1. Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest.
A. HCl
B. H2S
C. HBr
D. BH3
2. Without consulting the table of acid-dissociation constants, match the following acids to the given Ka1 values.
1. H2S
2. H2SO3
3. H2SO4
A. Kal = 1.7 x 10^-7
B. Kal = 1.7 x 10^-2
C. Kal = very large

Answer:

[tex]m_{Si}=37.2gSi[/tex]

Explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]

We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:

[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]

Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:

[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]

Best regards.

Answer:

See explanation

Explanation:

-) True or False: All R stereocenters are dextrorotatory.

The absolute configuration is based is in specific rules that are not related to the ability to deflect polarized light. FALSE

-) True or False: Chiral centers in organic molecules will have 4 different groups attached to a carbon.

A chiral carbon by definition is a carbon with 4 groups. TRUE

-) True or False: A racemic mixture has an optical activity of 0.

In a racemic mixture, we have equal amounts of enantiomers, and this cancels out the optical activity. TRUE

-) True or False: Normal linear amines can be chiral centers.

In primary amines, we have 2 hydrogens. Therefore all the groups can not be different. So, is TRUE

-)True or False: Compound C has an optical activity of 0.

We need to know the structure of the compound

-)True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree.

If we have the exact opposite (as we have in this case) the magnitud of the optical activity value will remain and the sign will change. TRUE

-)True or False: A CN is a higher priority than a CH2OH.

In this case, a carbon is directly bonded to the chiral carbon. The carbon on CN is bonded to a nitrogen atom and the carbon on CH2OH is bonded to an oxygen. So, the CH2OH will have more priority because O has a higher atomic number. FALSE

-)True or False: All molecules with chiral centers are optically active.

We can have for example mesocompounds in which the optical activity is canceled out due to symmetry planes. FALSE

-)True or False: To have an enantiomer a molecule must have at least two chiral centers.

A pair of enantiomers is made between at least 1 chiral carbon. Enantiomer R and enantiomer S. FALSE

-)True or False: Chiral molecules are always optically active.

We can have racemic mixtures or mesocompounds with chiral carbons but without optical activity. FALSE

-)True or False: A CH2CH2Br is higher priority than a CH2F.

The "Br" atom is bonded in the third carbon (respect to the chiral carbon) and the "F" atom is bonded to the second carbon. Therefore CH2F has more priority than CH2CH2Br. FALSE

-)True or False: Meso molecules with two stereocenters have a R,S configuration.

On the compounds with R and S configuration at the same time can have symmetry planes so, we will not have optical activity. TRUE

True or False: Diastereomers have the same physical properties except in a chiral environment.

All diastereomers have the same physical properties. TRUE

True or False: Compound H has an optical activity of 0.

We have to have the structure of the compound.

True or False: A C=C double bond is higher priority than a -CH(CH3)2.

In the case of C=C, we can say that is equivalent to two carbon bonds without hydrogens. Therefore C=C has higher priority. TRUE

Answer:

pH of the buffer is 7.48

Explanation:

The H₂PO₄⁻/HPO₄²⁻ buffer has a pKa of 7.21. You can find pH of this buffer following H-H equation:

pH = pKa + log [A⁻] / [HA]

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

Where [] represents molarity of each specie of the buffer and, as volume is 1.00L, also represents its moles.

Thus, to find pH of the buffer we need to calculate moles of each specie, thus

Moles of 18.0g of KH₂PO₄(Molar mass: 136.086g/mol) = moles of H₂PO₄⁻ are:

18.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.132 moles of KH₂PO₄= H₂PO₄⁻

Moles of 35.0g of Na₂HPO₄(Molar mass: 141.96g/mol) = moles of HPO₄²⁻ are:

35.0g Na₂HPO₄ ₓ (1mol / 141.96g) = 0.2465 moles of Na₂HPO₄= HPO₄²⁻

Replacing in H-H equation:

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

pH = 7.21 + log [0.2465] / [0.132]

pH = 7.48

Answer:

Ph3OH

Explanation:

The reaction is between ethyl benzoate (PhCOOC2H5) and a Grignard reagent PhMgBr(C6H5MgBr).

The first step in the reaction mechanism is that the ethyl benzoate is converted to PhCOPh by the first molecule of Grignard reagent.

The second molecule of Grignard reagent now converts PhCOPh to Ph3O^-. In the presence of acid, Ph3O^- is now protonated to yield Ph3OH which is the major organic product of the reaction.

See image attached for more details.

Answer:

The side chains of the amino acids alternate above and below the sheet

Explanation:

Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.

Explanation:

Combustion of butane gives following reaction

C4H10 + 13/2 O2 → 5H20 + 4CO2

Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.

17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.

