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Rama09 [41]
3 years ago
5

A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at 2 1.2 m/s , which

Chemistry
1 answer:
Phantasy [73]3 years ago
3 0
G. ………………………. . Z z… z z z z z z z z
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Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d
worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.  

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B\rightarrow Products

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

3.20\times 10^{-1}=k(1.50)^a(1.50)^b ....(1)

Expression for rate law for second observation:

3.20\times 10^{-1}=k(1.50)^a(2.50)^b ....(2)

Expression for rate law for third observation:

6.40\times 10^{-1}=k(3.00)^a(1.50)^b ....(3)

Dividing 1 from 2, we get:

\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0

Dividing 1 from 3, we get:

\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[A]^1[B]^0

\text{Rate}=k[A]

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0
3 years ago
A 3.06 gram sample of an unknown hydrocarbon with empirical formula CH2O was found to contain 0.0170 moles of the substance. Wha
Yanka [14]

Answer:

180 amu

C₆H₁₂O₆

Explanation:

Step 1: Determine the molecular mass of the compound

The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:

M = m/n = 3.06/0.0170 mol = 180 g/mol

Then, the molecular mass is 180 amu.

Step 2: Determine the molar mass of the empirical formula.

M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)

M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol

Step 3: Determine the molecular formula

First, we will determine "n" according to the following expression.

n = molar mass molecular formula / molar mass empirical formula

n = 180 g/mol / 30 g/mol = 6

The molecular formula is:

n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

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What is required for storing chemicals
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A chemical storing thing
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A titration involves adding a reactant of a known quantity to a solution of another reactant while monitoring the equilibrium co
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Answer: Be a big brain

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Complete the statement about the environment and its components.
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the environment is Healthy

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