Question 3 (11 points)
A gas has a volume of 690.0mL at -15.1°C and 392.0 mmHg. What would the volume of the gas be at
233.0°C and 0.700 atm of pressure? Answer with no decimal places.

Answer:

BF3

Explanation:

Hybridization can be defined as the mixing of two or more atomic pure orbitals. ( s, p ,  and d) to produce two or more hybrid atomic orbitals that are similar and identical in shape and energy e.g sp,sp²,sp³ ,sp³d, sp³d². Usually , the central atom of a covalent molecules or ion undergoes hybridization.

in BF3; Boron is the central atom. Here, A 2s  electron is excited from the ground state of boron ( 1s²2s²2p¹) to one empty orbitals of 2p.

The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals tat are trigonally arranged in the plane in order to minimize repulsion . Each of the three hybrid orbitals overlaps with p-orbital of fluorine atom to form three bonds of equal strength and with bond angles of 120⁰.

                                  Energy                                                                  

B     ⇵  ║   ⇅  ║    ↑     ----------->    *B    ⇵ ║     ↑  ║     ↑  ║   ↑

       1s      2s         2p                              1s       2s           2p  

    Ground state                                        Excited State

The above shows an illustrative example of how electrons  move from the ground state to the excited state.

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Explanation:

When lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, the following products are obtained:

Pb(NO3)2(aq) + NaCl(aq) —>

Aqueous Pb(NO3)2 will dissociate in solution as follow:

Pb(NO3)2(aq) —> Pb2+(aq) + 2NO3-(aq)

On the other hand, aqueous NaCl will dissociate as follow:

NaCl(aq) —> Na+(aq) + Cl-(aq)

The double displacement reaction will take place as follow:

Pb(NO3)2(aq) + NaCl(aq) —>

Pb2+(aq) + 2NO3-(aq) + Na+(aq) + Cl-(aq) —> Pb2+(aq) Cl-(aq) + Na+(aq) 2NO3-(aq)

Pb(NO3)2(aq) + NaCl(aq) —> PbCl2(s) + NaNO3(aq)

We simply balance the equation by putting 2 in front of NaCl and 2 in front of NaNO3 as shown below:

Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)

Answer: The empirical formula is [tex]PbO_2[/tex]

Explanation:

Mass of Pb =  4.33 g

Mass of O = (5.00-4.33) g = 0.67 g

Step 1 : convert given masses into moles

Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Pb = [tex]\frac{0.021}{0.021}=1[/tex]

For O = [tex]\frac{0.042}{0.021}=2[/tex]

The ratio of Pb O=  1: 2

Hence the empirical formula is [tex]PbO_2[/tex]

Answer:

Answer:

Mg, Sc, Ca

Explanation:

To figure out increasing atomic radii, we use Periodic Trends applied to the Elements of the Periodic Table to help us out. We know that the trend for atomic radii is increasing left and down. Since Ca is the furthest down and left of the 3, it has the largest atomic radius. Since Sc is next element to Ca, it would be the 2nd largest atomic radius of the 3. Since Mg is above Ca, it has the smallest atomic radius of the 3.

Answer:

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of formation of white precipitate of calcium carbonate.

Hope it helps you!

The milky substance formed is CO₂ gas.

The gas evolved because of reaction of acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas.

This happens because of formation of white precipitate of calcium carbonate.

Lime water turns milky when the gas liberated from an acidified carbonate solution is passed into it.

The liberated CO₂ reacts with lime water to give calcium bicarbonate as the precipitate.

Hence, the milky substance formed is CO₂ gas.

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Answer:

[tex]m_{Fe}=23.0gFe[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]FeO+Mg\rightarrow Fe+MgO[/tex]

Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:

[tex]n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO[/tex]

Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:

[tex]n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO[/tex]

Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:

[tex]m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\ \\m_{Fe}=23.0gFe[/tex]

Regards.

The question is incomplete, the solute was not given.

Let the solute be K₂CrO₄ and the solvent be water

Complete Question should be like this:

The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.

Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.  

