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DaniilM [7]
3 years ago
14

To begin the experiment, 1.5g of methane CH4 is burned in a bomb calorimeter containing 1,000 grams of water. The initial temper

ature of water is 25°C. The specific heat of water is 4.184 J/g°C. The heat capacity of the calorimeter is 695 J/°C . After the reaction the final temperature of the water is 37°C.
1. Using the formula q =m• C• ΔT ,calculate the heat absorbed by the water. Show work.

2. The total heat absorbed by the water and the calorimeter can be calculated by adding the heat calculated in steps 3 and 4. The amount of heat released by the reaction is equal to the amount of heat absorbed with the negative sign as this is an exothermic reaction. Using the formula ∆H = -(qcal + qwater), calculate the total heat of combustion. Show work.

3. Calculate the experimental molar heat of combustion in kJ/mol? Show work.

4. Given the formula: % error = ((accepted value-experimental value)/accepted value) * 100, calculate the percent error for the experiment. Show work
Chemistry
1 answer:
vredina [299]3 years ago
4 0

Mass of methane takne = 1.5g

moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles


mass of water = 1000 g

Initial temperature of water = 25 C

final temperature = 37 C

specific heat of water = 4.184 J /g C

1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) =  50208 Joules

2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J

3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules

This heat is released by 0.094 moles of methane

So heat released by one mole of methane =

                            - 622851.06 Joules = 622.85 kJ / mole

4) standard enthalpy of combustion = -882 kJ / mole

Error = (882-622.85) X 100 / 882 = 24.84 %

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3 years ago
A solution is made from ethanol, C2H5OH and water is 0.75 m. How many grams of ethanol are contained per 250. g of water?
kumpel [21]

Answer:

m_{solute}=8.6g

Explanation:

Hello,

In this case, the unit of concentration is molality which is defined by:

m=\frac{n_{solute}}{m_{solvent}}

Whereas the mass of the solvent is measured in kilograms. In such a way, with the given data, we first compute the kilograms of water:

m_{solvent}=250g*\frac{1kg}{1000g} =0.25kg

Then, we solve for the moles of the solute:

n_{solute}=0.75\frac{mol}{kg}*0.25kg =0.19mol

Finally since the molar mass of ethanol is 46 g/mol, we compute the grams for the given solution:

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8 0
3 years ago
What mass of Na2SO4is needed to make 2.5 L of 2.0 Msolution? (Na = 23 g; S = 32 g; O = 16 g)
ivanzaharov [21]

Answer:

mass (g) needed = 710.2 grams Na₂SO₄(s)

Explanation:

Needed is 2.5 Liters of 2.0M Na₂SO₄; formula wt Na₂SO₄ = 142.04g/mol.

mass (grams) of Na₂SO₄(s) = Molarity needed x Volume needed in Liters x Formula Wt of solute

mass (grams) of Na₂SO₄(s) = (2.5L)(2.0M)(142.04g/mol) = 710.2 grams Na₂SO₄(s)

Mixing: Transfer 710.4 grams Na₂SO₄ into mixing vessel and add water-solvent up to but not to exceed 2.5 Liters total volume. Mix until dissolved.

Gives 2.5 Liters of 2.0M Na₂SO₄(aq) solution.

5 0
3 years ago
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