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Mamont248 [21]
3 years ago
14

A gas has a volume of 30 L at 105 kPa. What pressure is required to compress the gas to 21 L?

Chemistry
1 answer:
Aloiza [94]3 years ago
8 0

Answer: 150 kPa

Explanation:

Given that,

Original volume of gas V1 = 30L

Original pressure of gas P1 = 105 kPa

New pressure of gas P2 = ?

New volume of gas V2 = 21L

Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

105 kPa x 30L = P2 x 21L

3150 kPa L = P2 x 21L

P2 = 3150 kPa L / 21 L

P2 = 150 kPa

Thus, 150 kPa of pressure is required to compress the gas

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Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang
balandron [24]

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

5 0
3 years ago
1N2 + 3H2 -->
Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

5 0
2 years ago
A ballon is filled with helium gas. Another balloon, of the same size, is filled with nitrogen gas. Explain why the ratio of the
LenaWriter [7]
Let's say say there are n1 mols of helium in the first balloon and n2 mols of nitrogen in the second one, which are equivalent to m1 grams of helium and m2 grams of nitrogen.

The molar mass of hydrogen is thus M1=m1/n1, same for nitrogen M2=m2/n2 hence the ratio of their masses is m1/m2=(M1n1)/(M2n2). Since both gases are rather similar, we can assume that n1~n2 hence m1/m2=M1/M2
4 0
3 years ago
How many moles are in 65 g of carbon dioxide (CO2)?
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The answer is 44.0095. We assume you are converting between grams CO2 and mole.
3 0
3 years ago
Read 2 more answers
The total concentration of ions in a 0.75 M solution of HCl is
EastWind [94]
<h3>Answer:</h3>

The total concentration of ions in a 0.75 M solution of HCl is 1.5 M

That is; 0.75 M H⁺ and 0.75 M Cl⁻

<h3>Explanation:</h3>
  • Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
  • The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
  • For instance, HCl dissociates to give H⁺ and Cl⁻

       HCl(aq) → H⁺(aq) + Cl⁻(aq)

  • Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
  • Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
4 0
3 years ago
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