Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.5 × 10^-4. Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is
4.5 × 10^-4.

Answers

Answer 1

Answer: HNO2 =====> H+ + NO2-
Initial concentration = 0.311
C = -x,x,x
E = 0.311-x,x,x

KNO2 ====>K+ + NO2-
Initial concentration = 0.189
C= -0.189,0.189,0.189
E = 0,0.189,0.189

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Chip stones should be added to a solution that is to be refluxed as they act as nucleation sites for formation of solvent bubbles.

This should be added when the temperature of the liquid is still low.

What are Boiling Chips?

These are usually added to a solution as they contain trappedair which

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The high surface area ensures it acts as nucleation sites and the best time

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The heat of combustion of propane, C3H8, is 2220 kJ/mol. The specific heat of copper is 0.385 J/g°C. How many grams of propane must be burned to raise the temperature of a 10.0 kg block of copper from 25.0°C to 65.0°C, assuming none of the heat is lost to the surroundings?

Answers

The heat of combustion ( $\Delta$ Hc0) is the amount of energy released as heat when a compound completely burns with oxygen under standard conditions.

3.05988g. grams of propane must be burned to raise the temperature of a 10.0 kg block of copper from 25.0°C to 65.0°C.

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The heat of combustion ( $\Delta$ Hc0) is the amount of energy released as heat when a compound completely burns with oxygen under standard conditions. In most cases, a hydrocarbon reacts with oxygen to produce carbon dioxide, water, and heat.
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q=m*c*(change of T)

q=10000g(0.385J/g*c)*(65.0C-25.0C)or (338.2 K-298.2K)

q=154000J

154000J*(1 mol/2220 KJ)=69.36936 x $10 ^-3$ mol

here's where I'm stuck

0.069369 mol

and i know that for every 1 mol there is 44.11g of C3H8.

0.069369 mol* (44.11g C3H8)/1mol = 3.05988g.

To learn more about : Heat of combustion

Ref : brainly.com/question/25109613

#SPJ2

Answer:

Explanation:

q = (mass) (temp change) (specific heat)

q = (10000 g) (40 °C) (0.385 J/g⋅°C) = 154000 J = 154 kJ

154 kJ / 2220 kJ/mol = 0.069369369 mol

0.069369369 mol times 44.0962 g/mol = 3.06 g (to three sig figs)

answer choice 4

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Answers

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I hope this helps. Let me know if anything is unclear.

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