Pvap = ________atm

Answer:

Pvap (of water above the solution) = 0.0306 atm

Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

Explanation:

Given

volume of solution = 1 Litre = 1000 mL of the solution

density of the solution = 1.063 g/mL

concentration of the solution= 0.438M

temperature of the solution= 298 K

vapour pressure of pure water = 0.0313atm

Recall: density = mass/volume

∴mass of solution = volume x density

m = 1000 x 1.063 = 1063 g

To calculate the moles of K₂CrO₄ = volume x concentration

= 1 x 0.438 = 0.438 mol

Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g

Mass of water = mass  of solution - mass of K₂CrO₄

= 1063 - 85.055 = 977.945 g

moles of water = mass/molar mass

∴ moles of water = 977.945/18.02 = 54.27 mol

 Dissolution of the solute

K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻

(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution

moles of ions = 3 x moles of K₂CrO₄

= 3 x 0.438 = 1.314 mol

Vapor pressure of solution = mole fraction of water x vapor pressure of water  

= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm

Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

The molar mass of Mg is 24.31 g/mol.

[tex]12g \times \frac{1mol}{24.31g} = 0.49mol[/tex]

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.

Answer:

First, write a balanced equation of the reaction.

A hydrocarbon reacting with oxygen usually gives out carbon dioxide and water (combustion)

2C2H6 + 7O2 ---> 4CO2 + 6H2O

From the equation, we can see the mole ratio of ethane : oxygen is 2:7, meaning 2 moles of ethane reacts with 7 moles of oxygen.

If there's 1 mole of ethane gas, we need y moles of oxygen gas for complete reaction.

[tex]\frac{2}{7} =\frac{1}{y}[/tex]

y= 3.5 moles

Since 5 moles of oxygen is more than the required 3.5 moles, we can deduce oxygen gas is in excess, meaning ethane is limiting.

(You can get the same results too if you take y as ethane gas required).

The no. of moles of product produced ALWAYS depend on the no. of moles of the limiting reactant, since they are the ones which reacts completely.

So, from the equation, since the mole ratio of ethane : water = 2:6,

hence, (take the no. of moles of water produced as z),

[tex]\frac{2}{6} =\frac{1}{z} \\z= 3[/tex]

Therefore, 3 moles of water is produced.

Ethane is the limiting reagent in the given reaction. The number of moles of water produced from the reaction is 3 moles.

A limiting reagent can be defined as that reactant in the chemical reaction which is consumed first among the other reactants during the completion of a chemical reaction.

The limiting reagent is the reactant which decides the yield of the product when the quantity of the reactants is not taken in stoichiometry.

Given, chemical reaction of ethane and oxygen is:

[tex]C_2H_6 + \frac{7}{2} O_2 \longrightarrow 2CO_2 +3H_2O[/tex]

Given, the number of moles of ethane = 1 mol

The number of moles of oxygen = 5 mol

One more of ethane reacts with moles of oxygen = 3.5 mol

Therefore, ethane is the limiting reagent in this reaction as the oxygen is given in excess.

From the balanced equation, one mole of ethane produces moles of water equal to 3 moles.

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Answer:

Gaseous

Explanation:

Gasses can move freely and do not form the shape of their containers

Liquids are more free than solids, but they conform to the shape of their container

Solids are not free

Hey there!

For this we can use the combined gas law:

[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]

We are only working with pressure and temperature so we can remove volume.

[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]

P₁ = 2 atm

T₁ = 27 C

P₂ = 2.2 atm

Plug these values in:

[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]

Solve for T₂.

[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]

[tex]2atm * T_{2}={2.2atm}*27C[/tex]

[tex]T_{2}={2.2atm}\div2atm*27C[/tex]

[tex]T_{2}=1.1*27C[/tex]

[tex]T_{2}=29.7C[/tex]

Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.

Hope this helps!

Answer:

Manganese (II) ion, Mn²⁺

Explanation:

Hello,

In this case, given the overall reaction:

[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]

Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺

Best regards.

Answer:

D. BCl₃

Explanation:

BCl₃ molecular geometry is trigonal planar and it has a bond angle of 120°.

Hope that helps.

According to the molecular  geometry, BCl₃ has trigonal planar geometry and a bond angle of 120°.

Molecular geometry can be defined as a three -dimensional arrangement of atoms  which constitute the molecule.It includes parameters like bond length,bond angle  and torsional angles.

It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.

They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information  about the distance between nuclei and electron density.

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Answer:

Hello

you're answer should be E.HOCH2CH2OH

hope this answer is correct

The substance with the highest surface tension - e. HOCH2CH2OH.