Weight % of O2 in Air = 20 %

Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %

The dew point temperature of H2O is :

D = (237.3×B)/(1-B)

B = ln(E/6.108)/17.27

E is vapor pressure at given temp.

At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars

B = 0.1114

D = 29.74°C

Answer:

0.0668g

Explanation:

Step 1: Given data

Concentration of lead: 668 ppm (mg/kg)

Mass of water: 100.0 g

Step 2: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]100.0g \times \frac{1kg}{1,000g} = 0.1000kg[/tex]

Step 3: Calculate the mass of lead in 0.1000 kg of water

There are 668 mg of Pb in 1 kg of water.

[tex]0.1000kgWater \times \frac{668mgPb}{1kgWater} = 66.8mgPb[/tex]

Step 4: Convert the mass of Pb to grams

We will use the relationship 1 g = 1,000 mg.

[tex]66.8mg \times \frac{1g}{1,000mg} = 0.0668g[/tex]

Answer:

Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.

However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.

In the qualitative analysis test of chloride upon addition of AgNO₃, presence of diffrerent chloride salts other than KCl will gives false test.

On adding halogens on the silver nitrate solution we will get the precipitate of diffrent colors of diffrent halides.

So, presence of diffrent chloride salt in the silver nitrate solution in addition with KCl will gives a false positive result for the test.

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Answer:

8.96g\ cm3

Explanation:

D = ( 89.6g \ 10cm3)

( 89.6\ 10) ( g\ cm3) = 8.96g\cm3

Answer:

Explanation:

The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.

Answer:

A. It shows that the number of each type of atom stays the same.

Explanation:

Though you may see a change in the way they are arranged, the same  number of atoms are present before and after. Balanced chemical equations show equal numbers of  atoms of each element on each side of the equation.

Answer:

Explanation:

 one mole of photon will contain  

6.02 x 10²³ no of photons

energy of one photon = h x f

= h c / λ

h is plank's constant , c is velocity of light and λ is wavelength

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 179.3 x 10⁻⁹

= .11 x 10⁻¹⁷ J

energy of one mole of photon

= 6.02 x 10²³ x .11 x 10⁻¹⁷

= .6622 x 10⁶ J

Answer:

pH = 10.87

Explanation:

You can find pH of a buffer (Mixture of a weak base, X, and its conjugate acid, HX) by using H-H equation:

pH = pKa + log [X] / [HX]

pKb is -log Kb = 3.13

Using Hint pKw = pKa + pKb (pKw = 14)

pKw = pKa + pKb

14 - 3.13 = pKa

10.87 = pKa

Replacing in H-H equation:

pH = 10.87 + log [0.42M] / [0.42]

pH = 10.87 + log 1

Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

Answer:

The correct answer is 1.60.

Explanation:

Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,  

Moles = volume * concentration of HCl

= 25/1000*0.10 = 0.0025 moles

Similarly the moles of NaOH added will be determined by using the formula,  

Moles of NaOH added = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

The reaction taking place in the given case is,  

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

= 0.001 / 0.040 = 0.025 M

pH = -log[H+]

= -log[0.025]  

= 1.60

The pH after 15 ml of NaOH has been volume is 1.60.

It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,

Moles is = volume * concentration of HCl

Then is = 25/1000*0.10 = 0.0025 moles

Besides, the moles of NaOH added will be determined by using the formula,

When the Moles of NaOH added is = volume * concentration of NaOH

= 15/1000 * 0.10 = 0.0015 moles

When The reaction taking place in the given case is,

HCl + NaOH = NaCl + H2O

Now the moles of excess H+ = moles of excess HCl

= 0.0025 - 0.0015 = 0.001 moles

Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L

[H+] = moles of H+/total volume

After that = 0.001 / 0.040 = 0.025 M

pH = -log[H+]

Then = -log[0.025]

Therefore, = 1.60

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Answer:

A. it gives up electrons

Explanation:

In a redox reaction, the reducing agent is the element or compound that undergoes oxidation and gives up electrons. The oxidizing agent is the element or compound that undergoes reduction and gains electrons.

Hope that helps.

Answer:

1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻

2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻

3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃

Hope this helps.

The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).

The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻

Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.

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Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = 1.40x10⁻³ moles of H₃C-CH(OH)-COOH

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

Answer:

1447584654 years or 1.44 billion years

Explanation:

From the formula;

0.693/t1/2=2.303/t log No/N

Where;

t1/2 = half life of the U-235 nuclides

No= amount of U-235 initially present

N= amount of U-235 present after a time t

t= time taken for N amount of U-235 to remain

Since N= 0.24No

Substituting values

0.693/703×10^6 = 2.303/t log No/0.24No

9.8578×10^-10 = 2.303/t log 1/0.24

9.8578×10^-10 = 1.427/t

t= 1.427/9.8578×10^-10

t= 1447584654 years

Answer:

The correct answer is 2.75 grams of HCl.

