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Triss [41]
3 years ago
6

A student completed a lab report. Which correctly describes the difference between the “Question” and “Hypothesis” sections of h

er report?
“Question” states what she is asking, and “Hypothesis” states the result of her experiment.
“Question” states what she is asking, and “Hypothesis” states what she thinks the answer to that question is in “if . . . then . . . because” format. “Question” describes what she is trying to find out, and "Hypothesis" states the procedures and methods of data collection.
“Question” describes what she is trying to find out, and “Hypothesis” states any additional information or prior knowledge about the question
Chemistry
2 answers:
valkas [14]3 years ago
6 0

Answer:

b

Explanation:

b

ValentinkaMS [17]3 years ago
4 0

The answer is:

“Question” states what she is asking, and “Hypothesis” states what she thinks the answer to that question is in “if . . . then . . . because” format.

I hope you find my answer helpful.

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THIS IS URGENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Pani-rosa [81]

Answer:

1- 1.54 mol.

2- 271.9 kPa.

3- Yes, the tires will burst.

4- 235.67 kPa.

5- As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in image 1 that represents the Maxwell’s Distribution of Speeds of molecules. "Kindly, see the explanation and the attached images".

<em>Explanation:</em>

<em>Q1- How many moles of nitrogen gas are in each tire?  </em>

  • To calculate the no. of moles of nitrogen gas in each tire, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the nitrogen gas (P = 247.0 kPa/101.325 = 2.44 atm),

V is the volume of the nitrogen gas (V = 15.2 L),

n is the no. of moles of the nitrogen gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the nitrogen gas (T = 21°C + 273 = 294 K).

∴ n = PV/RT = (2.44 atm)(15.2 L)/(0.082 L/atm/mol.K)(294.0 K) = 1.54 mol.

<em>Q2: What would the maximum tire pressure be at 50 degrees C?  </em>

  • Now, the temperature is raised to be 50°C (T = 50°C + 273 = 323 K).
  • The pressure can be calculated using the general gas law: PV = nRT.

<em>∴ P = nRT/V </em>= (1.54 atm)(0.082 L/atm/mol.K)(323.0 K)/(15.2 L) = 2.68 atm = <em>271.9 kPa.</em>

<em>Q3: Will the tires burst in Moses Lake? Explain.</em>

  • <em>Yes,</em> the tires will burst because the internal pressure be 271.9 kPa that exceeds 270 kPa, the pressure above which the tires will burst.

<em>Q4: If you must let nitrogen gas out of the tire before you go, to what pressure must you reduce the tires before you start your trip? (Assume no significant change in tire volume.)  </em>

  • To get the pressure that we must begin with:
  • Firstly, we should calculate the no. of moles at:

T = 55°C + 273 = 328 K,

Pressure = 270 kPa (the pressure above which the tires will burst). (P =270 kPa/101.325 = 2.66 atm).

V = 15.2 L, as there is no significant change in tire volume.

∴ n = PV/RT = (2.66 atm)(15.2 L)/(0.082 L.atm/mol.K)(328 K) = 1.5 mol.

  • 1.5562 moles of N₂ in the tires will give a pressure of 270 kPa at 55°C, so this is the minimum moles of N₂ that will make the tires burst.
  • Now, we can enter this number of moles into the original starting conditions to tell us what pressure the tires will be at if we start with this number of moles of N₂.

P = ???  

V = 15.6 L.

n = 1.5 mol

T = 21°C + 273 = 294.0 K  

R = 0.0821 L.atm/mol.K.

∴ P = nRT/V = (1.5 mol x 0.082 x 294.0 K) / (15.6 L) = 2.2325 atm = 235.67 kPa.

<em>So, the starting pressure needs to be 235.67 kPa or just under in order for the tires not to burst.</em>

<em />

<em>Q5: Create a drawing of the tire and show a molecular view of the air molecules in the tire at 247 kpa vs the molecular view of the air molecules after the tires have been heated. Be mindful of the number of molecules that you use in your drawing in the before and after scenarios. Use a caption to describe the average kinetic energy of the molecules in both scenarios.</em>

<em />

  • As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in “image 1” that represents the Maxwell’s Distribution of Speeds of molecules.
  • The no. of molecules that possess a critical K.E. of molecules increases due to increasing the temperature activate the motion of molecules with high velocity as
  • (K.E. = 3RT/2), K.E. directly proportional to the temperature of the molecules (see image 2).
  • Also, the average speed of molecules increases as the K.E of the molecules increases (see image 3).

3 0
2 years ago
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
2 years ago
Who described atoms as small spheres that could not be divided into anything smaller?
gtnhenbr [62]
John Dalton


"matter cannot be created nor destroyed or divided into smaller particles"
5 0
3 years ago
How many significant figures<br> are in this number?<br> 0.00708900
Norma-Jean [14]
Don’t know sorry 1.627
4 0
3 years ago
Read 2 more answers
To find the number of atoms look at the number known as the________. Please help.
Rufina [12.5K]

Answer:

molar mass or AMU

Explanation:

4 0
2 years ago
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