True or false? The formation of a new state of matter is evidence of a chemical reaction

Vadim26 [7]

The answer to this question is B

4 0

3 years ago

A 25.0 ml sample of 0.723 m hclo4 is titrated with a 0.27303 m koh solution. the h3o+ concentration after the addition of 66.2 m

tia_tia [17]

This doesn't need an ICE chart. Both will fully dissociate in water.

Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.

Step 1:

write out balanced equation for the reaction

HClO4+KOH ⇔ KClO4 + H2O

the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4

Step 2:

Determining the number of moles present in HClO4 and KOH

Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4

Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L

Remember:

M = moles/L so we have 0.025 L of 0.723 moles/L HClO4

Multiply the volume in L by the molar concentration to get:

0.025L x 0.723mol/L = 0.0181 moles HClO4.

Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH

Step 3:

Determine how much HClO4 remains after reacting with the KOH.

Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:

moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0

This means all of the HClO4 is used up in the reaction.

If all of the acid is fully reacted with the base, the pH will be neutral = 7.

Determine the H3O+ concentration:

pH = -log[H3O+]; [H3O+] = 10-pH = 10-7

The correct answer is 1.0x10-7.

3 0

3 years ago

Read 2 more answers

I need the answer fast plzzz!! NO LINKS

Arturiano [62]

Answer:

your answer is 12 hope it's correct answer

4 0

3 years ago

The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super

castortr0y [4]

Answer:

The pH of the solution is 8.0.

Explanation:

taking the test rn

6 0

3 years ago

Read 2 more answers

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago