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3 years ago

2. Identify the limiting reactant when 4.68 g of iron reacts with 2.88 g of sulfur to produce Fes.

Chemistry

1 answer:

Vinvika [58]3 years ago

6 0

Answer:

Iron is limiting reactant

Explanation:

Based on the reaction:

Fe + S → FeS

1 mole of iron reacts per mole of Sulfur

To solve this question we must convert the mass of each reactant to moles using molar masses of each reactant. As the reaction is 1:1, the reactant with the lower amount of moles is limiting reactant.

Moles Fe -Molar mass: 55.845g/mol-

4.68g * (1mol / 55.845g) = 0.0838 moles

Moles S -Molar mass: 32.065g/mol-

2.88g * (1mol / 32.065g) = 0.0898 moles

As the amount of moles of Fe < Moles S,

Iron is limiting reactant

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An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

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3 years ago

Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac

olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

$n=0.089 M\times 0.200 L=0.0178 mol$ of HCL

$HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)$

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

$\frac{1}{1}\times 0.0178 mol=0.0178 mol$ of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

8 0

3 years ago

A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo

Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

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3 years ago

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Simora [160]

6.28×1013+7.30×1011 this =13741.94

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3 years ago

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Troyanec [42]

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Cl- would give you 18 electrons
F- would give you 10 electrons
Li+ would give you 2 electrons
Na+ would give you 10 electrons

Cl- would be the correct answer

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3 years ago