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jeyben [28]
3 years ago
9

The density of water is 1.00 g/ml at 4°c. how many water molecules are present in 2.56 ml of water at this temperature?

Chemistry
1 answer:
o-na [289]3 years ago
5 0

Answer:

             8.55 × 10²² Molecules

Solution:

Step 1: Calculate the Mass of Water as;

                       Density  =  Mass ÷ Volume

Solving for Mass,

                       Mass  =  Density × Volume

Putting values,

                       Mass  =  1 g.mL⁻¹ × 2.56 mL

                       Mass  =  2.56 g

Step 2: Calculate Moles of Water as,

                       Moles  =  Mass ÷ M.Mass

Putting values,

                       Moles  =  2.56 g ÷ 18 g.mol⁻¹

                       Moles  =  0.142 mol

Step 3: Calculate Number of Molecules of Water as,

                       # of Molecules  =  Moles × 6.022 × 10²³

Putting value of mole,

                       # of Molecules  =  0.142 × 6.022 × 10²³

                       # of Molecules  =  8.55 × 10²² Molecules

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What volume (in L) of H2 would be collected at 22.0oC and a pressure of 713 torr if 2.65 g of zinc react according to the equati
ASHA 777 [7]

Answer: 83.74L

Explanation:

Temp. = 295K

P = 713torr = 0.938atm

Mass = 2.65g

PV = nRT

V = nRT/PV

n = Mass/Molar mass

Molar mass of Hydrogen gas = 1.00784*2= 2.0156g/mok

n = 2.65/2.0156 = 1.31469mol

V = 1.31469*0.08205*295/0.938

V = 83.74L

The volume = 83.74L

4 0
3 years ago
One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) techniq
frozen [14]

Answer:

a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

7 0
2 years ago
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).

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<span>
Evaporation occurred between the reaction of the hydrochloric acid and the arsenic group because there is a formation of gas. A liquid is cooled if condensation exceeds evaporation. This is because the cooling of the liquid decreases the kinetic energy of the molecules. Their movement is being restricted by the colder temperature. The molecules tend to be closer with each other. Also, because of their restricted movement, the liquid may turn into solid due the colder temperature. An example of this is the cooling of liquid water by placing it into the refrigerator into solid water (ice).</span>
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