15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Write the balanced equation and calculate the enthalpy change for this reaction.

Answer:

92.93 g

Explanation:

Number of half lives that have elapsed in eight days =8/14.3 = 0.559

Fraction of the radioactive nuclide that remains after 0.559 half lives is given by

N/No=(1/2)^0.559

Where N= mass of radioactive nuclides remaining after a time t

No= mass of radioactive nuclides originally present

N/No=(1/2)^0.559= 0.679

Mass of nuclides present eight days before= 63.1g/0.679

Mass of nuclides present eight days before=92.93 g

Answer:

wavelength, λ =  486.6 nm

frequency, f = 6.16 * 10¹⁴ Hz

Explanation:

a. Wavelength

Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)

Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s

1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)

1/λ = 2.055 * 10⁶ m

λ = 4.866 * 10⁻⁷ m

wavelength, λ =  486.6 nm

b.  Frequency

Using f = c/λ

f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m

frequency, f = 6.16 * 10¹⁴ Hz

Answer:

sodium

Explanation:

(Na) atom is paramagnetic and sodium is a na atom.

Answer:

A. Electron clouds or energy levels

Explanation:

Electrons can exist in two states:

Electrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.

Hence, the correct answer is A.

Answer:

Up and DOWN spin

Explanation:

Answer:

[tex]HCN~~Ka=4.9x10^-^1^0[/tex]

Explanation:

In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:

[tex]HA~->~H^+~+~A^-[/tex]    [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher

So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].

I hope helps!

HCN has the highest pH among all the acids listed in the question.

The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.

Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.

Learn more: https://brainly.com/question/6505878

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

Answer:

B and C. Just finished my lesson on Edge.

:Answer : The limiting reactant is  and the theoretical yield of methanol is, 0.96 grams.

Explanation :

First we have to calculate the moles of  and .

where,

= pressure of CO gas = 232 mmHg = 0.305 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of CO gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

and,

where,

= pressure of  gas = 374 mmHg = 0.492 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

= number of moles of  gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of  react with 1 mole of  

So, 0.0601 moles of  react with  moles of  

From this we conclude that,  is an excess reagent because the given moles are greater than the required moles and  is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of  

From the reaction, we conclude that

As, 2 mole of  react to give 1 mole of  

So, 0.0601 moles of  react with  moles of  

Now we have to calculate the mass of  

Therefore, the theoretical yield of methanol is, 0.96 grams.

The theoretical yield of methanol is 0.496 g of methanol.

The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).

From the partial pressures of each reactant, we can obtain the number of moles of reactants.

For CO;

P = 232 mmHg or 0.305 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.018 moles

For hydrogen;

P = 397 mmHg or 0.522 atm

V = 1.5 L

T = 305 K

n = ?

R = 0.082 atmL-1mol-1K-1

PV = nRT

n = PV/RT

n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K

n = 0.031 moles

From the reaction equation;

1 mole of CO reacted with 2 moles of H2

0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole

= 0.036 moles of H2

We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.

2 moles of H2 yields 1 mole of methanol

0.031 moles of H2 yields  0.031 moles × 1 moles/2 mole

= 0.0155 moles of methanol

Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol

= 0.496 g of methanol

Learn more: https://brainly.com/question/2286339

Answer:

Here's what I get  

Explanation:

pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44

The higher the pKₐ, the weaker the acid.

Thus, o-vanillin is the weaker acid and has a stronger conjugate base.

The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.

The equation for the equilibrium is

H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺

    weaker acid              weaker base          stronger  base        stronger acid

The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.

Thus, o-vanillin does not fully protonate p-toluidine.

O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.

The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.

The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that  o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.

Learn more: https://brainly.com/question/9743981

Answer:

gaining two electrons

Explanation:

electron configuration

2:6

so add two to 6 to get stable 2:8

Answer:

B. CH3Br

Explanation:

Dipole -Dipole interactions take place in polar molecules.

CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.

While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.

Hence, the correct option is B. CH3Br.

The compound that exhibits dipole -dipole forces is CH3Br

It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.

Therefore, The compound that exhibits dipole -dipole forces is CH3Br

Learn more about force here: https://brainly.com/question/14379797

Answer:

Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.

