carbons. So, for example, if we have 5 carbons and a ketone as the main functional group we will have a "keto-pentose". If we have for example 4 carbons and an aldehyde as the main functional group we will have a "tetra-aldose".
In this case, we have an aldohexose, so we will have 6 carbons and an aldehyde as main functional group. So, we can draw a structure with 6 carbons, in carbon 1 we have to put the aldehyde group and in the other carbons we have to put "OH" groups.
See figure 1
I hope it helps!
4. A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate as the tide goes out. If the volume of a puddle decreases to 23% of its original volume, what is the sodium chloride concentration if it was initially 0.449 M?
Answer:
0.583M NaCl
Explanation:
Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.
In the puddle, the solute is sodium chloride that is dissolved in water and you have 0.449 moles of NaCl per liter of water
When the solution begins to evaporate, amount of water decreases whereas moles of NaCl remain constant.
As 23% of the water evaporates, amount of water that remains is 100-23 = 77%, that means now you have 0.449 moles of NaCl per 77% of a liter, 0.770L. The new concentration is:
0.449 moles NaCl / 0.770L =
0.583M NaClwhat is the value of the equilibrium constant at 500k for a chemical equilivrium that has a delta h value of 250kj mol and s value of 48 j mol k
Explain with examples following characteristics of chemical reactions: a. Change of colour b. Evolution of gas c. Change of smell d. Change of state
Answer:
Explanation:
a. change of colour:
A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.
Eg: copper reactions with the elements
b. Evolution of gas:
A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.
Eg: ammonium hydroxide breaks down to water and ammonia gas.
c. Change of smell :
Production of an Odor Some chemical changes produce new smells. ... The formation of gas bubbles is another indicator that a chemical change may have occured.
Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.
d. Change of state:
A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.
Eg: candle wax (solid) melts initially to produce molten wax (liquid)
plz mark as brainliest!!!!
what bonding is similar to ionic bonding, except there are no high-electronegativity atosms present to accept any electrons that the present atoms are willing to donate.
Answer:
Metallic bonding
Explanation:
Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.
The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.
In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.
Considering the steps involved in dissolution, which of the following do you expect to speed up a dissolution process?
A. Sweeping all the solute particles into a pile within the solvent.
B. Stirring the solution vigorously.
C. Grinding the solute down into tiny particles.
D. Gently heating the solution.
Answer:
C. Grinding the solute down into tiny particles.
Explanation:
The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.
However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.
Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.
Answer: stirring the solution vigorously
Grinding the solute down into tiny particles
gently heating the solution
Explanation:
A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution
A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?
Answer:
[tex]V_2 = 4.87 * 10^3[/tex]
Explanation:
This question is an illustration of ideal Gas Law;
The given parameters are as follows;
Initial Temperature = 25C
Initial Volume = 4.5 * 10³L
Required
Calculate the volume when temperature is 50C
NB: Pressure remains constant;
Ideal Gas Law states that;
[tex]PV = nRT[/tex]
The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n
Divide both sides by PT
[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]
[tex]\frac{V}{T} = \frac{nR}{P}[/tex]
Represent [tex]\frac{nR}{P}[/tex] with k
[tex]\frac{V}{T} = k[/tex]
[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
At this point, we can solve for the required parameter using the following;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;
From the given parameters;
V1 = 4.5 * 10³L
T1 = 25C
T2 = 50C
Convert temperatures to degree kelvin
V1 = 4.5 * 10³L
T1 = 25 +273 = 298K
T2 = 50 + 273 = 323K
Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]
Multiply both sides by 323
[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]
[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]
[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]
[tex]V_2 = 4.87 * 10^3[/tex]
Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]
An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
Name the following alkanes. Please need help last assignment. The questions in the picture.
Answer:
Hydrogen, carbon reaction
According to solubility rules, which of the following compounds is insoluble in water?
A.Cal
2
B.NA SO
2 4
C.AgF
D.AI(OH)
3
Answer:
The correct answer is - Al(OH)3
Explanation:
At the point when a substance is blended in with a soluble, there are a few potential outcomes. The deciding variable for the outcome is the solubility of the substance, which is characterized as the maximum concentration of the solute. These rules help figure out which substances are solvent, and how much.
According to the 11 rules of solubility rules, the insoluble compound in water is - Al(OH)3
Answer:
Na2So4
Explanation:
If you consult a table of solubility rules, like the one below, you will see that sodium sulfate (Na2SO4) is soluble in water.
What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 100. mL of solution
Answer:
[tex]3.65~M[/tex]
Explanation:
We have to remember the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the conversion:
[tex]100~mL\frac{1~L}{1000~mL}=~0.1~L[/tex]
Now we can calculate the moles. For this we have to calculate the molar mass:
O: 16 g/mol
H: 1 g/mol
C: 12 g/mol
[tex](16*1)+(1*4)+(12*1)=32~g/mol[/tex]
With the molar mass value we can calculate the number of moles:
[tex]1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH[/tex]
Finally, we can calculate the molarity:
[tex]M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M[/tex]
I hope it helps!
4. In which pair of substances does the first underlined atom have a lower oxidation number than the second? A. NH3OH+ NH4– B. H2O H2O2 C. SO3 SO42– D. HCHO C
Answer:
Option B is correct.
Only this option has the first underlined element with a lower oxidation number than the second amongst the options.
Explanation:
Complete Question
In which pair of substances does the first underlined atom have a lower oxidation number than the second?
A. NH₃OH⁺, NH₄⁻ (N is underlined)
B. H₂O, H₂O₂ (O is underlined)
C. SO₃, SO₄²⁻ (S is underlined)
D. HCHO, C (C is underlined)
Solution
In determination of the oxidation number of an atom in a compound, we first name the unknown oxidation number x.
Then, the total oxidation number of the atoms in the compound is equal to the charge of on the compound (or radical).
So, elements in their neutral state have no charge and no oxidation number.
A. NH₃OH⁺, NH₄⁻ (N is underlined)
N in NH₃OH⁺
Oxidation number of N = x
Oxidation number of H = +1
Oxidation number of O = -2
x + (3×+1) + (-2) + (+1) = +1
x - 3 - 2 + 1 = 1
x = +5
N in NH₄⁻
Oxidation number of N = x
Oxidation number of H = +1
x + (4×1) = -1
x + 4 = -1
x = -1 - 4 = -5
First underlined element has a greater oxidation number than the second. So, this doesn't qualify.
B. H₂O, H₂O₂ (O is underlined)
O in H₂O
Oxidation number of H = +1
Oxidation number of O = x
(2×1) + x = 0
2 + x = 0
x = -2
H₂O₂
Oxidation number of H = +1
Oxidation number of O = x
(2×1) + (2×x) = 0
2 + 2x = 0
2x = -2
x = (-2/2) = -1.
First underlined element has a lower oxidation number than the second. So, this qualifies.
C. SO₃, SO₄²⁻ (S is underlined)
S in SO₃
Oxidation number of S = x
Oxidation number of O = -2
x + (3×-2) = 0
x - 6 = 0
x = +6
SO₄²⁻
Oxidation number of S = x
Oxidation number of O = -2
x + (4×-2) = -2
x - 8 = -2
x = 8 - 2 = +6
First underlined element has the same oxidation number as the second. So, this doesn't qualify.
D. HCHO, C (C is underlined)
C in HCHO
Oxidation number of H = +1
Oxidation number of C = x
Oxidation number of O = -2
+1 + x + 1 - 2 = 0
x = 0
C in C
Oxidation number of C = x
x = 0
First underlined element has the same oxidation number as the second. So, this doesn't qualify.
Hope this Helps!!!
An ion of a single pure element always has an oxidation number of ________.
Answer: An ion of a single pure element always has an oxidation number of
zero.
Explanation:
1. Natural gas is used as a cooking fuel in many restaurants and homes. The primary chemical components of natural gas are hydrocarbons known as alkanes. Research and list the primary alkanes found in natural gas, and explain why these compounds excel as fuel sources.
Answer:
The main component of natural gas is methane (CH4) at 60 to 90% followed by various combination of ethane, propane, and butane whose percentage can vary from 0 to 20% each. For each unit mass of alkanes, the combustion energy (energy released when the fuel reacts with oxygen) released is very high about 13 to 15 kcal/g, which is higher than even those generated by petrol or diesel. So, for heating or other energy generation purpose for household purposes, this source of energy is used.
The equation for combustion of methane is shown below. Upon combustion, carbondioxide and water is produced with simultaneous generation of heat which is the source of energy used for consumption.
CH4 + 2O2 --> CO2+ 2H2O + heat [ For methane, the combustion energy is ~ 6kcal/g]
As the CH2 units are increased in the alkanes, the combustion energy increases, for e.g., ethane has combustion energy of 7 kcal/g and propane has about 12 kcal/g.
Explanation: