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7nadin3 [17]
3 years ago
12

A sample of gas occupies 250 mL at STP. What is the temperature if the gas expands to 1500 mL at a pressure of 0.500 atm? *

Chemistry
1 answer:
3241004551 [841]3 years ago
6 0

Answer:

819 K

Explanation:

Given data

  • Initial pressure (P₁): 1 atm (standard pressure)
  • Initial volume (V₁): 250 mL
  • Initial temperature (T₁): 273.15 K (standard temperature)
  • Final pressure (P₂): 0.500 atm
  • Final volume (V₂): 1500 mL
  • Final temperature (T₂): ?

We can find the final temperature using the combined gas law.

\frac{P_1 \times V_1 }{T_1} = \frac{P_2 \times V_2 }{T_2}\\T_2 = \frac{P_2 \times V_2 \times T_1 }{P_1 \times V_1} = \frac{0.500atm \times 1500mL \times 273.15K }{1atm \times 250mL}=819K

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0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
3 years ago
How many grams of sulfur trioxide are produced 18 mol O2 react with sufficient sulfur? Show all your work S8 +12O 2> 8SO3
SOVA2 [1]

Answer:

m_{SO_3}=2.31x10^4gSO_3

Explanation:

Hello!

In this case, since the reaction between sulfur and oxygen is:

S_8 +\frac{1}{2} O_2 \rightarrow 8SO_3

Whereas there is a 1/2:8 mole ratio between oxygen and SO3, and we can compute the produced grams of product as shown below:

m_{SO_3}=18molO_2*\frac{8molSO_3}{1/2molO_2} *\frac{80.06gSO_3}{1molSO_3} \\\\m_{SO_3}=23,057.3gSO_3=2.31x10^4gSO_3

Best regards!

7 0
3 years ago
HELPPP
maxonik [38]

1 endothermic 2 yes and exothermic 3 endothermic

3 0
3 years ago
Need the answer please
WARRIOR [948]
I think A, I might be wrong though sorry
5 0
3 years ago
The diagrams below represent some respiratory structures in three organisms. The labeled structures in these organisms all have
Igoryamba

Answer:

D

Explanation:

Lungs, Moist skin , Gills is in the exchange of gases system

5 0
3 years ago
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