Aspirin is usually packaged with
A. acetic anhydride
B. salicylic acid
C. buffering agents ​

Answer:

3.011x10^24 molecules

Explanation:

1 mole=6.022x10^23 molecules

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

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:

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

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:

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:

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:

The ranks is

Ge: 3d10 4s2 4p2 (6 electrons in the outer shell)

Br: 3d10 4s2 4p5 (7 electrons in the outer shell)

Kr: 3d10 4s2 4p6 (8 electrons in the outer shell)

Explanation:

Electronic configuration reffers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It gives us the understanding of the shape and energy of its electrons. The electronic configuration explain the The electron affinity or propensity to attract electrons

It Should be noted that the most stable configuration in an electronic configuration is attributed to when the last shell is full, i.e. when the last shell has 8 electrons.

When an atom is closer to reach the 8 electrons in the outer shell, then it's electron affinity big.

Considering the given three configuration of the elements above, we can see that "Br"needs requires only 1 electron to have 8 electrons in the outer shell, therefore, it is considered to have the biggest electron affinity among them which is reffers to as the LEAST NEGATIVE.

Ge: with the electronic configuration 3d10 4s2 4p2 has 6 electrons in the outer shell which means it still requires 2 electrons to complete 8 electrons in its outer shell, so it can be deducted that it posses an atom that is more negative than Br.

Kr: with the electronic configuration 3d10 4s2 4p6 which is a noble gas has 8 electrons in the outer shell cannot add more electrons to its outer shell because the 8 electrons is complete posses the least electron affinity among the three elements and it is the MOST NEGATIVE

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

Answer:

C = 98%

Explanation:

Hello,

To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.

Mass of salt recovered = 1.47g

Initial mass of salt = 1.50g

Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100

Percentage of salt recovered = (1.47 / 1.50) × 100

Percentage of salt recovered = 0.98 × 100

Percentage of salt recovered = 98%

The percentage of salt recovered is equal to 98%

Answer:

Explanation:

You will investigate the breakdown of starch by amylase at different pHs.

The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.

The amylase will break down the starch.

A series of test tubes containing a mixture of starch and amylase is set up at different pHs.

A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.

A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.

When the starch solution is added:

For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.

The time taken for the disappearance of starch is not the rate of reaction.

It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.

We can calculate the rate of the reaction by calculating  \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.

A similar experiment can be carried out to investigate the effect of temperature on amylase activity.

Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.

Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.

good luck :)

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:

142 grams

Explanation:

To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:

2(31)+5(16)=62+80=142

Hope this helps!

Complete Question

The complete question is shown on the first uploaded image

Answer:

For reaction 1

    [tex]K_{eq} = 10^{29}[/tex]

     The equilibrium position is to the right

For reaction 2

        The equilibrium position is to the left

Explanation:

Generally  [tex]pKa[/tex] is mathematically evaluated as  

[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]

And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]

From the question we are told that

For reaction 1

         [tex]pKa_\ {left}} \ = 9[/tex]

        [tex]pKa_\ {right }} \ = 38[/tex]

So

       [tex]pKa = 9-38[/tex]

       [tex]pKa =-29[/tex]

So  [tex]K_{eq} = 10^{-(-29)}[/tex]

      [tex]K_a = 10^{29}[/tex]

This implies that the equilibrium position is to the right

   For reaction 2

       [tex]pKa_\ {left}} \ = 15.9[/tex]

       [tex]pKa_\ {right }} \ = 9.24[/tex]

So

       [tex]pKa = 15.9-9.24[/tex]

       [tex]pKa = 6.66[/tex]

So  [tex]K_{eq} = 10^{-(6.66)}[/tex]

      [tex]K_{eq} = 10^{-6.66}[/tex]

This implies that the equilibrium position is to the left

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:

Li>Na>K>Rb

Explanation:

Answer:

c. Li > Na > K > Rb

Explanation:

edge 2021

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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:

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

Answer:

5-methylhex-1-yne//5-methylhex-2-yne//2-methylhex-3-yne

Explanation:

We have to start with the information on the IR spectrum. The signal at 3300 is due to a C-H bend sp carbon and the signal in 2200 is due to the stretch carbon-carbon. Therefore we will have an alkyne. Now if we have 2-methylhexane as the product of hydrogenation we have several options to put the triple bond. Between carbons 1 and 2 (5-methylhex-1-yne), between carbons 2 and 3 (5-methylhex-2-yne) and between carbons 3 and 4 (2-methylhex-3-yne). On carbon 5 we have a tertiary carbon therefore we dont have any other options.

See figure 1

I hope it helps!

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:

The rates of decay of radioactive elements

Explanation:

The age of a rock in years is called its absolute age. Geologists find absolute ages by measuring the amount of certain radioactive elements in the rock. When rocks are formed, small amounts of radioactive elements usually get included.

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

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

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:

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.

Learn more about the in terms of charge distribution, why a molecule.

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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:

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:

1 particle of the gas would occupy a volume of 3.718*10⁻²³L

Explanation:

Hello,

1. The sample has a particle of 6.022×10²²particles

2. Volume of the sample = 22.4L

3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K

4. Pressure of the sample = 1.0atm

What volume would 1 particle of the gas occupy?

But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms

What would be the number of moles for 1 particule?

1 mole = 6.022×10²² particles

X moles = 1 particle

X = (1 × 1) / 6.022×10²² particles

X = 1.66×10⁻²⁴ moles

Therefore, 1 particle contains 1.66×10⁻²⁴ moles

Since we know our number of moles, we can proceed to use ideal gas equation,

Ideal gas equation holds for all ideal gas and is defined as

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles of the gas

R = ideal gas constant = 0.082 L.atm / mol.K

T = temperature of the gas

PV = nRT

Solving for V,

V = nRT/ P

We can now plug in our values into the above

equation.

V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1

V = 3.718*10⁻²³L

Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.

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:

CONVERT IT:

50°F is equal to 10°C

Answer:

10 degrees Celsius

Explanation:

(50°F − 32) × 5/9 = 10°C