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mart [117]
2 years ago
6

2H2(g)+2NO(g)→2H2O(g)+N2(g) Part A If the concentration of NO changed from 0.100 M to 0.025 M in the first 15 minutes of the rea

ction, what is the average rate of the reaction during this time interval?
Chemistry
1 answer:
andrew11 [14]2 years ago
6 0

Answer:

0.0025 M/min

Explanation:

The rate of a reaction can be calculated for an element, based on its stoichiometric coefficient. For a reaction:

aA + bB = cC + dD , the rate will be

r = -(1/a)x(Δ[A]/Δt) = -(1/b)x(Δ[B]/Δt) = (1/c)x(Δ[C]/Δt) = (1/d)x(Δ[D]/Δt)

Where Δ[X] is the variation of the concentration of the X compound, Δt is the time variation, and the signal of minus in the reagents compounds is because they are disappearing, so Δ[X] will be negative, and r must be positive. So, for the reaction given:

r = -(1/2)x(Δ[NO]/Δt)

r = -(1/2)x( (0.025 - 0.1)/15)

r = 0.0025 M/min

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What element has 42 neutrons
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Explanation:

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2 years ago
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A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas
Anna71 [15]

Answer:

\large \boxed{\text{528.7 g} }

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"  

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:                      32.00

              2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol:                                                                  7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}

2. Mass of O₂

\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}

3 0
3 years ago
Please help me like now please
777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

<u>Question 1</u>

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

<em>To balance the equation, count the number of atoms on both sides of the equation</em>

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

<em>Combining the pluses will give you the following</em>

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

<em>Both sides are the same, therefore the chemical equation is balanced (originally). </em>

From the equation, we can see that <u>1 mole of NaHCO3</u> produces <u>1 mole of CO2</u>.

So that means <u>0.5 mole of NaHCO3</u> would produce <u>0.5 mole of CO2</u>.

<u>Question 2</u>

C4H10 + O2 --> CO2 + H2O

<em>Again, count the number of atoms on both sides of the equation</em>

(4C, 10H, 2O) --> (1C, 2H, 3O)     <em>This time left does not equal right side</em>

<em>You now need to find </em><u><em>factors </em></u><em>that can make both sides equal. </em>

C4H10 + O2 --> <u>4</u>CO2 + H2O    <em>Now the C is balanced, let's recount </em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H)      H&O is still not balanced</em>

C4H10 + O2 --> 4CO2 + <u>5</u>H2O    <em>Now the H is balanced, let's recount</em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H)      O is still not balanced</em>

C4H10 + (<u>13/2</u>)O2 --> 4CO2 + 5H2O    <em>Now the O is balanced</em>

<em>(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)</em>

<em>But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).</em>

<u>2</u>C4H10 + <u>13</u>O2 --> <u>8</u>CO2 + <u>10</u>H2O    Now it's completely balanced!

<em>(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H)     Yayy! It's balanced.</em>

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0
2 years ago
Which of the following is true for the equilibrium constant of a reaction? (5 pc
Slav-nsk [51]

The equilibrium constant of the reaction is represented by the symbol K. Thus, option C is the correct and accurate statement about the equilibrium constant.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant is a representation of the concentration of the products and the reactants of the reaction that is raised to the powers through their stoichiometry coefficient.

Its value varies and changes at different temperatures and is not always less than 1. The equilibrium constant is the ratio of the coefficient of the products to reactants.

Therefore, option C. equilibrium constant is represented by K is true.

Learn more about the equilibrium constant here:

brainly.com/question/12429748

#SPJ1

8 0
2 years ago
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