CHEMISTRY COLLEGE

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.4Al(s) + 302(g) —> 2Al2O3(s)
Express your answer as a chemical formula.

1 mol Al; 1 mol O2
4 mol Al; 2.5 mol O2
12 mol Al; 10 mol O2
15.4 mol Al; 10.7 mol O2

Answers

Answer 1

Answer:

Answer:

1 mol Al; 1 mol O2

Explanation:ol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

or the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer aor the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

4Al(s) + 302(g) —> 2Al2O3(s)

Express your answer as a chemical formula.

1 mol Al; 1 mol O2

4 mol Al; 2.5 mol O2

12 mol Al; 10 mol O2

15.4 mol Al; 10.7 mol O2

Hold on, our servers are swamped. Wait for your answer to fully load.s a chemical formula.

Related Questions

According to the valence bond theory the triple bond in ethyne consists of
Please Please! help help! so stress
What is the most likely position for the hurricane indicated by the wind readings from the three weather stations shown?
15.2 grams of CO2 = ? molecules of CO2
Need help ASAP!!!!!!!

The amount of I − 3 ( aq ) in a solution can be determined by titration with a solution containing a known concentration of S 2 O 2 − 3 ( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S 2 O 2 − 3 ( aq ) + I − 3 ( aq ) ⟶ S 4 O 2 − 6 ( aq ) + 3 I − ( aq ) Given that it requires 29.4 mL of 0.380 M Na 2 S 2 O 3 ( aq ) to titrate a 30.0 mL sample of I − 3 ( aq ) , calculate the molarity of I − 3 ( aq ) in the solution.

Answers

Answer:

The molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

Explanation:

Given net ionic equation:

2S₂O₃²⁻ (aq) + I₃⁻ ( aq ) ⟶ S₄O₆²⁻ (aq) + 3I⁻ (aq)

Number of moles of S₂O₃²⁻: n₁ = 2, Number of moles of I₃⁻: n₂ = 1

Given- For S₂O₃²⁻ solution: Molarity: M₁ = 0.380 M, Volume: V₁ = 29.4 mL;

For I₃⁻ (aq) solution: Molarity: M₂ = ? M, Volume: V₂ = 30.0 mL

To calculate the molarity of I₃⁻ (aq) solution, we use the equation:

$(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))$

$((0.380 M)* (29.4 mL))/(2)=(M_(2)* (30.0 mL))/(1)$

$\Rightarrow M_(2) = ((0.380 M)* (29.4 mL))/((30.0 mL)* 2) = 0.186 M$

Therefore, the molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

brainly.com/question/7916557

I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

Hope this helps!

Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combined. Note: Sulfites follow the same solubility trends as sulfates.

Answers

Answer: The net ionic equation for the given reaction is $2H^+(aq.)+SO_3^(2-)(aq.)\rightarrow H_2O(l)+SO_2(g)$

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

$2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)$

Ionic form of the above equation follows:

$2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^(2-)(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)$

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$2H^+(aq.)+SO_3^(2-)(aq.)\rightarrow SO_2(g)+H_2O(l)$

Hence, the net ionic equation for the given reaction is written above.

Final answer:

The net ionic equation for the reaction between hydrochloric acid and potassium sulfite is H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq), following the solubility trends of sulfates and sulfites under standard conditions.

Explanation:

The reaction between excess hydrochloric acid (HCl) and potassium sulfite (K2SO3) is a typical acid-base neutralization reaction. In the initial step, potassium sulfite dissociates into its ions in the aqueous solution:

K2SO3 (aq) → 2K+ (aq) + SO3^2- (aq)

Hydrochloric acid, being a strong acid, also dissociates completely:

HCl (aq) → H+ (aq) + Cl- (aq)

The hydrogen ion from the acid then reacts with the sulfite ion to form sulfuric acid and water, creating a net ionic equation :

2H+ (aq) + SO3^2- (aq) → H2SO3 (aq)

Because of the solubility trends of sulfates and sulfites under standard conditions, the sulfuric acid produced also dissociates into ions:

H2SO3 (aq) → 2H+ (aq) + SO3^2- (aq)

Therefore, the overall net ionic equation is:

H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq)

Learn more about Net Ionic Equation here:

brainly.com/question/35304253

#SPJ3

Compare Dalton’s and Democritus’ ideas.

Answers

Answer:

please give me brainlist and follow

Explanation:

The key difference between Democritus and Dalton atomic theory is that the Democritus atomic theory is an ancient theory that scientists later refined and elaborated whereas Dalton atomic theory is a comparatively modern, scientific theory that we cannot discard due its important statements.

The voltage generated by the zinc concentration cell described by, zn(s)|zn2 (aq, 0.100 m)||zn2 (aq, ? m)|zn(s),is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

Answers

The concentration cell is:
Zn(s) \ Zn²⁺(aq,0.100 M) // Zn²⁺(aq, x M) \ Zn(s)
voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V
- In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.
- Oxidation takes place at the anode and reduction takes place at the cathode.
so [Zn²⁺]oxidation = 0.100 M
[Zn²⁺] reduction = x M
From Nernst equation:
Ecell = -0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction
Number of electrons, n = 2. Substitute and solve for x:
16 x 10⁻³ V = - 0.0592 / 2 log (0.100 /x)
log 0.100 / x = - 0.54
0.100 / x = 0.288
x = 0.347
So the concentration of Zn²⁺ at the cathode = 0.406

Other Questions

Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) → Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? addition of more tin metal to the anode compartment increasing the pressure of hydrogen gas in the cathode compartment lowering the pH in the cathode compartment increasing the tin (II) ion concentration in the anode compartment Any of the above will change the measured cell potential. Request Answer

Which of the following statements about compounds is true?A- Each Compound contains only one element.A- Each Compound contains only one element.B- Compound can be classified as either heterogenous or homogenous.C- A Compound has a defined ratio by mass of the elements that it containsD- Compounds Vary in chemical composition depending on the sample size.

Name the physical properties used in seperating kerosene and petrol

Calculate the mass percent of oxygen in KMnO4.