In what unit do we usually measure the force of the earth gravity? Acceleration due to gravity is 9.8/s^2

Answer:

A. The cell will undergo crenation

B. The cell will undergo hemolysis

C. The cell will undergo hemolysis

D. The cell will undergo crenation.

E. The cell will undergo neither crenation nor hemolysis

Explanation:

A hypotonic solution is a solution in which the concentration of solutes is greater inside the cell than outside the cell.

A hypertonic solution is a solution in which the concentration of solutes is greater outside the cell than inside the cell.

An isotonic solution is a solution in which the concentration of solution is the same outside and inside of the cell. A solution with 5% (m/v) glucose and 0.9% (m/v) NaCl is an isotonic solution.

When a red blood cell is placed in a hypotonic solution, it will swell and burst. This is known as hemolysis.

When placed in a hypertonic solution, a red blood cell will lose water and shrivel. This is is known as cremation.

When a red blood cell is placed in an isotonic solution, neither hemolysis or crenation occurs as there is no net movement of water across the cell's membrane.

A: 3.75 % (m/v) NaCl Solution is a hypertonic solution. The cell will undergo crenation

B: 1.92 % (m/v) glucose Solution is hypotonic. The cell will undergo hemolysis

C: Distilled H2O Solution is a hypotonic solution. The cell will undergo hemolysis

D: 9.03 % (m/v) glucose Solution is a hypertonic solution. The cell will undergo crenation

E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl are both isotonic solutions. The cell will undergo neither hemolysis not crenation.

Solution A (3.75% NaCl): Crenation

To find out if a red blood cell will shrink, burst, or stay the same when placed in a solution, we have to think about how concentrated the solution is compared to the red blood cell.

A red blood cell has a normal concentration of about 0. 9% salt or 0. 3% sugar Solutions that have more concentrated substances are called hypertonic, while solutions that have less concentrated substances are called hypotonic.

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Answer:

not 100% but i think its 1.57x10^20

Explanation:

5.25x10^-4g / 2.016g

2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules

this equation is balanced.

Answer:

Nothing to do, is already balanced.

Explanation:

MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)

Left Side

Mn =1

Cl = 2

N = 2

H = 8

C = 1

O = 3

Right Side

Mn =1

Cl = 2

N = 2

H = 8

C = 1

O = 3

Answer:

2.7 °C

Explanation:

78.4-75.7=2.7

The student's result is 2.7° C less than the expected result.

We have a student conducting an experiment to find the boiling point of ethanol.

We have to determine how far off from the accepted value is the student's result.

Measurement Error (also called Observational Error) is the difference between a measured quantity and its true value.

According to the question, we have -

True value of the boiling point of ethanol [T] = 78.40°C.

Measured value of the boiling point of ethanol [M] = 75.70°C.

The Error (E) in the measurement will be -

E = T - M

E = 78.40° - 75.70° = 2.7° C

Hence, the student's result is 2.7° C less than the expected result.

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Answer:

Explanation:

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas.

The covalent bond between two atoms can be polar or nonpolar. If the atoms are equal, the bond will be nonpolar (since no atom attracts electrons more strongly). But, if the atoms are different, the bond will be polarized towards the most electronegative atom, because it will be the atom that attracts the electron pair with more force. Then it will be polar.

It can occur in a molecule that the bonds are polar and the molecule is nonpolar. This occurs because of the geometry of the molecule, which causes them to cancel the different equal polar bonds of the molecule.

In carbon tetrachloride the bonds are polar, but the tetrahedral geometry of the molecule causes all four dipoles to cancel out and the molecule to be apolar.

The carbon tetrachloride s CCL4 is a carbon molecule and four chloride molecule's. The carbon tetrachloride is a nonpolar, as the dipole movement of the molecules ae evenly spaced around the central carbon atom.

Hence the carbon tetrachloride happens to be a nonpolar molecular.

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The given question is not complete, the complete question is:

Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100mL. The ratio of these quantities is equal to the partition coefficient, k, which equals What weight of malononitrile would be recovered by extraction of (a) three 100-mL portions of ether and (b) one 300-mL portion of ether? SHOW WORK (Can be written in pen and attached to report). Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In part (a), for the first of the three extractions, the concentration of malononitrile in the ether layer is x/100 and in the water layer is (30-x)/100.

