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Stells [14]
2 years ago
13

In the PhET simulation window, click the radio button labeled Mystery in the Blocks menu on the right-hand side of the screen. U

sing the scale and water tank, determine the mass of each block using the scale and the volume of each block by submerging it in water. Use these values to calculate the density. Rank the mystery blocks in order from least dense to most dense. Rank from least dense to most dense. To rank items as equivalent, overlap them.
Tank A 103.38 L, B 100.64 L, C 104.08 L, D 103.10 L, E 101.00 L

Scale A 65.14 kg, B 0.64 kg, C 4.08 kg, D 3.10 kg, E 3.53 kg
Chemistry
1 answer:
Yuliya22 [10]2 years ago
4 0

Answer:

B, D, E, C, A

Explanation:

We have 5 blocks with their respective masses and volumes.

Block            Mass            Volume

  A                65.14 kg       103.38 L

  B                0.64 kg         100.64 L

  C                4.08 kg         104.08 L

  D                3.10 kg          103.10 L

  E                 3.53 kg         101.00 L

The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V

ρA = 65.14 kg / 103.38 L = 0.6301 kg/L

ρB = 0.64 kg / 100.64 L = 0.0064 kg/L

ρC = 4.08 kg / 104.08 L = 0.0392 kg/L

ρD = 3.10 kg / 103.10 L = 0.0301 kg/L

ρE = 3.53 kg / 101.00 L = 0.0350 kg/L

The order from least dense to most dense is B, D, E, C, A

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Assuming that nitrogen gas is ideal, we can use the equation PV = nRT to relate first conditions to the second condition. At constant temperature, pressure and volume are indirectly related as follows:

P = k / V

k is equal nRT

P1V1 = P2V2
P2 = 101.325 ( 4.65 ) / .480 = 981.586 kPa
6 0
3 years ago
What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

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2 years ago
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Answer:

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Explanation:

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