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Maslowich
2 years ago
11

A tank of gas is found to exert 8.6 atm at 38°C. What would be the required

Chemistry
1 answer:
Vesna [10]2 years ago
6 0

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

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 iron + bromine ⟶ product

2.0 g +     <em>x</em> g     ⟶   11.0 g

According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.

∴2.0 + <em>x</em> = 11.0

<em>x</em> = 11.0 – 2.0 = 9.0

The reaction uses 9.0 g Br₂.

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2 years ago
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6 0
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Read 2 more answers
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
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I hope this helps.

7 0
3 years ago
A hot gas flowing through a pipeline can be considered as a:________
k0ka [10]

Answer:

B) irreversible process

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The process given here is irreversible.

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