Answer:

The correct answer is 10 grams and 18 grams.

Explanation:

Based on the given question, 20 gram per 100 ml is the solubility of malononitrile in ether, and 13.3 gram per 100 ml is the solubility of malononitrile in water.  

Thus, the ration of the solubility is,  

Solubility in water/solubility in ether = 20/13.3 = 1.50

a) Let w be the weight of malononitrile extracted into water in every extraction. Then the concentration of the ether layer will be w/100. The concentration in the water layer will be 30-w/300. Now the ratio will be,  

Ratio = w/100 / (30-w)/300

1.50 = w/100 * 300 (30-w)

w = 10

Hence, the weight of malononitrile recovered by extraction is 10 grams.  

b) The concentration in the ether layer will be w/300. The concentration in the water layer will be (30-w) / 300. Now the ratio will be,  

Ratio = w/300 / (30-w) / 300

1.50 = w/300 * 300 (30-w)

w = 18

Hence, 18 grams is the weight of malononitrile recovered by extraction.  

Answer:

If you're ever shaving in the bathroom, turn the water off.

Explanation:

If you do this, you can save at least 3-4 pounds of water.

Answer:

The following activities can help conserve water while taking showers:

1) Lower shower time

2) Don't leave shower running.

3) Check for leaks

Answer:

The class of this compound is aldehyde or ketone (i).

Explanation:

Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.

Further information on molecular formula would be required for structural elucidation.

Answer:

Diamagnetic have paired electrons while paramagnetic have at least one unpaired electron.

Explanation:

F2, C2 and N2 are diamagnetic while O2 and B2 are paramagnetic. Diamagnetic are those atoms which have paired electrons while paramagnetic are those atoms which contain at least one unpaired electron so we can say that F2, C2 and N2 have paired electrons while O2 and B2 have unpaired electrons. When diamagnetic materials are allowed to contact with external magnetic field so they will be repelled while paramagnetic materials are attracted by magnetic field due to the presence of unpaired electrons.

Iron oxide (FeO) is the strongest paramagnetic material having the value of 720.

Answer:

Approximately [tex]21\; \rm g[/tex].

Explanation:

[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:

[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].

In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:

[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].

Look up the relative atomic mass data on a modern periodic table:

Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:

[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].

[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].

Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].

Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].

Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:

[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].

When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of

[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)

Answer: D. They would device an experiment that could test the two scientists conclusions.

Explanation:

The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.

In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.

The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.

Answer:d

Explanation:

Answer:

A. [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.  [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

Explanation:

The equation for this chemical reaction is :

[tex]2NO_{(g)} +O_{2(g)} \to 2NO_{2(g)}[/tex]

Using the following relation to calculate [tex]\Delta G^0[/tex];

[tex]\Delta G^0 = [2(\Delta G^0_{NO_{2(g)}}] - [1(\Delta G^0_{O_{2(g)}})+ 2(\Delta G^0_{NO_{g}})][/tex]

At 298 K; the standard Gibbs Free Energy for the formation are as follows:

[tex]\Delta G^0_{NO_{2(g)}} = 51.2 \ kJ/mol[/tex]

[tex]\Delta G^0_{O_{2(g)}} = 0[/tex]

[tex]\Delta G^0_{NO_{g}}= 87.6 \ kJ/mol[/tex]

Replacing them into the above equation;

[tex]\Delta G^0 = [2(51.2 \ kJ/mol}] - [1(0)+ 2(87.6 \ kJ/mol})][/tex]

[tex]\Delta G^0 = [102.4 \ kJ/mol}] - [175.2 \ kJ/mol})][/tex]

[tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex]

Thus; [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.

