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Free_Kalibri [48]
3 years ago
10

The volume of a sample of gas, initially at 25 °C and 158 mL, increases to

Chemistry
1 answer:
Artist 52 [7]3 years ago
5 0

Answer:

The final temperature = 848.7 K or 575.7 °C

Explanation:

Step 1: Data given

Temperature = 25.0 °C = 298 K

Volume = 158 mL = 0.158 L

The volume increases to 450 mL = 0.450 L

The pressure is constant

Step 2: Calculate the final temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.158 L

⇒with T1 = the initial temperature = 298 K

⇒with V2 = the increased volume = 0.450 L

⇒with T2 = the new temperature = TO BE DETERMINED

0.158 L / 298 K = 0.450 L / T2

T2 = 0.450 L / (0.158L/298K)

T2 = 848.7 T

The final temperature = 848.7 K or 575.7 °C

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Be sure to answer all parts. Compounds a and b are isomers having molecular formula c5h12. Heating a with cl2 gives a single pro
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Homework 1
Bond [772]

Answer:

average for silk =141"

average for cotton =96"

average for nylon = 70"

if you desire a slower falling parachute to protect the body from damage,

silk is the best

if you desire a faster falling parachute to escape enemy bullets,

nylon is the best

Explanation:

Homework 1

Problem Solving

1. Kelvin and Xavier were doing an investigation on parachutes. One of them suggested that the type of material the parachute was made from had an effect on how long it took to reach the ground. Their results are given in

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(a) Calculate the average time in seconds for each material.

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MATERIAL

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2nd TRY

3rd TRY

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144

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Answer:

Explanation:

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<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

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<em></em>

<em><u>Using cross multiplication:</u></em>

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0.48 mol of O₂ occupies → ??? L.

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