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Rainbow [258]
3 years ago
7

When 8.5 g of methane (ch4) is burned in a bomb calorimeter (heat capacity = 2.677 × 103 j/°c), the temperature rises from 24.00

to 27.08°c. how much heat is absorbed by the calorimeter?ch4(g) + 2o2(g) → co2(g) + 2h2o(l); ∆h° = –1283.8 kj?
Chemistry
2 answers:
Scrat [10]3 years ago
8 0
The molar mass of methane is 16 g/mol. The heat absorbed by the calorimeter is the sensible heat, which is the heat gained or lost during a temperature change.

Sensible heat = CΔT = (<span>2.677 kJ/°C)(27.08 - 24</span>°C)
<em>Sensible heat = 8.24 kJ</em>
Oliga [24]3 years ago
3 0

Answer is: 8.24 kJ of heat is absorbed by the calorimeter.

ΔT= 27.08°C - 24.00°C.

ΔT = 3.08°C; temperature change of reaction.

m = 8.5 g; mass of methane.

cp= 2.677 kJ/°C, specific heat capacity of the bomb calorimeter.  

Qcal = ΔT(methane) · cp(methane).  

Qcal = 3.08°C · 2.677 kJ/°C.  

Qcal = 8.245 kJ; amount of heat absorbed.

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The common laboratory solvent ethanol is often used to purify substances dissolved in it. The vapor pressure of ethanol , CH3CH2
OverLord2011 [107]

Answer:

Molar mass of solute is 183.4g/mol

Explanation:

Using Raoult's law it is possible to find moles of solute, thus:

P_{solution} = X_{solvent}P_{solvent}^0

Where pressure of solution is 53.15 mmHg, pressure of pure solvent is 54.68 mmHg and mole fraction is:

53.15 mmHg = X 54.68 mmHg

<em>0.9720 = X</em>

<em />

Mole fraction of solvent is defined as moles of solvent / total moles.

Moles of solvent are:

286.8g × (1mol / 46.07g) = 6.225 moles of ethanol.

That is:

0.9720 = \frac{6.225mol}{Y+6.225mol}

<em>Where Y are moles of solute.</em>

6.051 + 0.9720Y = 6.225

0.9720Y = 0.174

Y = 0.179 moles of solute

As mass of solute dissolved was 32.83g. Molar mass of solute is:

32.83g / 0.179mol = <em>183.4g/mol</em>

4 0
3 years ago
How does heat transfer between objects
pentagon [3]

Answer:

Explanation:

       Heat can transfer between objects in two different ways. Generally, heat will travel from places of higher heat to places of lower heat.

       The first is conduction. This is when the object being heated and releasing heat are in direct contact. Not as much heat is lost in this process, since the thermal energy has nowhere else to go except for the object it is touching. An example would be putting a kettle on a hot stove, but it could also be grabbing a cold pole with your relatively warm hands.

       The second is convection. This is where heat is radiated into the air, and thus, transferred by the air, to another object. The actual heat that you feel is actually electromagnetic waves, and its transfer from an object is called electromagnetic radiation. Convection is the heat you feel from a near fire or a space heater. This is also why wind is present in our atmosphere.

       There is also radiation. This is caused from the burning or breaking down of a substance. This might come from the sun.

I hope I did enough to deserve the 45 points!

8 0
3 years ago
How do I find percent mass composition
castortr0y [4]

Answer:

1. Find the molar mass of all the elements in the compound in grams per mole.

2. Find the molecular mass of the entire compound.

3. Divide the component's molar mass by the entire molecular mass.

4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

Explanation:

5 0
3 years ago
How many liters of C3H6O are present in a sample weighing 25.6 grams?
lawyer [7]

To Find :

Number of moles of C₃H₆O present in a sample weighing 25.6 grams.

Solution :

Molecular mass of C₃H₆O is :

M = (6×12) + (6×1) + (16×1) grams

M = 94 grams/mol

We know, number of moles of 25.6 grams of C₃H₆O is :

n = \dfrac{Given \ Mass \ Of \ C_3H_6O }{Molar\ Mass \ Of \ C_3H_6O }\\\\n = \dfrac{25.6}{94}\ mole\\\\n = 0.27 \ mole

Hence, this is the required solution.

4 0
3 years ago
Checking if my answer is correct: I got LARQY, but I'm pretty sure it's wrong. Can somebody please help?
babunello [35]

Answer:

you are right

Explanation:

and i do not need to explain it because you did

8 0
3 years ago
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