Using the same above chemical equation;

The relation used for calculating [tex]\mathbf{\Delta G^0}[/tex] of the reaction when the temperature is 702 K is:

[tex]\Delta G^0_{702 \ K} = \Delta H^0_{xn} - T \Delta S^0_{rxn}[/tex]

where;

[tex]\Delta G^0_{702 \ K} =[/tex] Gibbs free energy of the reaction at 702 K

[tex]\Delta H^0_{xn}[/tex] = standard enthalpy of the reaction = -116.2 kJ/mol

[tex]\Delta S^0_{rxn}[/tex] = standard entropy of the reaction = -146.6 J/mol/K

Temperature T = 702 K

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( -146.6 \ J/mol. K (\dfrac{1 \ kJ }{1000 \ J})[/tex]

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( 0.1466 \ kJ/mol.K})[/tex]

[tex]\Delta G^0_{702 \ K} = -13.2868 \ kJ/mol.K}[/tex]

[tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex]

Thus [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

Answer:

2. Precipitation Reaction

Explanation:

Na3PO4(aq) + FecCl2(aq) = Fe3(PO4)2(s) + NaCl(aq)

Fe3(PO4)2(s) - solid, it means it will precipitate.

It is a precipitation reaction.

A precipitate is a solid that forms out of a solution. A common example is that of the mixing of two clear solutions: (1) silver nitrate (AgNO3) and (2) sodium chloride (NaCl): The reaction is. The precipitate forms because the solid (AgCl) is insoluble in water.

A precipitate is a solid formed in a chemical reaction that is different from either of the reactants. This can occur when solutions containing ionic compounds are mixed and an insoluble product is formed. The identity of the precipitate can often be determined by examining solubility rules.

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Answer:

Explanation:

There are 35 protons.

The number of electrons = 36 electrons gives a -1 charge.

Where did all the other minus charges go?

They must be balanced by 35 protons.

Answer:

p = 1.714 g/cm3

Explanation:

Density Equation

p=mV

Where:

p = density

m = mass  = 60g

V = volume = 35cm3

p = 60g x 35cm3

p = 1.714 g/cm3

p=1.714g/cm³

Explanation:

v=35cm³

m=60g

P=mass/volume (density formula)

=60/35

=1.714g/cm³

Answer:

3.011x10^24 molecules

Explanation:

1 mole=6.022x10^23 molecules

5 moles*6.022x10^23 molecules/mole=3.011x10^24 molecules

the second statement is the correct one quarks are needed to balance charges in all subatomic particles such as neutrons, protons and electrons

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

Answer:

pH = 8.75

Explanation:

100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.

The reaction is:

HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO

Moles of butanoic acid are:

0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic

For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:

0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.

Total volume in equilibrium is 1.56L + 0.10L = 1.66L

That means concentration of butanoic acid is:

0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂

At equivalence point, there is just C₃H₇CO⁻ in solution

Kb of butanoic acid is:

C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻

Kb = Kw / Ka

Ka = 10^-pKa

Ka = 1.51x10⁻⁵

Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰

The equilibrium of Kb is:

Kb = 6.61x10⁻¹⁰ =  [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]

As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:

[C₃H₇CO⁻] = 0.04884M - X

[HC₃H₇CO₂] = X

[OH⁻] = X

Replacing in Kb:

6.61x10⁻¹⁰ =  X² / [0.04884M - X]

0 =  X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹

Solving for X:

X = -5.68x10⁻⁶ → False solution. There is no negative concentrations

X = 5.683x10⁻⁶ → Right solution.

As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.

pOH = - log [OH⁻]

pOH = 5.245

pH = 14 - pOH

Answer:

3) 310 g/mol

Explanation:

Hello,

In this case, for calcium carbonate, we are able to compute its gram-formula mass by considering the atomic mass of each element composing it and their subscripts as shown below:

[tex]M=3*m_{Ca}+2*m_P+4*2*m_O\\[/tex]

Thus, we compute:

[tex]M=3*40g/mol+2*31g/mol+4*2*16g/mol\\\\M=310g/mol[/tex]

Hence answer is 3) 310 g/mol . Remember this is also known as the molar mass of the mentioned compound.

Best regards.

Answer:

An orbital is a region in space where there is a high probability of finding an electron.

Explanation:

The orbital is a concept that developed in quantum mechanics. Recall that Neils Bohr postulated that the electron occupied stationary states which he called energy levels. Electrons emit radiation when the move from a higher to a lower energy level. Similarly, energy is absorbed by an electron to move from a lower to a higher orbit.

