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3 years ago

After the solution reaches equilibrium, what concentration of zn2 (aq remains?

1 answer:

8 0

According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

N76 [4]

Answer:

Oxidizing agent - CrO4^2-

Reducing agent- N2O

Explanation:

Let us look at the equation closely;

CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]

The reduction half equation is;

CrO4^2- (aq) + 3e -------->Cr^3+ (aq)

Oxidation half equation is;

3N2O(g) ------>3 NO(g) +3 e

Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.

8 0

3 years ago

What volume of carbon dioxide is produced when 6.40 g of methane gas, CH4 (g), reacts with excess oxygen? All gases are at 35.0C

Mandarinka [93]

Answer:

V = 10.3 L

Explanation:

Given data:

Mass of methane = 6.40 g

Volume of CO₂ produced = ?

Temperature = 35°C (35+273 = 308 K)

Pressure = 100.0 KPa (100.0/101 = 0.98 atm)

Solution:

Chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 6.40 g/ 16 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of CO₂ with CH₄.

CH₄ : CO₂

1 : 1

0.4 : 0.4

Volume of CO₂:

Formula:

PV = nRT

0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K

0.98 atm ×V = 10.11 atm.L

V = 10.11 atm.L /0.98 atm

V = 10.3 L

3 0

3 years ago

If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent?

agasfer [191]

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,

$4Fe(s) + 3O_{2}(g) ---> 2Fe_{2}O_{3}(s)$

Mass of Fe = 227.8 g

Moles of Fe = $227.8 g Fe * \frac{1 mol Fe}{55.85 g Fe} = 4.079 mol Fe$

Mass of oxygen = 128 g

Moles of $O_{2} = 128 g O_{2}*\frac{1 mol O_{2}}{32 g O_{2}}= 4mol O_{2}$

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = $4.079 mol Fe * \frac{2 mol Fe_{2}O_{3}}{4 mol Fe} *\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 325.7 g Fe_{2}O_{3}$

Mass of iron (III) oxide produced from oxygen= $4 mol O_{2}*\frac{2 molFe_{2}O_{3}}{3 mol O_{2}}*\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 425.84 g Fe_{2}O_{3}$

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

7 0

3 years ago

Living organisms need energy to carry on their life processes. Photosynthesis is a process that allows for photosynthesizing org

Ad libitum [116K]

the correct combo is C

We know sunlight is key factor in photosynthesis so A and D are out.

we also know plants get their bulk by breaking up co2 and using the carbon which releases oxygen as a byproduct so B is also out of the question.

3 0

3 years ago

A beaker of nitric acid is neutralized with dilute aqueous solution of calcium hydroxide.Write a net ionic equation for this rea

Soloha48 [4]

Complete balanced equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

Ionized equation (with spectator ions):
2H⁺ + 2NO₃⁻ + Ca²⁺ + 2OH⁻   → Ca²⁺ + 2NO₃⁻ + 2H₂O

By eliminating the ions that are the same of both sides of the equation (spectator ions):

2H⁺ + 2OH⁻   → 2H₂O [Net Ionic Equation]

3 0

3 years ago

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