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Assoli18 [71]
2 years ago
10

Classify the following substances as an element,a mixture or a compound1.limestone​

Chemistry
1 answer:
vovangra [49]2 years ago
3 0

Answer: what’s your color. I wanna know oh oh oh

Explanation: Hahhhahhanahhahhajja

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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
Why the ph of glycine increases when 0.1 M NaOH is added dropwise​
shtirl [24]

Answer:

The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

Explanation:

The glycine is an amino acid with the following chemical formula:

NH₂CH₂COOH  

The COOH functional group is what gives the acid properties in the molecule.      

Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:

NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O

The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.  

Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.  

I hope it helps you!  

7 0
2 years ago
What is the independent variable, and the dependent variable in this?
Doss [256]

Answer:

The salt and sugar as well as the baby power and granulated sugar are the independent variables.

The dependent variable is how it reacts (will it dissolve or not)

Explanation:

5 0
2 years ago
Read 2 more answers
(Direct answer=Upvote cause I aint smart xD)
Vedmedyk [2.9K]
Paper is shredded, the paper is physically being shredded, iron rusting is due to oxygen reacting with it, gas is burned releases and odor which is a sign of chemical, shirt being bleached is chemical reaction.
5 0
2 years ago
A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
Andrej [43]

Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

1 μg = 0.001 mg

Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

5 0
2 years ago
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