Name the type of reaction involved in the conversation of ethanol to Ethan

Chemistry

1 answer:

Anestetic [448]2 years ago

3 0

The conversion of ethanol to ethanoic acid is an oxidation reaction.

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For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac

barxatty [35]

First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3:

when 245.2 KClO3 gives → 96 g O2

2.72 g KClO3 gives → X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

= 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

= 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

83.6 Kg KClO3 gives → X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

= 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

= 8.8 mg of O2

7 0

3 years ago

A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field

maksim [4K]

Answer:

the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C