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3 years ago

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The rate law for the reaction, 2no(g o2(g ? 2no2(g, is found to be: rate of reaction = k [ no]2 [o2 ]2. the first, fast step is

which best represents the rate equation for the rate-determining step of this reaction?

1 answer:

True [87]3 years ago

5 0

The Answer is D Why Because<span> clearly, including a scenario of the second and third steps as proof!!</span>

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CH4 + 202 → CO2 + 2H2O<br> How many moles O2 needed to produce 4 moles of H2O?

shepuryov [24]

<h3>Answer:</h3>

4 mol O₂

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 4 mol H₂O

[Solve] x mol O₂

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

[RxN] 2 mol H₂O → 2 mol O₂

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 4 \ mol \ H_2O(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})$
Multiply/Divide: $\displaystyle 4 \ mol \ O_2$

7 0

3 years ago

Read 2 more answers

Help plz will make brainliest

Jobisdone [24]

Answer:

G

Explanation:

5 0

3 years ago

Each of the given statements describes a type of column chromatography. Match the statements to the type of chromatography they

vlada-n [284]

Answer:

The proper matching is given below.

Explanation:

a Separate molecules by size size exclusion chromatography

b Separate molecules by charge Ion exchange chromatography

c The stationary phase has a covalently bound group to which a protein in the mobile phase can bind. Affinity chromatography

d uses mobile phase and stationary phase to separate protein Size exclusion chromatography

e The stationary phase contain cross linked polymers with different pore size

Size exclusion chromatography

f can separate molecules based on protein ligand binding Affinity chromatography

g The stationary phase may contain negatively or positively charged groups

ion exchange chromatography

6 0

3 years ago

A 10.0-milliliter sample of NaOH(aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH(aq)?

Ludmilka [50]

$\frac{10 mL}{1000 mL} = 0.01 L$

$\frac{40mL}{1000mL} =0.04L$

$M _{1} V_{1}=M_{2}V_{2}$

$M_{1}(0.01)=(0.50)(0.04)$

$M_{1}(0.01)=0.02$

$M_{1}= \frac{0.02}{0.01}$

$M_{1}=2$

3 0

4 years ago

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

3 years ago