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attashe74 [19]
3 years ago
11

Four students all do the same experiment for the science fair. They test reaction times for pushing a button when a specific col

or is shown. Their data is as follows:
⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢Student 1Student 2Student 3Student 4Reaction Time (seconds)Trial 10.440.350.410.42Trial 20.410.430.520.43Trial 30.470.380.570.41Trial 40.390.390.460.42Trial 50.460.420.550.41Trial 60.420.440.490.42⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥
[
Reaction Time(seconds) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 Student 1 0.44 0.41 0.47 0.39 0.46 0.42 Student 2 0.35 0.43 0.38 0.39 0.42 0.44 Student 3 0.41 0.52 0.57 0.46 0.55 0.49 Student 4 0.42 0.43 0.41 0.42 0.41 0.42
]



Put the students in order from most to least reliable data.

A. Student 1, Student 2, Student 3, Student 4
B. Student 4, Student 1, Student 2, Student 3
C. Student 2, Student 1, Student 4, Student 3
D. Student 3, Student 2, Student 1, Student 4
Chemistry
1 answer:
emmasim [6.3K]3 years ago
3 0
Sorry, Cant really tell how to differentiate the students and trial times
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1249.88 mol.

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<em>∴ no. of moles of Fe </em>= (6.98 x 10⁴ g)/(55.845 g/mol) = <em>1249.88 mol.</em>

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3 years ago
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
tekilochka [14]

Answer:

See explanation

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First, let's write the balanced equation again:

2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)

Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:

Ptotal = Pwat + PCO2 (1)

This is the dalton's law for partial pressures.

The pressure can be also be relationed with the moles

Ratio of mole = Ratio of pressure

so, taking this in consideration we can say the following:

Pwater/PCO2 = moles water / moles CO2

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT     using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)    MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

n = 1.19 moles

so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

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3 years ago
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