Answer:
The entropy of the final solution decreases, as the reaction disorder is less.
Explanation:
The higher the temperature, the greater the heat of the reaction and the greater the disorder it has, so the entropy will increase ... But this is not the case, since the solution cools, decreasing the entropy proportionally.
Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)
Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.
The balanced chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Calculate the pH of mixing 24 mL of 1M acetic acid with 76 mL of 1M sodium acetate. For the purpose of this calculation, assume the Ka of acetic acid is 1.8 X 10-5. You must include units to obtain full credit. You must show all your work to obtain any credit.
Answer:
pH = 5.24
Explanation:
Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.
Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:
[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M
[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M
Replacing in H-H equation:
pH = 4.74 + log₁₀ [0.76M] / [0.24M]
pH = 5.24
A warm can of soda has more spray when opened than a cold one. Drag the terms on the left to the appropriate blanks on the right to complete the sentences.
1. more
2. less
3. increased
4. decreased
A. The solubility of a gaseous solute is_____at a higher temperature, and the CO2 pressure in the can is______.
B. When the can of warm soda is opened, ______CO2 is released, producing______spray.
Answer:
A. The solubility of a gaseous solute is decreases at a higher temperature, and the CO₂ pressure in the can is increased.
B. When the can of warm soda is opened, more CO₂ is released, producing more spray.
Explanation:
We know that , solubility of the gases is inverse proportional to the temperature and directly proportional to the pressure .
We also know that , the kinetic energy of the gases is directly proportional to the temperature of the gas. If the temperature of the gas increase then the kinetic energy of gas will increases that increasing in the kinetic energy of the gas leads to decrease in the solubility of the gases.
When the pressure of the gas increases then the solubility of the gases also will increases because molecule of the gas try to settle in the small area.
From the above we can filled the below blank places
A. The solubility of a gaseous solute is decreases at a higher temperature, and the CO₂ pressure in the can is increased.
B. When the can of warm soda is opened, more CO₂ is released, producing more spray.
The concept of resonance describes molecular structures Question 17 options: that have several different geometric arrangements. that have delocalized bonding. that are formed from hybridized orbitals. that have different molecular formulas. that have electrons resonating.
The concept of resonance illustrates molecular structure which possess delocalized bonding.
• The resonating structures are required when a single structure is not adequate to illustrate all the characteristics of a molecule.
• The number of resonating structures relies on the existence of delocalized bonds or delocalized electrons.
• For example, benzene exhibits two resonance structures as it possess delocalized bonds while cyclohex-1-ene exhibits no resonance structure as it does not possess delocalized bond.
Thus, the resonance concept illustrates molecular composition, which possess delocalized bonds.
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Add distilled water to the beaker until the volume
totals 15 mL.
Record the amount of oil that dissolved.
Answer:
i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances
Explanation:
Answer:
None
Explanation:
Answer on Edge 2022
What do we call temperature changes caused by changes in air pressure?
Answer:
Fronts
Explanation:
For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!
Hope this helps! Please mark as brainiest!
Identify the elements that correspond to the following generalized electron configuration: (noble gas]ns2(n − 2) f6
Express your answer as the element symbol. If there is more than one answer, separate them by a comma.
Answer:
Samarium:
Electron configuration:
Samarium
Explanation:
Samarium is a chemical element that belongs to the lanthanoid series. The lanthanoids are the chemical elements that follow lanthanum. They are all known to possess 4f orbitals. The 4f electrons are found in the antepenultimate shell of the elements of the lanthanoid series and they do not take part in chemical bonding. They are neither removed in bonding nor do they take part in crystal field stabilization of lanthanoid complexes.
The electronic configuration of samarium is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6 while the condensed, short hand electronic configuration is [Xe] 4f6 6s2. This corresponds to (noble gas]ns2(n − 2) f6 as required by the question, hence the answer provided above.
A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?
Answer:
3.07 × 10⁻⁴
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
[tex]pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M[/tex]
Step 2: Calculate the concentration of HY
5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:
[tex]Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M[/tex]
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}[/tex]
Which of the following best describes a salt bridge? a) A pathway composed of salt water that ions pass through. b) A pathway between the cathode and anode in which ions are reduced. c) A pathway by which counterions can flow between the half-cells with the solutions in the half-cell completely mixing.
Answer: A
Explanation:
How many moles of ammonia are in 0.175 L of a 6.50 M aqueous ammonia solution?
Answer:
Concentration (C) = number of moles (n) / volume (v)
Therefore
number of moles (n) = concentration × volume
Concentration = 6.50M
Volume =0.175 L = 0.175dm³
n = 6.50 × 0.175
n = 1.138 moles
Moles of ammonia in the solution is
1.138 moles.
Hope this helps
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:
ΔT = kf×m×i
Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = 0.253m
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
The freezing point of the solution is -1.4°C
Answer:
THE FREEZING POINT IS -1.41 °C
Explanation:
Using the formula of change in freezing point:
ΔTf = i Kf m
i = 3 (1 Ca, 2 Br)
i is the number of the individual elements in the molecules
Kf of water = 1.86 °C/m
mass of CaBr2 = 189.9 g
Calculate the Molar mass of CaBr2:
Molar mass = ( 40 + 80*2) = 200 g/mol
Calculatee the molarity:
molarity = 189.9 g * 1 mole / 200 g/mol / 3.75 kg of water
molarity = 0.2532 M
So therefore, the change in freezing point is:
ΔTf = 1 Kf * M
ΔTf = 3 * 1.86 * 0.2532
ΔTf = 1.41 °C
The freezing point = old freezing point - change in freezing point
The freezing point = 0 - 1.41 °C = - 1.41 °C
The freezing point therefore is -1.41 °C
If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
[tex]Q=4154J[/tex]
Explanation:
Hello,
In this case, the involved heat in this heating process is considered to be computed via:
[tex]Q=nCp\Delta T[/tex]
Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:
[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]
Regards.
The reaction: A + 3 B → D + F was studied and the following mechanism was determined. A + B C (fast) C + B → D + E (slow) E + B → F (very fast) The species, C, is properly described as
Answer:
Intermediate.
Explanation:
Hello,
In this case, we can rewrite the steps as:
[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]
Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).
Best regards.
Which of the following would have a fixed shape and volume? (3 points) nitrogen gas solid wood liquid water neon gas
Answer:
Solid Wood
Explanation:
Wood is like a solid block, whereas gases flow freely and liquids spread to fill the shape of their container.
Please let me know if I misunderstood the question, by the way.
If the pKaof HCHO2is 3.74 and the pH of an HCHO2/NaCHO2solution is 3.11, which of the following is true?
A. [HCHO2] < [NaCHO2]
B. [HCHO2] = [NaCHO2]
C. [HCHO2] << [NaCHO2]
D. [HCHO2] > [NaCHO2]
E. It is not possible to make a buffer of this pH from HCHO2 and NaCHO2
Answer:
D. [HCHO₂] > [NaCHO₂]
Explanation:
Formic acid, HCHO₂, is a weak acid that, in presence of its conjugate base, NaCHO₂ (CHO₂⁻), produce a buffer following H-H equation:
pH = pKa + log [CHO₂⁻] / [HCHO₂]
As pKa of the acid is 3.74 and pH of the solution is 3.11:
3.11 = 3.74 + log [CHO₂⁻] / [HCHO₂]
-0.63 = log [CHO₂⁻] / [HCHO₂]
0.2344 = [CHO₂⁻] / [HCHO₂]
A ratio [CHO₂⁻] / [HCHO₂] < 1, means:
[HCHO₂] > [CHO₂⁻]If a solute in water has a concentration of 0.12 mg/mL and has a concentration in ether of 0.60 mg/mL, what is the distribution coefficient, K, between ether and water
Answer:
K = 5
Explanation:
Distribution coefficient, K, is defined as the ratio between the concentration of a solute in two inmiscible solvents.
K = Concentration in solvent 1 / Concentration in solvent 2
Usually solvent 2 is water.
Replacing with the concentrations in the problem:
K = Concentration in ether / Concentration in water
K = 0.60mg/mL / 0.12mg/mL
K = 5a. When comparing the two elements Ca and As , the more metallic element is:__________ based on periodic trends alone.
b. When comparing the two elements Ca and As , the more metallic element is: ______________ based on periodic trends alone.
c. When comparing the two elements Sb and Pb , the more metallic element is:___________ based on periodic trends alone.
d. When comparing the two elements S b and P b , the more metallic element is: ________________ based on periodic trends alone.
