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Dmitriy789 [7]
2 years ago
8

a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp

erature increased by 26.0 ∘c. based on this, how much heat energy was required to dissolve the sample of koh? assume the specific heat of the solution is 4.184 j/g⋅°c.
Chemistry
1 answer:
Snowcat [4.5K]2 years ago
4 0

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

#SPJ1

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A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c
gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat  capacity and the temperature change. Working with this equation, and assuming no heat is lost to  the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat.  </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write  the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

3 0
3 years ago
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6 0
3 years ago
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How many moles of Al are needed to react completely with 1.8 mol of FeO?
Tju [1.3M]
<span>1.8 × (2/3) = 1.2

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3 years ago
How do carrier proteins transport substances across cell membranes?
stepan [7]
<em>Transport of substances across cell membrane:</em>
<em>Carrier proteins facilitate transport of material across a membrane by changing its shape to allow a substance to pass through its membrane. It transports substances from high concentration to lower concentration levels.</em>
7 0
3 years ago
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
2 years ago
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