Explanation:

The given balanced equation is:  

CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams

The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,  

= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.  

Answer:

Sn2 mechanism

Explanation:

In this case, our nucleophile is the "OH" on (2R,3R)-3,5-dimethylhexan-2-ol. The alcohol group will attack the [tex]PCl_3[/tex] to produce a new bond between O and P with a positive charge in the oxygen. Additionally, when the OH attacks a Br atom leaves the molecule producing a bromide ion.

In the next step, the bromide ion produced will attack the carbon bonded to the OH that now is bonded to [tex]PCl_2[/tex]. An Sn2 reaction takes place and the substitution would be made in only one step. Due to this, we will have an inversion in the stereochemistry and the absolute configuration on carbon 2 will change from "R" to "S" to produce (2S,3R)-2-bromo-3,5-dimethylhexane.

I hope it helps!

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

Answer:

To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:

delta(m)= mass of the products - mass of reactants

Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu;  atomic mass of  U-235= 235.043922 amu

Therefore:

delta(m)=  (143.9385 + 89.907738) - (235.043922)  = -1.197682 amu

Recall that to calculate energy in joules, we use the formula:

Energy = mc^2

Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2

= -1.7898104 x 10^-10 J

= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)

= -4.586 x 10^11 J per gram of  energy released

Explanation:

To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:

delta(m)= mass of the products - mass of reactants

Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu;  atomic mass of  U-235= 235.043922 amu

Therefore:

delta(m)=  (143.9385 + 89.907738) - (235.043922)  = -1.197682 amu

Recall that to calculate energy in joules, we use the formula:

Energy = mc^2

Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2

= -1.7898104 x 10^-10 J

= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)

= -4.586 x 10^11 J per gram of  energy released

Answer:

The correct answer is - alternating and direct, in order.

Explanation:

Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.

Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.

Thus, the correct answer is -  alternating and direct, in order.

Answer:

1. alternating

2. direct

3. The sun heats up the atmosphere as Earth spins, creating areas of high and low temperature. This temperature difference causes wind to start moving through convection, which can then drive a wind turbine to produce electricity.

Explanation:

From Penn

Answer:

See figure 1

Explanation:

For this question, we have to remember that a hydrogen bond is an interaction in which we have a partial attraction between a positive dipole and a negative dipole, and in this attraction, we have in the middle a hydrogen atom.

In this interaction, we can have a donor (positive dipole) or a receptor (negative dipole). The receptor is a heteroatom (an atom different to carbon or hydrogen) with high electronegativity. The donor is usually hydrogen atom bonded to the heteroatom.  

I hope it helps!

Answer:

1-remain the same

2- remain the same

3-decrease

--------------------------

- decrease

- As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in Kc.

--------------------------------

1- decrease

2-increase

3-decrease

4-remain the same

5-decrease

Explanation:

According to Le Chateliers principle, an increase in the volume of a gaseous system at equilibrium will shift the equilibrium position towards the side in which there are less volumes. Hence the answers written. When there is no change in volume, the number of moles of products remain the same.

-------------------------------

For an exothermic reaction, increasing the temperature shifts the equilibrium to the lefthand side.

When temperature is decreased, the equilibrium position will shift towards the right according to Le Chateliers principle

---------------------

Addition of a catalyst aids the reaction in which CO is consumed to proceed faster hence CO decreases in the system.

Since the reaction is exothermic, according to Le Chateliers principle, when the temperature is increased, the equilibrium position shifts towards the lefthand side and more CO is now present in the system.

When the pressure of the system is increased, the equilibrium position will shift towards the right hand side and more CO is converted to products hence its concentration in the system decreases.

Addition of argon gas has no effect on the equilibrium position since it does not participate in the reaction. However, addition of the reactant gases increases the rate of reaction and shifts the equilibrium position towards the right hand side thus decreasing the concentration of CO in the system.

Answer:

[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q  = It  

[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.  

(a) Calculate q

[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]

Note: The answer can have only one significant figure because that is all you gave for the time.

Answer:

Q = 6.68x10⁻⁶

Explanation:

The initial solutions are Pb(NO₃)₂ and NaBr. As sodium and nitrates salts are soluble, the insoluble product must be the formed from the other ions, PbBr₂.

The soluble product equilibrium is written as:

PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)

Where Q is defined as:

Q = [Pb²⁺] [Br⁻]²

As:

[Pb²⁺] = 0.0630M

[Br⁻] = 0.0103M

Q = [Pb²⁺] [Br⁻]²

Q = [0.0630M] [0.0103M]²

Answer:

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