Surface tension is the elastic tendency of a fluid, caused by the attraction of particles in the surface which makes it acquire the least surface area.  

Thus, The substance with the highest surface tension - e. HOCH2CH2OH.

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Answer:

Explanation:

0,44

Answer:

HPO₄⁻² predominates at pH 9.3

Explanation:

These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.

H₃PO₄ + H₂O   ⇄   H₂PO₄⁻   +   H₃O⁺   pKa 2.14

H₂PO₄⁻  +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺  pKa 6.86

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺  pKa 12.38

The H₂PO₄⁻  works as amphoterous, it can be a base and acid, according to these equilibriums.

H₂PO₄⁻ +  H₂O   ⇄   HPO₄⁻²  +  H₃O⁺

H₂PO₄⁻ +  H₂O   ⇄   H₃PO₄  +  OH⁻

pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:

HPO₄⁻ +  H₂O   ⇄   H₂PO₄⁻  +  OH⁻

HPO₄⁻²  +  H₂O   ⇄   PO₄⁻³   +   H₃O⁺

The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:

(6.86 + 12.38) /2 =  9.62

Above this pH, [PO₄⁻³] > [HPO₄⁻²].

In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]

Answer:

density = 5.1g/m³

Explanation:

use ideal gas equation

Pv=nRT

28.26 x 3.51 = m/v x 0.08206 x 237

m/v = (28.26 x3.51)/(237 x 0.08206) = 5.1g/m³

Answer:

C7H5F3

Explanation:

The following data were obtained from the question:

Mass of Carbon (C) = 57.53g

Mass of Hydrogen (H) = 3.45g

Mass of Fluorine (F) = 39.01g

The empirical formula of the compound can be obtained as follow:

C = 57.53g

H = 3.45g

F= 39.01g

Divide each by their molar mass

C = 57.53/12 = 4.79

H = 3.45/1 = 3.45

F = 39.01/19 = 2.05

Divide each by the smallest

C = 4.79/2.05 = 2.3

H = 3.45/2.05 = 1.7

F = 2.05/2.05 = 1

Multiply through by 3 to express in whole number

C = 2.3 x 3 = 7

H = 1.7 x 3 = 5

F = 1 x 3 = 3

Therefore, the empirical formula for the compound is C7H5F3

Answer:

Ecell = +0.25V

Explanation:

the half-cell reactions for a voltanic cell

cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)

anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻

we have the standard cell potential E⁺cell = 0.18V at 80C respectively

Q = [H⁺]/[Cl⁻]

sub for [H+] = 0.10M and [Cl-] = 1.5M

Q= 0.1M/1.5M

Q = 0.067

Ecell = E⁺cell - [tex]\frac{0.059}{n}[/tex] logQ

= 0.18 - [tex]\frac{0.056}{1}[/tex] log 0.067

0.18- 0.059(-1.174)

Ecell = +0.25V

Answer:

NaH(s) + H₂O(aq) ⇒ NaOH (aq) + H₂ (g)

Al(s) + Cl₂(g) ⇒ AlCl₃(s)

Explanation:

NaH - sodium hydride

H₂O - water

NaOH - sodium hydroxide

Al - aluminium

Cl₂ - Chlorine

AlCl₃ - aluminium chloride

Answer:

Check Explanation.

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

NaBr (s)

The Standard formation reaction is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

Using appendix C, the standard heat of formation of NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The Standard formation reaction is

S (s) + (3/2) O₂ (g) → SO₃ (g)

Using appendix C, the standard heat of formation of SO₃(g) is

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The Standard formation reaction is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

Hope this Helps!!!

Answer:

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

Answer:

Oxidation by FAD  

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;                  E°´ =  -0.031 V  

NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺;                      E°´ = -0.320 V

Succinate + NAD⁺ ⇌ Fumarate  + NADH + H⁺; E°' =  -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;        E°´ = -0.031 V  

FAD + 2H⁺ + 2e⁻ ⇌ FADH₂;                    E°´  = -0.219 V

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.  

Answer:

1.41 M

Explanation:

First we must use the information provided to determine the concentration of the aluminum hydroxide.