Answer:
Ca.
Sb.
Explanation:
a. When comparing the two elements Ca and As , the more metallic element is:__________ based on periodic trends alone.
The correct option is Ca. Ca belongs to group 2 which is more easily give off it's electrons, compared to As which is a group 5 member.
c. When comparing the two elements Sb and Pb , the more metallic element is:___________ based on periodic trends alone.
The correct option here is Pb. Thhs is because ionization energy decreases down the group. Pb has a lower ionization enthalpy than Sb
Answer:
a) Ca
b) Ca
c) Pb
d) Pb
Explanation:
We must have it behind our minds that metallic properties decreases as we tend towards the right and side of the periodic table. This implies that we find more metallic species at the left hand did of the periodic table and as we tend towards the right hand side, elements become less metallic as indicated by periodic trends.
Calcium is a group 2 metal with full metallic properties while arsenic is a metalloid. Similarly, lead is a group 14 metal with full metal properties while antimony is a metalloid. Hence the answers given above.
An aqueous solution of nitric acid is standardized by titration with a 0.110 M solution of calcium hydroxide. If 21.1 mL of base are required to neutralize 23.8 mL of the acid, what is the molarity of the nitric acid solution? ? M Nitric acid
Answer:
[tex]\large \boxed{\text{0.195 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O
(b) Moles of Ca(OH)₂
[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]
(c) Moles of HNO₃
The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂
[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]
(d) Molar concentration of HNO₃
[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]
Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)
Answer:
pH = 4.05
Explanation:
The pH of the benzoic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.187
pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]
Where [] can be understood as moles of each specie.
Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.
Initial moles:
Initial moles of benzoic acid and sodium benzoate are:
Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid
Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate
Moles after reaction:
Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:
HCl + C₆H₅COONa → C₆H₅COOH + NaCl
That means after reaction moles of both species are:
Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles
Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles
Replacing in H-H equation:
pH = 4.187 + log [0.0475] / [0.065]
pH = 4.05
40.0 mL of a 0.65 M solution of HF is titrated with 0.100 M NaOH. After 0.400 L of the NaOH solution has been added, the resultant solution is:
Answer:
The solution is basic and after the equivalence point
Explanation:
HF reacts with NaOH as follows:
HF + NaOH → H₂O + NaF
1 mole of acid reacts per mole of base.
The solution of HF contains:
40.0mL = 0.0400L × (0.65mol / L) = 0.026 moles of HF,
The titration will reach equivalence point when you add 0.026 moles of NaOH. If more NaOH is added the solution will be basic because NaOH (A base) is in excess doing the solution basic.
Moles added of NaOH are:
0.400L × (0.100 mol / L) = 0.0400 moles of NaOH
As the addition of NaOH is > 0.026 moles,
The solution is basic and after the equivalence pointA triacylglycerol consists of _______ and ______ fatty acids. A glycerophospholipid is different in that it also consists of _______ but has ______ fatty acids. The hydroxyl group on its ______ carbon is attached by a phosphate bond to _______, which is attached by a phosphate ester bond to an ionized amino alcohol.
Answer:
1. Glycerol
2. Three
3. Bilayer membrane
4. diglyceride
5. 1 and 2 (C1 and C2)
6. C3
Explanation:
A triglyceride is defined as an ester which consists of glycerol and three fatty acids.
Glycerophospholipid are different from triglyceride as they are also consist of the membrane bilayer, which make a tough cell membrane and help them to hold the definite shape. Glycerophospholipids carry two or diglyceride fatty acid.
Hydroxyl group in glycerophospholipid is present in C1 and C2 and a phosphate group is attached at C3 (Carbon-3) with phosphodiester bond.
Hence, the correct answers are:
1. Glycerol
2. Three
3. Bilayer membrane
4. diglyceride
5. 1 and 2 (C1 and C2)
6. C3
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol
The given question is incomplete, the complete question is:
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol
A) –309 kJ/mol
B) –329 kJ/mol
C) None of the above
D) –349 kJ/mol
E) –369 kJ/mol
Answer:
The correct answer is option D, that is, -349 kJ/mol.
