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borishaifa [10]
3 years ago
5

Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H20).

Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Answer: The equilibrium concentration of CH_4 , expressed in scientific notation is 5.9\times 10^{-2}M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

                          CO(g)+3H_2(g)\rightleftharpoons CH_4(g)+H_2O(g)

At eqm. conc.     (0.30) M     (0.10) M      (x) M   (0.020) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CH_4]\times [H_2O]}{[CO]\times [H_2]^3}

Now put all the given values in this expression, we get :

3.90=\frac{x\times (0.020)}{(0.30)\times (0.10)^3}

By solving the term 'x', we get :

x = 0.059 M=  5.9\times 10^{-2}M

Thus, the concentrations of CH_4 at equilibrium is :

Concentration of CH_4 = (x) M  = 5.9\times 10^{-2}M

The equilibrium concentration of CH_4 , expressed in scientific notation is 5.9\times 10^{-2}M

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Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti
djverab [1.8K]
<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

                        = 80,000 g

Therefore;

Moles of water = \frac{80,000g}{18.02 g/mol}

                = 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O  available = 65.0 kg or 65,000 g

Molar mass Li₂O  = 29.88 g/mol

Moles of Li₂O  = 65,000 g ÷ 29.88 g/mol

          = 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

  • Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
  • However, the number of moles of water to be removed is 4.44 × 10³ moles  but only 2.175 × 10³ moles will react with the available Li₂O.
  • This means, Li₂O  is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water =  4.44 × 10³ moles  - 2.175 × 10³ moles

                                           = 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

                                          = 4.0815 × 10⁴ g or

                                          = 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0
3 years ago
Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.
loris [4]

<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

<u>Explanation:</u>

The IUPAC nomenclature of alkanes are given as follows:

  • Select the longest possible carbon chain.
  • For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
  • A suffix '-ane' is added at the end of the name.
  • If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

4 0
3 years ago
1.) Activation energy is _____.
balandron [24]
To answer the questions,

(1) Activation energy is the amount of energy that is needed for the reaction to proceed, converting the reactant to products. The answer is letter B.

(2) The rate of chemical reaction normally increases as the reactant concentration is increased. The answer is letter C. 

6 0
3 years ago
Read 2 more answers
How many moles are in 3.24 x 1020 particles of lead? (show work plz)
lesya [120]
<h3>Answer:</h3>

0.000538 mol Pb

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

3.24 × 10²⁰ particles Pb (lead)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                                  \displaystyle 3.24 \cdot 10^{20} \ particles \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ particles \ Pb})
  2. Multiply/Divide:                                                                                               \displaystyle 0.000538 \ mol \ Pb

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

Our final answer is in 3 sig figs, no need to round.

7 0
3 years ago
During an experiment, the percent yield of calcium chloride from a reaction was 82.38%. Theoretically, the expected amount shoul
gregori [183]

Answer:

Actual yield = 86.5g

Explanation:

Percent yield = 82.38%

Theoretical yield = 105g

Actual yield = x

Equation of reaction,

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

Percentage yield = (actual yield / theoretical yield) * 100

82.38% = actual yield / theoretical yield

82.38 / 100 = x / 105

Cross multiply and make x the subject of formula

X = (105 * 82.38) / 100

X = 86.499g

X = 86.5g

Actual yield of CaCl₂ is 86.5g

7 0
3 years ago
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