When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *
Answer:
if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!
Explanation:
Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.
Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.
Oxidation half equation;
Mg(s)-----> Mg^2+(aq) + 2e
Reduction half equation;
O2(g) + 2e ------> 2O^2-(aq)
From the balanced reaction equation, two moles of electrons is transferred.
Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.
A growing concern in agricultural and food chemistry is the presence of residues in food. We use many forms of organic chemicals in agriculture and food chemistry and there is growing concern as to how safe these materials are. Choose an organic chemical used in agricultural of food chemistry and report on the functional groups contained in your compound, the uses of the compound, and the safety of that compound.
Answer:
Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.
Explanation:
Monosodium Glutamate (MSG) is a chemical that is used in types of different food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
Best regards.
Answer:
B. Mixing a solute and a solvent
Explanation:
ap3x
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
Which is true regarding a water molecule?
Answer:
Has many answers, but one is that it consists of small polar v shaped molecules with a molecular formula H20.
Explanation:
Water molecules consists of 2 hydrogen atoms bonded with on oxygen atom. Each molecule is electrically neutral but polar, with the center of positive and negative charges located in different places.
Each hydrogen atom has a nucleus consisting of a single positively-charged proton surrounded by a 'cloud' of a single negatively-charged electron and the oxygen atom has a nucleus consisting of eight positively-charged protons and eight uncharged neutrons surrounded by a 'cloud' of eight negatively-charged electrons.
Hoped this helped!
Pyruvate is the end product of glycolysis. Its further metabolism depends on the organism and on the presence or absence of oxygen. Draw the structure of the product from each reaction as it would exist at pH 7. Include the appropriate hydrogen atoms. Reaction A: aerobic conditions in humans or yeast
The given question is incomplete. The image present in the question for Reaction A is attached below along with the answer.
Explanation:
Pyruvate molecule reacts with Coenzyme A in the presence of oxygen and it results in the formation of acetyl Coenzyme A and carbon dioxide.
The enzyme pyruvae dehydrogenase helps in catalyzing this reaction. As in this biochemical reaction [tex]NAD^{+}[/tex] gets converted into NADH.
This reaction is shown in the image attached below.
Which process absorbs the greatest amount of heat?
a. the cooling of 10 g of liquid water from 100°C to 0°C.
b. the heating of 10 g of liquid water from 0°C to 100°C.
c. the freezing of 10 g of liquid water the melting of 10 g of ice.
d. the condensation of 10 g of gaseous water.
Answer:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Explanation:
Hello,
In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.
Regards.
The process having the greatest amount of heat is:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Looking at all the options:The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.
The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.
Find more information about Heat here:
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calculate the moles of 25.2 g Na2S2O8
Answer:
To calculate the moles we must first find the molar mass M
M (Na2S2O8) = (23*2) + (32*2) + (16*8)
= 46 + 64 + 168
= 278g/mol
Molar mass = mass/moles
moles =mass / molar mass
= 25.2/278
= 0.0906mol
Hope this helps.
A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added
Answer:
pH = 9.08
Explanation:
Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:
Q + H₂O ⇄ QH⁺ + OH⁻
Where pKb is 5.10
Using H-H equation for weak bases:
pOH = pKb + log₁₀ [QH⁺] / [Q]
The reaction of quinine with HCl is:
Q + HCl → QH⁺ + Cl⁻
Initial moles of quinine are 0.125 moles and moles added of HCl are:
0.05000L × (1.00mol / L) = 0.05000moles.
That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:
Q = 0.125mol - 0.050mol = 0.075 moles
QH⁺ = 0.050 moles
Replacing in H-H equation:
pOH = 5.10 + log₁₀ [0.050] / [0.075]
pOH = 4.92
As pH = 14 - pOJ
pH = 9.08differentiate between sol,aerosol and solid soluti
Answer:
Sol is a colloidal suspension with solid particles in a liquid. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas. While solid solution contain solid as solute in either solid, liquid or gas.
What is cell culturing?
a technique that uses specific antibodies to visualize features of cells
a technique that visualizes how specific genes are used within a cell
a technique in which cells are purposefully grown under specific conditions
an imaging technology used to study features smaller than the human eye can see
Answer:
a technique in which cells are purposefully grown under specific conditions
Explanation:
Answer:
its c
Explanation:
correct edge2020
Photochromic lenses contain Group of answer choices both AgCl and CuCl embedded in the glass. only AgCl embedded in the glass. neither AgCl nor CuCl embedded in the glass. only CuCl embedded in the glass.