This idea was upturned by the Heisenberg uncertainty principle. This principle state that the momentum and position of a particle can not be simultaneously measured with precision.

Instead of defining a 'fixed position' for the electron, we define a region in space where there is a possibility of finding an electron with a certain amount of energy. This orbital is identified by a set of quantum numbers.

Answer:

three - dimensional space that shows the probability where an electron is most likely to be found

Answer: they blow up

Explanation: add them together and they will blow up

Answer:

Magnesium nitrate Reactions

Magnesium nitrate has a high affinity towards water. Therefore, heating it results to decompose into magnesium oxide, nitrogen oxides, and oxygen. 2 Mg(NO3)2 → 2 MgO + 4 NO2 + O2.

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.

Answer:

a technique in which cells are purposefully grown under specific conditions

Explanation:

Answer:

its c

Explanation:

correct edge2020

Answer:

A. Solubility of calcium sulfite increases

B. Solubility of calcium fluoride increases

C. Solubility of Silver bromide decreases

Explanation:

The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.

Case 1. Calcium sulfite

The dissociation reaction of CaSO3 is given below:

CaSO3 ----> Ca²+ + SO3²-

SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;

H+ + SO3²- ---> HSO3-

This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.

Case 2. Calcium fluoride

The dissociation reaction of calcium fluoride (CaF2) is shown below.

CaF2 ----> Ca²+ + 2F-

Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:

H+ + F- ---->. HF

Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.

Case 3: Silver bromide

The dissociation reaction of AgBr is as follows:

AgBr ----> Ag+ + Br-

The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.

The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.

Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.

Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.

However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.

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Answer:

u

uric acid is a useful diagnostic tool as screening for most of purine metabolic disorders. The importance of uric acid measurement in plasma and urine with respect of metabolic disorders is highlighted. Not only gout and renal stones are indications to send blood to the laboratory for uric acid examination

Answer:

No additional particle was produced during the decay.

Explanation:

The equation of decay is given as;

¹⁰₆C  + ⁰₋₁ e → ¹⁰₅B + x

To identify x, we have to calculate its atomic and mass number.

In the reactants side;

Atomic Number = 6 + (-1) = 5

Mass number = 10 + 0 = 10

In the products side;

Atomic Number = 5 + x

Mass Number = 10 + x

Generally, reactant = product

Atomic Number;

5 = 5 + x

x = 5 - 5 = 0

Mass Number;

10 = 10 + x

x = 10 - 10 = 0

This means no additional particle was produced during the decay.

Answer:

Lithium

Explanation:

The specific geat capacity of a substance is the energy required to raise 1 unit of that substance by one degree.

Heat energy (Q) = mc∇t

Q = heat energy

M = mass of the substance

c = specific heat capacity

∇t = change in temperature of the substance

Generally, increase in the specific heat capacity will lead to a lower final temperature likewise decrease in the specific heat capacity will lead to increase the final temperature of the substance.

From the data above, we can take just two specific heat capacity and test this theory.

Assuming we have a

Mass = 25g

Heat energy applied (Q) = 1 J

Initial temperature (T1) = 10°C

Final temperature (T2) = ?

Q = mc∇t

Q = mc (T2 - T1)

For Lithium, specific heat capacity = 3.58J/g°C

1 = 25 × 3.58 (T2 - 10)

Solve for T2

1 = 89.5 (T2 - 10)

1 = 89.5T2 - 895

89.5T2 = 896

T2 = 896 / 89.5

T2 = 10.011°C

For Magnesium (Mg) specific heat capacity = 1.02J/g°C

Q = mc∇t

1 = 25 × 1.02 × (T2 - 10)

1 = 25.5 (T2 - 10)

1 = 25.5T2 - 255

Solve for T2

25.5T2 = 256

T2 = 10.039°C

Notice the trend that decrease in the specific heat capacity leads to increase in the final temperature.

Try and continue for the elements and see how it works.

Answer:

See explanation below

Explanation:

This problem is a little long so I'm gonna be as clear as possible.

a) In this case we have two acids, HBr and HCHO2. Between these two acids, the HBr is the strongest, and does not have a Ka value to dissociate, while HCHO2 do.