Answer:

The  energy change is  [tex]E = 9.0 *10^{9}\ J[/tex]

Explanation:

   From the question we are told that

          Mass loss  is  [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]

  Generally the energy change that  would accompany this loss  is mathematically represented as

     [tex]E = m * c^2[/tex]

Where  c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]

     [tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]

     [tex]E = 9.0 *10^{9}\ J[/tex]

Answer:

afshkkyfugutuiryfyi

Answer:

A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.

b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules

c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.

d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.

e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids

Explanation:

Answer:

2–Ethyl–3–methlypentanal.

Explanation:

To name the compound given in the question above, we must observe the following:

1. The functional group of the compound is Alkanal i.e Aldehyde,

—CHO and it is located at carbon 1.

Note: the aldehyde functional group is always at carbon 1 and there will be no need to state it's position in the compound.

2. The longest continuous carbon chain is 5 i.e pentane. But the presence of the functional group will replace the –e at the end of pentane with –al, making the name to the pentanal.

3. The substituents attached are:

a. Ethyl, CH2CH3 at carbon 2.

b. Methyl, CH3 at carbon 3.

4. Combine the above to get the name of the compound.

Therefore, the name of the compound is:

2–Ethyl–3–methlypentanal.

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.

Learn more: https://brainly.com/question/16108775

Answer:

B

Explanation:

The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.

The element you hit is Argon.

The answer is B

Answer:

B ar

Explanation:

pen foster answer

Answer:

8.00L of ammonia can be produced

Explanation:

The reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g)

Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.

As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:

1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia

Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:

12L H₂(g) ₓ (2L NH₃(g)  /  3L H₂(g)) =

Answer:

The correct answer is -2878 kJ/mol.

Explanation:

The reaction that takes place at the time of the oxidation of glucose is,  

C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)

The standard free energy change for the oxidation of glucose can be determined by using the formula,  

ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)

The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.  

Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)

ΔG°rxn = -2878 kJ/mol

Answer:

negative charge electrode

Explanation:

In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.

Answer:

The electrode that removes ions from the solution :) a p e x

Answer:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Explanation:

To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.

Nitrite ion (NO₂⁻)

1 × N + 2 × O = -1

1 × N + 2 × (-2) = -1

N = +3

Nitrous oxide (NO)

1 × N + 1 × O = 0

1 × N + 1 × (-2) = 0

N = +2

Nitrate ion (NO₃⁻)

1 × N + 3 × O = -1

1 × N + 3 × (-2) = -1

N = +5

Ammonia (NH₃)

1 × N + 3 × H = 0

1 × N + 3 × (+1) = 0

N = -3

Nitrogen gas (N₂)

2 × N = 0

N = 0

The order of increasing nitrogen oxidation state is:

C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate

Answer:

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)

Explanation:

All the following are oxidation–reduction reactions except:________

a. H₂(g) + F₂(g) → 2HF(g).  Redox. H is oxidized and F is reduced.

b. Ca(s) + H₂(g) → CaH₂(s).  Redox. Ca is oxidized and H is reduced.

c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).  Redox. K is oxidized and H is reduced.

d. 6Li(s) + N₂(g) → 2Li₃N(s).  Redox. Li is oxidized and N is reduced.

e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number

The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.

Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.

The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).

Learn more: https://brainly.com/question/8646601

Answer:

O-H stretch signal at 3300 cm-1

Explanation:

In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).

In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).

I hope it helps!

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

[tex]p^{OH}[/tex]=14-56.17

      =8.823

The volume of the [tex]NH_{3}[/tex] = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated [tex]NH_{3}[/tex] =0.10 M

The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M

The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml

convert into the liter= 0.030L  

The value of concentrated [tex]H_2So_4[/tex]=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]

calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:

[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]

create the ICE table:    

[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]

Answer:

0.1M solution of NaOH

Explanation:

1 mole of NaOH - 40g

? moles - 1 g = 1/40 = 0.025 moles.

Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume

= 0.025/0.25

= 0.1M.

Answer and Explanation:

Given the following chemical equation:

Cl₂(g) + F₂(g) ⇒ 2ClF(g)

The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:

On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).

On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).

Answer:

Final product: cyclopent-1-en-1-ylbenzene

Explanation:

In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).

See figure 1

I hope it helps!

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of Al produce 2 moles of Al₂O₃

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

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:

[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.