Explanation:
Based on the given information, the reaction is:
NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.
Therefore, there is a need to reverse the reaction and change the sign on ΔG.
Now the reaction will become,
Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).
A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250. g of water freezes at –2.34°C. Calculate the molar mass (in units of g/mol) of the solute. K f of water is 1.86°C/m
Answer:
[tex]M=174.87g/mol[/tex]
Explanation:
Hello,
In this case, we consider the depression in the freezing point as:
[tex]\Delta T=-i*m*Kf[/tex]
Whereas the van't hoff factor for ascorbic acid is 1 since it is covalent, thus, we solve for the molality as shown below:
[tex](-2.34-0)\°C=-m*1.86\°C/m\\\\m=\frac{-2.34\°C}{-1.86\°C/m} =1.26m[/tex]
Next, since molal units are mol/kg, we can compute the present moles of ascorbic acid in the 250 g (0.25kg) of water:
[tex]n=1.26mol/kg*0.25kg=0.315mol[/tex]
Finally, the molar mass with the given 55.0 g of ascorbic acid:
[tex]M=\frac{55.0g}{0.315mol}\\ \\M=174.87g/mol[/tex]
Regards.
The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?
Answer:
The correct answer would be - 2.4KJ or, 2400J
Explanation:
Given:
heat capacity of liquid water - 4.18 J/g·°C
heat of vaporization - 40.7 kJ/mol
Mass of water = 1g
Moles of water = mass/molar mass
= 1g/18.016g
= 0.055 moles
Then,
Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)
= m *s*dt + moles * heat of vaporization
= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ
= 137.94J + 2.26KJ
=0.138KJ + 2.26KJ
=2.4KJ or, 2400J
Thus, the correct answer would be - 2.4KJ or, 2400J
a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary
A certain phospholipid molecule contains two fatty acid chains of eight carbon atoms each having no carbon-carbon double bonds in the chain. The phosphate ion is attached to a polar
organic molecule, X, just as highlighted in yellow in the figure in the introduction.
Complete the structure of the phospholipid molecule using the structure in the introduction. To include a X group in your structure, draw another atom, hover over the atom with the
mouse, and then press the X key on your keyboard.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.
Answer:
Here's what I get
Explanation:
The fatty acids form ester C-O bonds with the OH groups of glycerol and the phosphoric acid forms phosphate P-O bonds.
The phosphate also forms P-O ester linkages with the OH groups in other alcohols such as choline.
The structure below shows the phospholipid with two saturated C₈ fatty acids and the phosphate group.
Which statement best describes covalent bonding?
Answer:
Option C. Electrons are shared between two atoms
Explanation:
Covalent bonding is a type of bonding which exist between two non metals.
In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.
This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:
H + Cl —> HCl
Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.
Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.
During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.
C. Electrons are shared between two atoms.
What is Covalent Bonding?
A covalent bond is fashioned among non-metals which have comparable electronegativities. Neither atom is "strong" sufficient to draw electrons from the other.It is formed when pairs of electrons are shared by atoms.Atoms will covalently bond with different atoms that allows you to benefit extra stability, that is gained through forming a complete electron shell. By sharing their outer most (valence) electrons, atoms can replenish their outer electron shell and gain stability.For example:
In H₂ molecule; there is a covalent bond formation between two hydrogen atoms as the electron from each hydrogen atom is shared leading to the formation of hydrogen molecule.
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Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
A student uses gravimetric determination to find the water of hydration of a hydrated sample of cobalt (II) sulfate. The student’s sample contained 0.0098459 moles of anhydrous cobalt(II) sulfate (CoSO4) and 0.068921 moles of water (H2O). What is the formula of the cobalt(II) sulfate hydrate? (1 point)
Answer:
CoSO₄.7H₂O
Explanation:
Some salts are in its hydrate form to stabilize them. The hydrated form of CoSO₄ is:
CoSO₄.XH₂O
To find the hydration of the COSO₄ you must find the ratio of H₂O and CoSO₄ that is:
0.068921 moles H₂O / 0.0098459 moles CoSO₄ = 7
That means you have 7 moles of water per mol of CoSO₄ and the formula is:
CoSO₄.7H₂O