In order to calculate pH we need the [H₃O⁺], and in this case, as HBr is stronger, the contribution of the weaker acid can be negligible, therefore, the pH of this mixture will be:

pH = -log[H₃O⁺]

pH = -log(0.115)

pH = 0.93

b) In this case it happens the same thing as part a) HNO₃ is the strongest acid, so the contribution of the HNO₂ which is a weak acid is negligible too, therefore the pH of this mixture will be:

pH = -log(0.085)

pH = 1.07

c) Now in this case, HCHO2 and HC2H3O2 are both weak acids, so to determine which is stronger, we need to see their Ka values. In the case of HCHO2 the Ka is 1.8x10⁻⁴ and for the HC2H3O2 the Ka is 1.8x10⁻⁵. Note that the difference between the two values of Ka is just 10¹ order, so, we can neglect the concentration of either the first or the second acid. We need to see the contribution of each acid, let's begin with the stronger acid first, which is the HCHO2, we will write an ICE chart to determine the value of the [H₃O⁺] and then, use this value to determine the same concentration for the second acid and finally the pH:

        HCHO₂ + H₂O <-------> CHO₂⁻ + H₃O⁺     Ka = 1.8*10⁻⁴

i)        0.185                                0          0

c)           -x                                 +x        +x

e)       0.185-x                             x           x

1.8*10⁻⁴ = x² / 0.185-x      

As Ka is small, we can assume that "x is small" too, therefore the (0.185-x) can be rounded to just 0.185 so:

1.8*10⁻⁴ = x²/0.185

1.8*10⁻⁴ * 0.185 = x²

x² = 3.33*10⁻⁵

x = 5.77*10⁻³ M = [H₃O⁺]

Now that we have this concentration, let's write an ICE chart for the other acid, but taking account this concentration of [H₃O⁺] as innitial in the chart, and solve for the new concentration of [H₃O⁺] (In this case i will use "y" instead of "x" to make a difference from the above):

        HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.225                                  0           5.77x10⁻⁶

c)            -y                                     +y            +y

e)        0.225-y                                y           5-77x10⁻³+y

1.8x10⁻⁵ = y(5.77x10⁻³+y) / 0.225-y   ---> once again, y is small so:

1.8x10⁻⁵ = 5.77x10⁻³y + y² / 0.225

1.8x10⁻⁵ * 0.225 = 5.77x10⁻³y + y²

y² + 5.77x10⁻³y - 4.05x10⁻⁶ = 0

Solving for y:

y = -5.77x10⁻³ ±√(5.77x10⁻³)² + 4*4.05x10⁻⁶ / 2

y = -5.77x10⁻³ ±√4.95x10⁻⁵ / 2

y = -5.77x10⁻³ ± 7.04x10⁻³ / 2

y₁ = 6.35x10⁻⁴ M

y₂ = -6.41x10⁻³ M

We will take y₁ as the value, so the concentration of hydronium will be:

[H₃O⁺] = 5.77x10⁻³ + 6.35x10⁻⁴ = 6.41x10⁻³ M

Finally the pH for this mixture is:

pH = -log(6.41x10⁻³)

pH = 2.19

d) In this case, we have the same as part c, however the Ka values differ this time. The Ka for acetic acid is 1.8x10⁻⁵  while for HCN is 4.9x10⁻¹⁰. In this ocassion, we the difference in their ka is 10⁵ order, so we can neglect the HCN concentration and focus in the acetic acid. Let's do an ICE chart and then, with the hydronium concentration we will calculate pH:

         HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.050                                  0              0

c)            -y                                     +y            +y

e)        0.050-y                                y              y

1.8*10⁻⁵ = y² / 0.050-y      

As Ka is small, we can assume that "y is small" too

1.8*10⁻⁵ = y²/0.050

1.8*10⁻⁵ * 0.050 = y²

y² = 9*10⁻⁷

y = 9.45*10⁻⁵ M = [H₃O⁺]

Finally the pH:

pH = -log(9.45x10⁻⁵)

pH = 3.02