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frez [133]
3 years ago
15

A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise

d by 74 K. Calculate the heat lost or gained by the system. Calculate your answer to three significant figures and express you answer in J. Heat capacity for Helium = 20.8 J/(mol K) Molecular weight for Helium = 4 g/mol
Chemistry
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

<u>Step 1: </u>Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

<u>Step 2:</u> Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

<u>Step 3:</u> Calculate heat

Q = n*c*ΔT  or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

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Ionic Lewis Dot Structure for BaF2, hope this helps.

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3 years ago
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Just post the question on here
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2 years ago
For the following reaction, 6.94 grams of water are mixed with excess sulfur dioxide . Assume that the percent yield of sulfurou
Alexxx [7]
<h3>Answer:</h3>

#a. Theoretical yield = 31.6 g

#b. Actual yield = 25.72 g

<h3>Explanation:</h3>

The equation for the reaction between sulfur dioxide and water to form sulfurous acid is given by the equation;

SO₂(g) + H₂O(l) → H₂SO₃(aq)

The percent yield of H₂SO₃ is 81.4%

Mass of water that reacted is 6.94 g

#a. To get the theoretical yield of H₂SO₃ we need to follow the following steps

Step 1: Calculate the moles of water

Molar mass of water = 18.02 g/mol

Mass of water = 6.94 g

But, moles = Mass/molar mass

Moles of water = 6.94 g ÷ 18.02 g/mol

                        = 0.385 mol

Step 2: Calculate moles of H₂SO₃

From the equation, the mole ratio of water to H₂SO₃ is 1 : 1

Therefore, moles of water = moles of H₂SO₃

Hence, moles of H₂SO₃ = 0.385 mol

Step 3: Theoretical mass of H₂SO₃

Mass = moles × Molar mass

Molar mass of H₂SO₃ = 82.08 g/mol

Number of moles of H₂SO₃ = 0.385 mol

Therefore;

Theoretical mass of H₂SO₃ = 0.385 mol ×  82.08 g/mol

                                             = 31.60 g

Thus, the theoretical yield of H₂SO₃ is 31.6 g

<h3>#b. Calculating the actual yield</h3>

We need to calculate the actual yield

Percent yield of H₂SO₃ is 81.4%

Theoretical yield is 31.60 g

But; Percent yield = (Actual yield/theoretical yield)×100

Therefore;

Actual yield = Percent yield × theoretical yield)÷ 100

                   = (81.4 % × 31.6) ÷ 100

                  = 25.72 g

The percent yield of H₂SO₃ is 25.72 g

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3 years ago
Question 5 (1 point)
Marianna [84]

Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

 n= number of moles

 R= Avogadro constant = 0.0821

 T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

P= 12.15

7 0
3 years ago
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