why Al is a member of group 13 rather than group 3?

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:

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:

Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)

Explanation:

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:

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:

- N2 does not exist as a liquid at pressures below 0.127 atm.

- N2 is a solid at 16.7 atm and 56.5 K.

- N2 is a liquid at 1.00 atm and 73.9 K

- N2 is a gas at 0.127 atm and 84.0 K.

Explanation:

Hello,

At first, we organize the information:

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Best regards.

Answer:

Explanation:

0,44

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:

if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half

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:

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:

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:

using ideal gas equation =12.4576atm to 4.significant figure

using vander Waals equation = 12.3504

The differences is 0.10atm

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

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:

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.  

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: 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:

[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.

Answer:

0.0031 moles

Explanation:[tex]Molarity=\frac{molSolute}{LitreSolution}\\ 0.0125M=\frac{molRNA}{0.25L} \\molRNA=0.0125*0.25=0.0031 mol[/tex]

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.

The question is incomplete as the options are missing, however, the correct complete question is attached.

Answer:

The correct answer is option A. ( check image)

Explanation:

The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.

Thus, the correct answer is option A. ( check image)

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.

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:

Option D.

Explanation:

This compound has in its formula:

- Eight bromines

- Four phosphorous

8 → octa prefix

4 → tetra prefix

Right answer is Octabromine tetraphosphide

It can not be option A, tetraphosphate is P4O7

It can not be B and C, we do not have B (boron)

Answer:

Octabromine tetraphosphide

Explanation:

Answer via Educere/ Founder's Education

Answer:

296.1g of Ag is the maximum amount of silver

Explanation:

A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:

Mass of solution:

Assuming a density of 1g/mL:

[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]

If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:

4700L × (6.3g Ag / 100g) =

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