Answer:
both AgCl and CuCl embedded in the glass
Explanation:
Photochromic lenses contain both AgCl and CuCl embedded in the glass.
They are light-sensitive lenses that adapt to environmental changes. They appear clear when in an apartment or a building and automatically darken when outside as a result of exposure to sunlight. The darkening is activated by the UV component of the sunlight.
Photochromic lenses are otherwise known as light-adaptive or intelligent lenses and they are formed by coating lenses with silver chloride compounds whose concentration ranges from 0.01 to 0.001 %. Copper (I) chloride is also included in addition to the silver halide.
In summary, photochromic lenses contain both AgCl and CuCl.
The standard free energy change, ΔG°', for this reaction is +6.7 kJ/mol. However, the observed free energy change (ΔG) for this reaction in pig heart mitochondria is +0.8 kJ/mol. What is the ratio of [isocitrate]/[citrate] in these mitochondria at 25.0 °C?
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
(a) Titration curve for the titration of 5.00 mL 0.010 M NaOH(aq) with 0.005 M HCl(aq), indicating the pH of the initial and final solutions and the pH at the stoichiometric point.
What volume of HCl has been added at
(b) the stoichiometric point
(c) the halfway point of the titration?
Answer:
AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L
AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L
Explanation:
In solving titration problems, you must remember this formula;
MaVa = MbVb
Since M a= 0.005 M
Mb = 0.010 M
Vb = 5 mL = 5 /1000 = 0.005 L
Va = unknown.
Solving for Va, we have:
Va = MbVb / Ma
Va = 0.010 * 0.005 / 0.005
Va = 0.01 L
So therefore, the volume of acid added at:
1. the stoichiometric point is 0.01 L
2. half-way point of titration is 0.01 /2 = 0.0050 L
For the pH:
Since HCl is a strong acid, it dissociate into {H30}+ ion.
First calculate the number of moles of hydronium ion
number of mole = concentration of hydronium ion {H30}+ * Volume
n = 0.005 * 0.01 = 0.00005 moles
A. At initial point of the titration, the volume of base added is 0 L
{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M
pH = - log {0.005}
pH = 2.3
B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.
n(H+) = n(OH^-)
pH = 7
Le Chatelier's Principle. For the reaction below, if the equilibrium concentrations were NH3 = 2 x 10-4, H3O+ = 2 x 10-4M and NH4+ = 18.0M, what is the equilibrium constant for the reaction and what would happen if you were to add some acid to this reaction? NH3 + H3O+ --> NH4+ + H2O
Answer:
Explanation:
NH3 + H3O+ --> NH4+ + H2O
equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]
=
by inserting thier respecive values can you calcaulte, by the way coniseder [ H2O] =1 ,
Give the IUPAC name for the following compound: Multiple Choice (1R,3R)-1-ethyl-3-methylcyclohexane (1S,3S)-1-ethyl-3-methylcyclohexane (1R,3S)-1-ethyl-3-methylcyclohexane (1S,3R)-1-ethyl-3-methylcyclohexane
Answer:
(1R,3R)-1-ethyl-3-methylcyclohexane.
Explanation:
NOTE: The question is not complete since we do not have the diagram to the chemical structure in the question. Kindly check the attached picture for the diagram of the chemical structure.
So, in order to name Enantiomers or chemical structure through the use of the R,S system requires series of rules and regulations to follow for the proper naming.
There is an ethyl attached to the compound as the first substituents and methyl at the third which are the secondary prefix.
=> The longest chain is 6, thus the compound has hexane as the root compound.
=> It is (1R,3R) because when we draw from the highest substituents to the lower substituents, this is done in a clockwise direction.
A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI
did the student use to make the sample?
Use M,V;-MV
3.70 mL
16.7 ml
30.0 mL
83.3 mL
Mark this and return
Save and Exit
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Submit
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
C on edg 2021Explanation:
i dont like reading
The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
Learn more about thermochemical equations here: https://brainly.com/question/25164433
Round to 3 significant figures.
1.4593
Answer : The correct answer is 1.46
Explanation :
The following rules are used to round off a number to the required number of significant figures:
(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.
(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.
(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.
(4) The same procedure is follow for decimal values.