Mass of aluminum hydroxide= 320mg = 0.32 g

Molar mass of aluminum hydroxide= 78 g/mol

Volume of the solution= 5.00 ml

From;

m/M= CV

Where;

m= mass of aluminum hydroxide= 0.32 g

M= molar mass of aluminum hydroxide = 78 g/mol

C= concentration of aluminum hydroxide solution = the unknown

V= volume of aluminum hydroxide solution = 5.0 ml

0.32 g/78 g/mol = C × 5/1000

C = 4.1×10^-3/5×10^-3

C= 0.82 M

Reaction equation;

Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)

Concentration of base CB= 0.82 M

Volume of base VB= 1.60 ml

Concentration of acid CA= the unknown

Volume of acid VA= 2.80 ml

Number of moles of acid NA = 3

Number of moles of base NB= 1

Using;

CA VA/CB VB = NA/NB

CAVANB = CBVBNA

CA= CB VB NA/VA NB

CA= 0.82 × 1.60 × 3/ 2.80 ×1

CA= 1.41 M

Therefore the concentration of HCl is 1.41 M

Answer:

2.29x10⁻¹² is Ksp of the salt

Explanation:

The Ksp of the metal hydroxide is:

M(OH)₂(s) ⇄ M²⁺ + 2OH⁻

Ksp = [M²⁺] [OH⁻]²

As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:

pOH = 14- pH

pOH = 14 - 10.22

pOH = 3.78

pOH = -log[OH⁻]

1.66x10⁻⁴ = [OH⁻]

And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵

Replacing in Ksp formula:

Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²

Ksp = 2.29x10⁻¹² is Ksp of the salt

Answer:

1661μL of a 6M HCl you need to add

Explanation:

pH is defined as -log[H⁺] ([H⁺] =10^{-pH}), the initial and final concentrations of [H⁺] you need are:

Initial [H⁺] = 10^{-3.5} = 3.16x10⁻⁴M H⁺

Final [H⁺] = 10^{-1} = 0.1M H⁺

In moles, knowing volume of the solution is 0.1L:

Initial [H⁺] = 0.1L ₓ (3.16x10⁻⁴mol H⁺ / L) = 3.16x10⁻⁵moles H⁺

Final [H⁺] = 0.1L ₓ (0.1mol H⁺ / L) = 0.01 moles H⁺.

That means, moles of H⁺ you need to add to the solution is:

0.01mol - 3.16x10⁻⁵moles = 9.9684x10⁻³ moles of H⁺.

A solution of HCl dissociates in H⁺ and Cl⁻ ions, that means moles of HCl added are equal to moles of H⁺. As you need to add 9.9684x10⁻³ moles of H⁺ = 9.9684x10⁻³ moles of HCl:

9.9684x10⁻³ moles of HCl ₓ (1L / 6mol) = 1.6614x10⁻³L

In μL:

1.661x10⁻³L × (1x10⁶μL / 1L) =

Answer:

14.6 g

Explanation:

Fe₂O₃  +  3 C  →  2 Fe  +  3 CO₃

First, convert grams of Fe₂O₃ to moles.  The molar mass is 159.69 g/mol.

(64.6 g)/(159.69 g/mol) = 0.4045 mol

Now, use stoichiometry to convert moles of Fe₂O₃ to moles of C.  By looking at the chemical equation, you can see that for every 1 mole of Fe₂O₃, you need 3 moles of C.  Use this relationship.

(0.4045 mol Fe₂O₃) × (3 mol C/1 mol Fe₂O₃) = 1.214 mol C

Now, convert moles of C to grams.  The molar mass is 12.01 g/mol.

(1.214 mol C) × (12.01 g/mol) = 14.58 g

You need to use significant figures.

14.58 g ≈ 14.6 g

Answer:

Le Châtelier's principle states that when a chemical system at equilibrium is distributed by a change in  conditions, the equilibrium position will shift in a direction that tends to counteract  the change.

Therefore, when there is a change in pressure, the equilibrium will  counteract  the change by reducing/increasing the pressure through adjusting the no. of moles of gas.

Note: At constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.

For example, when there's an increase in total pressure, the equilibrium position will shift to the side with a smaller no. of moles of gas so as to reduce the pressure.

Answer:

The equilibrium would shift to reduce the pressure change

Explanation:

Answer:

[tex]t=2.08s[/tex]

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]

Thus, we compute the time as shown below:

[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]

Best regards.