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Olegator [25]
3 years ago
11

A sample of nitrogen gas expands in volume from 1.3 to 5.7 L at constant temperature. Calculate the work done in joules if the g

as expands (a) against a vacuum: J (b) against a constant pressure of 0.60 atm: × 10 J Enter your answer in scientific notation. (c) against a constant pressure of 4.7 atm: × 10 J Enter your answer in scientific notation.
Chemistry
1 answer:
just olya [345]3 years ago
7 0

Answer:

a) 0 J

b)  -2.67x10² J

c) -2.09x10³ J

Explanation:

For an isothermic expansion (with constant temperature) the work (W) is :

W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).

a) In vacuum, the relative pressure is 0 atm, so the work:

W = -0x(5.7 - 1.3)

W = 0 J

b) For a constant pressure of 0.60 atm

W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm

1 L.atm = 101.3 J

W = -2.64x101.3 = -2.67x10² J

c) For a pressure of 4.7 atm

W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L

1 atm.L = 101.3 J

W = -20.68x101.3 = -2.09x10³ J

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Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO
sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂   → CaCO₃

Number of moles of CaO:

Number of moles  = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles  = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

                  CO₂         :                CaCO₃  

                  1               :                 1

                 0.31           :              0.31

                CaO           :               CaCO₃  

                 1                :                 1

                 0.26         :              0.26

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ =  26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

7 0
4 years ago
What states of matter does water naturally exist on Earth
amid [387]
Solid (ice caps)
Liquid (oceans, rivers, lakes, etc)
Gas (clouds)
5 0
3 years ago
Read 2 more answers
1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this
LenKa [72]
<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0
3 years ago
During a volcanic eruption, large amounts of poisonous gases and particles are released into the atmosphere. How do some of thes
xz_007 [3.2K]
A through rain because it is:)
3 0
3 years ago
How many grams of C3H8 would be needed to produce 6.39 grams of CO2?
just olya [345]

Answer:

2.13g

Explanation:

Atomic mass of CO2 =  12 + 32 = 44g/Mol

Atomic mass of C3H8 = 36 + 8 = 44g/Mol

Reaction

C3H8 + 5O2 --> 3CO2 + 4H2O

3CO2 = 6.39g

Required C3H8 = (6.39/(44 x 3)) x 44 = 2.13g  

8 0
3 years ago
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