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3 years ago

What is melting point of impure ice

Chemistry

2 answers:

lions [1.4K]3 years ago

8 0

<h2>what is melting point of impure ice </h2>

The temperature of the reverse change from liquid to solid is called the freezing point or crystallization point. For most substances the melting point and freezing points are equal. For ice, the melting point is 0°C or 273K.

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MrMuchimi3 years ago

3 0

Answer:

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Stearic acid (C18H36O2) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. I

Paha777 [63]

Answer:

There is 9671 Kj of heat released

Explanation:

Step 1: The balanced equation:

C18H36O2(s) + 26O2(g) --> 18CO2(g)+18H2O(g)

This means for 1 mole of C18H36O2 consumed there is 26 moles of O2 needed to produce 18 moles of CO2 and 18 moles of H2O.

Step 2: Calculate the heat of combustion

ΔH (combustion) = [18*(ΔHf of CO2) + 18*(ΔHf of H2O)] - [1*(ΔHf of C18H36O2) + 26*(ΔHf of O2)]

ΔH (combustion) = [18*(-394 kJ/mol) + 18*(-242 kJ/mol)] - [1*(-948 kJ/mol) + 26*(0 kJ/mol)]

ΔH (combustion) = [(-7092 kJ/mol) + (-4356 kJ/mol)] - (-948 kJ/mol)

= -10500 kJ/mol

ΔH (combustion) = heat released / number of moles of stearic acid

Step 3: Calculate moles of stearic acid

moles of stearic acid = mass / Molar mass of stearic acid

moles of stearic acid = 262g / 284.48 g/mole = 0.921 moles

Step 4: Calculate moles of oxygen

moles of O2 = 914.5 / 32g/mole

moles of O2 = 28.578125 moles

Stearic acid is the limiting reactant: it will completely react

There will react 26*0.921 mole = 23.946 mole of O2

This means there will remain 4.63 moles of O2

Step 5: Calculate heat released

q = (ΔH combustion) * (moles of stearic acid) = (-10500 kJ/mol) * (0.921 moles) = 9671 Kj

There is 9671 Kj of heat released

6 0

3 years ago

Solve the problem for the moles of oxygen. mol O2 DONE

Elza [17]

Answer: The answer is 0.245

Explanation: That is what it said the answer was

8 0

3 years ago

The same reaction is begun with an initial concentration of 0.05 M O3 and 0.02 M NO. Under these conditions, the reaction reache

Ivahew [28]

The balanced equation of the reaction is:

O3(g) + NO (g) → O2 (g) + NO2 (g)

Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2

If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.

The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.

And the rate of reaction is change in concetration divided by the time.

The change in concentration in O3 is 0.02 M

Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s

8 0

3 years ago

What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M

Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0

3 years ago

How many fluoride ion (F-1) are needed to balance a lead (IV) ion (Pb+4)?

salantis [7]

3

Pb2F3

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3 0

3 years ago

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What is melting point of impure ice ​

What is melting point of impure ice