As we are given, 1.4593
In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.
According to the rules, round off the given measurement in three significant figures as 1.46
Therefore, the correct answer is 1.46
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 × 10–4 mol L–1. Calculate the concentration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.
Answer:
The correct answer is 5.30 * 10^-4 mol per L.
Explanation:
Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.
Therefore, it can be written as,
C2/C1 = P2/P1
Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,
C2/(6.8*10^-4) = 0.78/1
C2 = 0.78 * 6.8*10^-4
C2 = 5.30 * 10^-4 mol per L.
Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.
ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]
Answer:
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Explanation:
Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]
Photon momentum = h/wavelength
[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]
Mass of H2 molecule m = molar mass/Avogadros number
[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]
Since momentum is conserved:
Photon momentum = H2 molecule momentum = mass x velocity of H2
[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]
velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]
Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2
[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
The fluoride ion is the conjugate base of the weak acid hydrofluoric acid. The value of Kb for F-, is 1.39×10-11. Write the equation for the reaction that goes with this equilibrium constant.
Answer:
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺ ions. In this sense, hydrofluoric acid is an acid according to the following equation.
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)
According to Brönsted-Lowry acid-base theory, a base is a substance that accepts H⁺ ions. In this sense, the fluoride ion is a base according to the following equation.
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
The equilibrium constant for this reaction is Kb = 1.39 × 10⁻¹¹.
Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).
Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
A chemistry student is given of a clear aqueous solution at . He is told an unknown amount of a certain compound is dissolved in the solution. The student allows the solution to cool to . At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitates, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 50,0 g.
Using only the information above, can you calculate the solubility of X in water at 16. C. If you said yes, calculate it.
Answer:
Solubility cannot be calculated.
Explanation:
To calculate the solubility of X it is necessary to know the value of the mass of the solute (X) that can be dissolved in 100 g of water.
[tex]Solubility = \frac{Solute mass}{100 grams of water}[/tex]
Taking into account that we do not know the value of the mass of the solution, therefore the value of the solubility of the compound cannot be determined.
Who proposed the plum pudding model and what does it say about the structure of the atom
Answer:
J. J. Thomson
Explanation:
First proposed by J. J. Thomson in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.
When pressure is increased on the following equilibrium, where will the shift be? 3H2 + N2 2NH3
Answer:
Explanation:
it is based on le chatliers principles
the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,
so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.
CAN SOMEONE HELP ME!!
Solutions of each of the hypothetical acids in the following table are prepared with an initial concentration of 0.100 M. Which of the four solutions will have the lowest pH and be most acidic? Explain please.
Acid pKa
HA 4.00
HB 7.00
HC 10.00
HD 11.00
a. HA
b. HB
c. HC
d. HD
e. All will have the same pH because the concentrations are the same.
Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Explanation:
We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.
[tex]pK_a = -log K_a[/tex]
This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.
As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.
In the given options, HA has the smallest [tex]pK_a[/tex] value.
Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
Answer:
Here's what I get
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
Your structure has two problems:
It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.Here's one way to draw a Lewis structure.
1. Draw a trial structure
Make F and O terminal atoms and give each one an octet (Fig. 1).
2. Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
3. Check the number of valence electrons available
1 S = 1 × 6 = 6 electrons
1 O = 1 × 6 = 6
4 F = 4 × 7 = 28
TOTAL = 40 electrons
The trial structure has the correct number of electrons.
4. Determine the formal charge on each atom.
To get the formal charges, we cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On S
VE = 6
BE = 5 bonding electrons = 5
FC = 6 - 5 = +1
(b) On O:
VE = 6
BE = 3 LP(six electrons) + 1 bonding electron = 7
FC = 6 - 7 = -1
(c) On F:
VE = 6
BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7
FC = 7 - 7 = 0
5. Minimize the formal charges
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
Move a lone pair from the O to make an S=O double bond (Fig. 2).
6. Recalculate the formal charges
(a) On S
VE = 6
BE = (3 bonding electrons) = 6
FC = 6 - 6 = 0
(b) On O:
VE = 6
BE = 2 LP(four electrons) + 2 bonding electrons = 6
FC = 6 - 6 = 0
Fig. 2 shows the Lewis structure in which all atoms have a formal charge of zero.
The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.
The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.
The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.
The lewis structure with the central S atom has been attached.
For more information about lewis structure, refer to the link:
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