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3 years ago

Lab worksheets oh what a tangled web we weave

1 answer:

8 0

<span>Proper format and submission of lab worksheets is helps develop professional standards. Please follow all format and submission guidelines carefully. The guidelines are meant to provide all employees with a standard organizational format for information from in progress and completed lab work. Also providing easy work flow should you have stop any where in process of a lab.</span>

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Consider the following reaction

Fynjy0 [20]

Answer:

92.72 kJ

Explanation:

2 N₂ (g) + O₂ (g) —-> 2 N₂O

According to question , one mole of N₂O requires 163.2 kJ of heat

Molecular weight of N₂O = 44 gm

25 g N₂O = 25 / 44 mole

25 / 44 mole will require 163.2 x 25 / 44 kJ

= 92.72 kJ

6 0

2 years ago

What is true about ionic compounds?

bezimeni [28]

It’s D I am pretty sure.

6 0

3 years ago

According to Newton's 1st law of motion, what is required to make an object slow down?

attashe74 [19]

Newton’s first law state that an object in motion stays in motion unless acted upon force so the answer would be RESISTANCE

7 0

3 years ago

Read 2 more answers

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

babymother [125]

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

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

The order of reaction with respect to C 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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

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

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Thus,

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

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

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

5 0

3 years ago

Will these changes increase, decrease, or have no effect on the mean free path of the molecules in a gas sample?

lisabon 2012 [21]

According to the kinetic theory, the mean free path is the average distance a single atom or molecule of an element or compound travels with respect with the other atoms during a collision. The greater the mean free path, the more ideal the behavior of a gas molecule is because intermolecular forces are minimum. To understand which factors affect the mean free path, the equation is written below.

l = μ/P * √(πkT/2m), where
l is the mean free path
μ is the viscosity of the fluid
P is the pressure
k is the Boltzmann's constant
T is the absolute temperature
m is the molar mass

So, here are the general effects of the factors on the mean free path:

Mean free path increases when:
1. The fluid is viscous (↑μ)
2. At low pressures (↓P)
3. At high temperatures (↑T)
4. Very light masses (↓m)

The opposite is also true for when the mean free path decreases. Factors that are not found here have little or no effect.

6 0

3 years ago