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3 years ago

Why is it harder to break an ionic bond than a covalent? Need answer asap!

Chemistry

1 answer:

Pavel [41]3 years ago

5 0

Answer:

The greater the energy required to break a bond, the stronger that bond must be.

Explanation:

Most iconic bonds are considerably stronger, more difficult to break, than covalent bonds.

Hope this helps! :) x

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Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)

Gelneren [198K]

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

Fe₂O₃ + 3C → 2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

ΔS = ∑S $_{products}$ - ∑S $_{reactants}$

Therefore:

ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

ΔS = +541.3Jmol⁻¹K⁻¹

5 0

2 years ago

Which type of molecule forms structures in cells and stores energy? (5 points)

Brums [2.3K]

Just to make sure I’m right, is number 1 miss spelled??

4 0

2 years ago

Which term names the action that is happening when an acid eats away part of a metal bar

dangina [55]

Answer:

corrosion

Explanation:

8 0

3 years ago

Someone please help! this is the last question<br>I only need help with B.<br><br>

ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

6 0

3 years ago

what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?

MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻ + H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF, the [ OH ⁻] from the given pH, and finally the mass needed to produce that OH⁻ concentration.

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is 6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14 ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

F⁻ HF OH⁻

Equilibrium X 2.51 x 10⁻⁶ 2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X = 1.5 x 10⁻¹¹ ⇒ X = (2.51 x 10⁻⁶)² / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L * 0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF * 42.0 g/mol NaF = 6.03 g NaF

7 0

2 years ago