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drek231 [11]
3 years ago
13

Given the reaction 2NaOH + H2SO4 = Na2SO4 + 2H2O, what is the total number of grams of NaOH needed to react completely with 196

grams of H2SO4
Chemistry
1 answer:
ollegr [7]3 years ago
3 0

Answer:

160 g of NaOH

Explanation:

Reaction: 2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

When a base and an acid react, the produce water and a salt. → Neutralization

In this reaction 2 moles of hydroxide need 1 mol of sulfuric acid to react, it is the same to say that 1 mol of sulfuric acid needs 2 moles of NaOH to react. We convert the mass of sulfuric to moles and we propose the rule of three:

196 g / 98 g/mol = 2 moles

1 mol of sulfuric acid needs 2 moles of NaOH to react

Then 2 moles of sulfuric acid will react with (2 . 2) /1 = 4 moles of NaOH

We convert the moles of base to grams → 4 mol . 40 g /1mol = 160 g

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To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

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3 years ago
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What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?
frutty [35]

Answer:

6.93

Explanation:

Step 1: Given data

  • Standard Gibbs free energy (∆G°): -5.20 kJ
  • Temperature (T): 50°C
  • Equilibrium constant (K): ?

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

5 0
3 years ago
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