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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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Calculate the density of carbon dioxide at STP
Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

5 0
3 years ago
A certain isotope has 53 protons 78 neutrons and 53 electrons. What is its atomic number? What is the mass number of this atom?
statuscvo [17]

Answer:

  1. 53 protons
  2. 131g
  3. Iodine
  4. Halogens

Explanation:

atomic no. = no. of protons

= 53 proton

mass = no. of protons + no. of

neutrons

= 53 + 78

= 131

5 0
3 years ago
Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p
Oduvanchick [21]

Answer:

Procedure (2)  

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

\dfrac{5}{4000} = \dfrac{1}{800}

Procedure (2)

For the first dialysis, the factor is

\dfrac{5}{1000} = \dfrac{1}{200}

After a second dialysis, the original concentration of NaCl will be reduced by a factor of  

\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}

Procedure (2) is more efficient by a factor of  

\dfrac{40000}{800} = \mathbf{50}

4 0
3 years ago
. One of the essential minerals in the human body is salt. How much salt (NaCl) is in the average adult human body?
son4ous [18]

Answer:

200g or 40 teaspoons

Explanation:

An average human, weighing about 50 pounds, has about 200 g or 40 tps of NACl

4 0
3 years ago
I don't know how to do this can I get the answers plz it's due in 1 hour
katrin [286]
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
4 0
3 years ago
Read 2 more answers
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